United States 

Environmental Protection 

Agency 

 Office of Environmental 

 Information

 Washington, DC 20460

 EPA/240/B-06/003 

February 2006 

Data Quality Assessment: 
 

Statistical Methods for 

Practitioners 

EPA QA/G-9S 

EPA QA/G-9S February 2006 iii

FOREWORD 

This document, Data Quality Assessment: Statistical Methods for Practitioners, provides 

general guidance to organizations on assessing data quality criteria and performance 

specifications. The Environmental Protection Agency (EPA) has developed the Data Quality 

Assessment (DQA) Process for project managers and planners to determine whether the type, 

quantity, and quality of data needed to support Agency decision making has been achieved. This 

guidance is the culmination of experiences in the design and statistical analyses of environmental 

data in different Program Offices at the EPA. Many elements of prior guidance, statistics, and 

scientific planning have been incorporated into this document. 

This document is one of a series of quality management guidance documents that the 

EPA Quality Staff has prepared to assist users in implementing the Agency-wide Quality 

System. Other related documents include: 

EPA QA/G-4 Guidance on Systematic Planning Using the Data Quality 

Objectives Process 

EPA QA/G-5S Guidance on Choosing a Sampling Design for Environmental Data 

Collection 

EPA QA/G-9R Data Quality Assessment: A Reviewer’s Guide 

This document provides guidance to EPA program managers and planning teams as well 

as to the general public as appropriate. It does not impose legally binding requirements and may 

not apply to a particular situation based on the circumstances. EPA retains the discretion to 

adopt approaches on a case-by-case basis that differ from this guidance where appropriate. 

This guidance is one of the U.S Environmental Protection Agency Quality System Series

documents. These documents describe the EPA policies and procedures for planning, 

implementing, and assessing the effectiveness of the Quality System. These documents are 

updated periodically to incorporate new topics and revision or refinements to existing 

procedures. Comments received on this, the 2006 version, will be considered for inclusion in 

subsequent versions. Please send your comments to: 

Quality Staff (2811R) 

U.S. Environmental Protection Agency 

1200 Pennsylvania Avenue, NW 

Washington, DC 20460 

Phone: (202) 564-6830 

Fax: (202) 565-2441 

E-mail: quality@epa.gov

Copies of the EPA Quality System documents may be downloaded from the Quality Staff Home 

Page: www.epa.gov.quality. 
EPA QA/G-9S February 2006 iv

PREFACE 

Data Quality Assessment: Statistical Methods for Practitioners describes the statistical 

methods used in Data Quality Assessment (DQA) in evaluating environmental data sets. DQA is 

the scientific and statistical evaluation of environmental data to determine if they meet the 

planning objectives of the project, and thus are of the right type, quality, and quantity to support 

their intended use. This guidance applies DQA to environmental decision-making (e.g., 

compliance determinations) and also addresses DQA in environmental estimation (e.g., 

monitoring programs). 

This document is distinctly different from other guidance documents in that it is not 

intended to be read in a linear or continuous fashion. Instead, it is intended to be used as a 

"tool-box" of useful techniques in assessing the quality of data. The overall structure of the 

document will enable the analyst to investigate many problems using a systematic methodology. 

Each statistical technique examined in the text is demonstrated separately in the form of a series 

of steps to be taken. The technique is then illustrated with a practical example following the 

steps described. 

This document is intended for all EPA and extramural organizations that have quality 

systems based on EPA policies and specifications, and that may periodically assess these quality 

systems (or have them assessed by EPA) for compliance to the specifications. In addition, this 

guidance may be used by other organizations that assess quality systems applied to specific 

environmental programs. 

The guidance provided herein is non-mandatory and is intended to help personnel who 

have minimal experience with statistical terminology to understand how a technique works and 

how it may be applied to a problem. An explanation of DQA in plain English may be found in 

the companion guidance document, Data Quality Assessment: A Reviewer’s Guide (EPA 

QA/G-9R) (U.S. EPA, 2006b). 
EPA QA/G-9S February 2006 v

TABLE OF CONTENTS 

Page

INTRODUCTION......................................................................................................................... 1

STEP 1: REVIEW DQOs AND THE SAMPLING DESIGN.................................................. 5

1.1 OVERVIEW AND ACTIVITIES............................................................................6 

STEP 2: CONDUCT A PRELIMINARY DATA REVIEW.................................................... 9

2.1 OVERVIEW AND ACTIVITIES..........................................................................12 

2.2 STATISTICAL QUANTITIES .............................................................................12 

2.2.1 Measures of Relative Standing ..................................................................12 

2.2.2 Measures of Central Tendency ..................................................................13 

2.2.3 Measures of Dispersion..............................................................................15 

2.2.4 Measures of Association............................................................................16 

2.2.4.1 Pearson’s Correlation Coefficient................................................16 

2.2.4.2 Spearman’s Rank Correlation......................................................18 

2.2.4.3 Serial Correlation.........................................................................18 

2.3 GRAPHICAL REPRESENTATIONS...................................................................20 

2.3.1 Histogram...................................................................................................20 

2.3.2 Stem-and-Leaf Plot ....................................................................................21 

2.3.3 Box-and-Whiskers Plot..............................................................................23 

2.3.4 Quantile Plot and Ranked Data Plots.........................................................25 

2.3.5 Quantile-Quantile Plots and Probability Plots ...........................................26 

2.3.6 Plots for Two or More Variables ...............................................................27 

2.3.6.1 Scatterplot ....................................................................................29 

2.3.6.2 Extensions of the Scatterplot .......................................................30 

2.3.6.3 Empirical Quantile-Quantile Plot ................................................31 

2.3.7 Plots for Temporal Data.............................................................................32 

2.3.7.1 Time Plot......................................................................................33 

2.3.7.2 Lag Plot........................................................................................34 

2.3.7.3 Plot of the Autocorrelation Function (Correlogram) ...................34 

2.3.7.4 Multiple Observations in a Time Period......................................35 

2.3.7.5 Four-Plot ......................................................................................36 

2.3.8 Plots for Spatial Data .................................................................................37 

2.3.8.1 Posting Plots ................................................................................38 

2.3.8.2 Symbol Plots and Bubble Plots....................................................39 

2.3.8.3 Other Spatial Graphical Representations.....................................39 

2.4 PROBABILITY DISTRIBUTIONS......................................................................40 

2.4.1 The Normal Distribution............................................................................40 

2.4.2 The t-Distribution.......................................................................................41 

2.4.3 The Lognormal Distribution ......................................................................41 

2.4.4 Central Limit Theorem ..............................................................................41 

STEP 3: SELECT THE STATISTICAL METHOD.............................................................. 43

3.1 OVERVIEW AND ACTIVITIES..........................................................................46 

3.2 METHODS FOR A SINGLE POPULATION ......................................................46 

3.2.1 Parametric Methods ...................................................................................48 

3.2.1.1 The One-Sample t-test and Confidence Interval or Limit ...........48 
EPA QA/G-9S February 2006 vi

Page

3.2.1.2 The One-Sample Tolerance Interval or Limit..............................48 

3.2.1.3 Stratified Random Sampling........................................................51 

3.2.1.4 The Chen Test..............................................................................52 

3.2.1.5 Land’s Method for Lognormally Distributed Data......................57 

3.2.1.6 The One-Sample Proportion Test and Confidence Interval.........58 

3.2.2 Nonparametric Methods.............................................................................60 

3.2.2.1 The Sign Test ...............................................................................60 

3.2.2.2 The Wilcoxon Signed Rank Test .................................................61 

3.3 COMPARING TWO POPULATIONS .................................................................63 

3.3.1 Parametric Methods ...................................................................................65 

3.3.1.1 Independent Samples ...................................................................66 

3.3.1.2 Paired Samples.............................................................................71 

3.3.2 Nonparametric Methods.............................................................................75 

3.3.2.1 Independent Samples ...................................................................75 

3.3.2.2 Paired Samples.............................................................................81 

3.4 COMPARING SEVERAL POPULATIONS SIMULTANEOUSLY...................86 

3.4.1 Parametric Methods ...................................................................................88 

3.4.1.1 Dunnett’s Test..............................................................................88 

3.4.2 Nonparametric Methods.............................................................................89 

3.4.2.1 The Fligner-Wolfe Test................................................................89 

STEP 4: VERIFY THE ASSUMPTIONS OF THE STATISTICAL METHOD ................ 93

4.1 OVERVIEW AND ACTIVITIES..........................................................................96 

4.2 TESTS FOR DISTRIBUTIONAL ASSUMPTIONS............................................96 

4.2.1 Graphical Methods.....................................................................................98 

4.2.2 Normal Probability Plot Tests....................................................................98 

4.2.3 Coefficient of Skewness/Coefficient of Kurtosis Tests.............................98 

4.2.4 Range Tests................................................................................................99 

4.2.4.1 The Studentized Range Test ........................................................99 

4.2.4.2 Geary’s Test...............................................................................100 

4.2.5 Goodness-of-Fit Tests..............................................................................100 

4.2.5.1 Chi-Square Test .........................................................................100 

4.2.5.2 Tests Based on the Empirical Distribution Function .................102 

4.2.6 Recommendations....................................................................................102 

4.3 TESTS FOR TRENDS ........................................................................................102 

4.3.1 Introduction..............................................................................................102 

4.3.2 Regression-Based Methods for Estimating and Testing for Trends ........103 

4.3.2.1 Estimating a Trend Using the Slope of the Regression Line.....103 

4.3.2.2 Testing for Trends Using Regression Methods .........................104 

4.3.3 General Trend Estimation Methods.........................................................104 

4.3.3.1 Sen's Slope Estimator.................................................................104 

4.3.3.2 Seasonal Kendall Slope Estimator.............................................104 

4.3.4 Hypothesis Tests for Detecting Trends....................................................104 

4.3.4.1 One Observation per Time Period for One Sampling Location 106
EPA QA/G-9S February 2006 vii

Page

4.3.4.2 Multiple Observations per Time Period for One Sampling 

Location .....................................................................................109 

4.3.4.3 Multiple Sampling Locations with Multiple Observations........110 

4.3.4.4 One Observation for One Station with Multiple Seasons..........113 

4.3.5 A Discussion on Tests for Trends............................................................113 

4.3.6 Testing for Trends in Sequences of Data.................................................114 

4.4 OUTLIERS ..........................................................................................................115 

4.4.1 Background..............................................................................................115 

4.4.2 Selection of a Statistical Test for Outliers ...............................................116 

4.4.3 Extreme Value Test (Dixon's Test)..........................................................117 

4.4.4 Discordance Test......................................................................................117 

4.4.5 Rosner's Test ............................................................................................119 

4.4.6 Walsh's Test .............................................................................................120 

4.4.7 Multivariate Outliers................................................................................120 

4.5 TESTS FOR DISPERSIONS...............................................................................122 

4.5.1 Confidence Intervals for a Single Variance.............................................122 

4.5.2 The F-Test for the Equality of Two Variances........................................123 

4.5.3 Bartlett's Test for the Equality of Two or More Variances......................123 

4.5.4 Levene's Test for the Equality of Two or More Variances......................123 

4.6 TRANSFORMATIONS ......................................................................................126 

4.6.1 Types of Data Transformations ...............................................................127 

4.6.2 Reasons for Transforming Data...............................................................128 

4.7 VALUES BELOW DETECTION LIMITS.........................................................130 

4.7.1 Approximately less than 15% Non-detects - Substitution Methods ........131 

4.7.1.1 Aitchison’s Method....................................................................131 

4.7.2 Between Approximately 15% - 50% Non-detects...................................132 

4.7.2.1 Cohen's Method .........................................................................132 

4.7.2.2 Selecting Between Aitchison’s Method and Cohen’s Method ..132 

4.7.3 Greater than Approximately 50% Non-detects - Test of Proportions......134 

4.7.4 Greater than Approximately 90% Non-detects........................................135 

4.7.5 Recommendations....................................................................................135 

4.8 INDEPENDENCE ...............................................................................................136 

STEP 5: DRAW CONCLUSIONS FROM THE DATA ...................................................... 139

5.1 OVERVIEW AND ACTIVITIES........................................................................141 

5.2 PERFORM THE STATISTICAL METHOD......................................................141 

5.3 DRAW STUDY CONCLUSIONS......................................................................141 

5.3.1 Hypothesis Tests ......................................................................................141 

5.3.2 Confidence Intervals or Limits ................................................................142 

5.3.3 Tolerance Intervals or Limits...................................................................143 

5.4 EVALUATE PERFORMANCE OF THE SAMPLING DESIGN......................143 

5.5 INTERPRET AND COMMUNICATE THE RESULTS....................................144 

APPENDIX A: STATISTICAL TABLES............................................................................. 145

APPENDIX B: REFERENCES.............................................................................................. 183

If applicable, a list of tables, figures, and boxes is listed at the beginning of each chapter. 
EPA QA/G-9S February 2006 viii
EPA QA/G-9S February 2006 1

INTRODUCTION 

Data Quality Assessment (DQA) is the scientific and statistical evaluation of 

environmental data to determine if they meet the planning objectives of the project, and thus are 

of the right type, quality, and quantity to support their intended use. DQA is built on a 

fundamental premise: data quality is meaningful only when it relates to the intended use of the 

data. This guidance describes the technical aspects of DQA in evaluating environmental data 

sets. A conceptual presentation of the DQA process is contained in Data Quality Assessment: 

A Reviewer’s Guide (EPA QA/G-9R) (U.S. EPA 2004). 

By using DQA, a reviewer can answer four important questions: 

1. Can a decision (or estimate) be made with the desired level of certainty, given the 

quality of the data? 

2. How well did the sampling design perform? 

3. If the same sampling design strategy is used again for a similar study, would the 

data be expected to support the same intended use with the desired level of 

certainty? 

4. Is it likely that sufficient samples were taken to enable the reviewer to see an 

effect if it was really present? 

The first question addresses the reviewer’s immediate needs. For example, if the data are 

being used for decision-making and provide evidence strongly in favor of one course of action 

over another, then the decision maker can proceed knowing that the decision will be supported 

by unambiguous data. However, if the data do not show sufficiently strong evidence to favor 

one alternative, then the data analysis alerts the decision maker to this uncertainty. The decision 

maker now is in a position to make an informed choice about how to proceed (such as collect 

more or different data before making the decision, or proceed with the decision despite the 

relatively high, but tolerable, chance of drawing an erroneous conclusion). 

The second question addresses how robust this sampling design is with respect to slightly 

changing conditions. If the design is very sensitive to potentially disturbing influences, then 

interpretation of the results may be difficult. By addressing the second question, the reviewer 

guards against the possibility of a spurious result arising from a unique set of circumstances. 

The third question addresses the problem of whether this could be considered a unique 

situation where the results of this DQA only apply only to this situation and cannot be 

extrapolated to other situations. It also addresses the suitability of using this data collection 

design again for future projects. For example, if it is intended to use a certain sampling design at 

a different location from where the design was first used, it should be determined how well the 

design can be expected to perform given that the outcomes and environmental conditions of this 

sampling event will be different from those of the original event. As environmental conditions 
EPA QA/G-9S February 2006 2

will vary from one location or one time to another, the adequacy of the sampling design should 

be evaluated over a broad range of possible outcomes and conditions. 

The final question addresses the issue of whether sufficient resources were used in the 

study. For example, in an epidemiological investigation, was it likely the effect of interest could 

be reliably observed given the limited number of samples actually obtained. 

The data life cycle comprises three steps: planning, implementation, and assessment. 

During the planning phase, the Data Quality Objectives (DQO) Process (or any other systematic 

planning procedure) is used to define criteria for determining the number, location, and timing of 

samples to be collected in order to produce a result with the desired level of certainty. This, 

along with the sampling methods, analytical procedures, and appropriate quality assurance (QA) 

and quality control (QC) procedures, is documented in the QA Project Plan. The implementation 

phase consists of collecting data following the QA Project Plan specifications. At the outset of 

the assessment phase, the data are validated and verified to ensure that the sampling and analysis 

protocols specified in the QA Project Plan were followed, and that the measurement systems 

performed in accordance with the criteria specified in the QA Project Plan. Then, DQA 

completes the data life cycle through determining if the performance and acceptance criterion 

from the DQO planning objectives were achieved and to draw conclusions from the data. 

DQA involves five steps that begin with a review of the planning documentation and end 

with an answer to the question posed during the planning phase of the study. These steps 

roughly parallel the actions of an environmental statistician when analyzing a set of data. The 

five steps, which are described in more detail in the following chapters of this guidance, are: 

1. Review the project objectives and sampling design. 

2. Conduct a preliminary data review. 

3. Select the statistical method. 

4. Verify the assumptions of the statistical method. 

5. Draw conclusions from the data. 

These five steps are presented in a linear sequence, but DQA is actually an iterative 

process. For example, if the preliminary data review reveals patterns or anomalies in the data set 

that are inconsistent with the project objectives, then some aspects of the study planning may 

have to be reconsidered in Step 1. Likewise, if the underlying assumptions of the statistical 

method are not supported by the data, then previous steps of the DQA may have to be revisited. 

This guidance is written for a broad audience of potential data users, data analysts, and 

data generators. Data users (such as project managers or risk assessors who are responsible for 

making decisions or producing estimates regarding environmental characteristics) should find 

this guidance useful for understanding and directing the technical work of others who produce 

and analyze data. Data analysts should find this guidance to be a convenient compendium of 

basic assessment tools. Data generators (such as analytical chemists or field sampling specialists 

responsible for collecting and analyzing environmental samples and reporting the resulting data 

values) should find this guidance useful for understanding how their work will be used and for 
EPA QA/G-9S February 2006 3

providing a foundation for improving the efficiency and effectiveness of the data generation 

process. 

This guidance presents background information and statistical tools for performing DQA; 

a non-statistical discussion of DQA is to be found in the companion document Data Quality 

Assessment: A Reviewer’s Guide (EPA QA/G-9R) (U.S. EPA 2004). Each chapter corresponds 

to a step in the DQA and begins with an overview of the activities to be performed for that step. 

Following the overviews in Chapters 1, 2, 3, and 4, specific graphical or statistical tools are 

described and step-by-step procedures are provided along with examples. Chapter 5 gives some 

advice on the interpretation of statistical tests. Appendix A contains statistical tables, and 

Appendix B provides references and useful publications for in-depth statistical analyses. 
EPA QA/G-9S February 2006 4
EPA QA/G-9S February 2006 5

CHAPTER 1 

STEP 1: REVIEW DQOs AND THE SAMPLING DESIGN 

Step 1: Review DQOs and Sampling Design

• Review the objectives of the study. 

 If DQOs were developed, then review the outputs from the DQO Process. 

 If DQOs have not been developed, then ascertain what these objectives were. 

• Translate the data user's objectives into a statement of the primary statistical hypothesis. 

 If DQOs were developed, translate them into a statement of the primary hypothesis. 

 If DQOs have not been developed, then ascertain what hypotheses or estimates were 

developed. 

• Translate the data user's objectives into limits on Type I or Type II decision errors. 

 If DQOs have not been developed, document the data user's probable tolerable limits on 

decision errors, width of gray region, and estimated preliminary values. 

 If DQOs were developed, confirm the limits on decision errors. 

• Review the sampling design and note any special features or potential problems. 

 Review the sampling design for any potentially serious deviations.

THE DATA QUALITY ASSESSMENT PROCESS

Review DQOs and Sampling Design

Conduct Preliminary Data Review 

Select the Statistical Method 

Verify the Assumptions 

Draw Conclusions from the Data 

REVIEW DQOs AND SAMPLING DESIGN

Purpose

Review the DQO outputs, the sampling design, and 

any data collection documentation for consistency. If

DQOs have not been developed, define the statistical 

method and specify tolerable limits on decision errors. 

 Activities

Review Study Objectives

Translate Objectives into Statistical Hypothesis

Develop Limits on Decision Errors

Review Sampling Design

Tools

Statements of hypotheses

Sampling design concepts
EPA QA/G-9S February 2006 6

CHAPTER 1 

STEP 1: REVIEW DQOs AND THE SAMPLING DESIGN 

1.1 OVERVIEW AND ACTIVITIES 

DQA begins by reviewing the key outputs from the planning phase of the data life cycle 

such as the Data Quality Objectives, the QA Project Plan, and any associated documents. The 

study objectives provide the context for understanding the purpose of the data collection effort 

and establish the qualitative and quantitative basis for assessing the quality of the data set for the 

intended use. The sampling design (documented in the QA Project Plan) provides important 

information about how to interpret the data. By studying the sampling design, the analyst can 

gain an understanding of the assumptions under which the design was developed, as well as the 

relationship between these assumptions and the objectives. 

In the unfortunate instances when project 

objectives have not been developed during the 

planning phase of the study, it is necessary to 

develop statements of the data user’s objectives prior 

to conducting the DQA. The primary purpose of 

stating the data user’s objectives prior to analyzing 

the data is to establish appropriate criteria for 

evaluating the quality of the data with respect to their 

intended use. Analysts who are not familiar with the 

DQO Process should refer to the Guidance for the 

Data Quality Objectives Process (EPA QA/G-4)

(U.S. EPA 2000), a book on statistical planning and 

analysis, or a consulting statistician. The seven steps 

of the DQO Process are illustrated in Figure 1-1. 

If the project has been framed as a hypothesis 

test, then the uncertainty limits can be expressed as 

the data user's tolerance for committing false 

rejection (Type I or a false positive) or false 

acceptance (Type II or a false negative) decision 

errors. A false rejection error occurs when the null 

hypothesis is rejected when it is in fact true. A false 

acceptance error occurs when the null hypothesis is 

not rejected when it is in fact false. Other related 

phrases in common use include "level of 

significance" which is equal to the probability of a 

Type I error, and "power" which is equal to 1 minus probability of a Type II error. Statistical 

power really is a function that describes a “power curve” over a range of Type II errors. The 

characteristics of the power curve can be a great importance in choosing the appropriate 

statistical test. For detailed information on how to develop false rejection and false acceptance 

Step 1. State the Problem

Define the problem that motivates the 

Identify the planning team; examine budget, schedule.

Step 2. Identify the Goal of the Study

State how environmental data will be used in solving the

problem; identify study questions; define alternative outcomes.

Step 3. Identify Information Inputs

Step 4. Define the Boundaries of the Study

Specify the target population and characteristics of interest;

define spatial and temporal limits, scale of inference.

Step 5. Develop the Analytic Approach

Define the parameter of interest; specify the type of inference 

and develop logic for drawing conclusions from the findings.

Develop performance criteria for new data being collected, 

Select the most resourceStatistical Estimation and other 

analytical approaches

Step 1. State the Problem

Define the problem that motivates the study;

Identify the planning team; examine budget, s

Step 2. Identify the Goal of the Study

State how environmental data will be used in solving the

Step 3. Identify Information Inputs

Identify data and information needed to answer study questions.

Step 4. Define the Boundaries of the Study

Specify the target population and characteristics of interest;

define spatial and temporal limits, scale of inference.

Step 5. Develop the Analytic Approach

Define the parameter of interest

and develop logic for drawing conclusions from the findings.

Step 6. Specify Performance or Acceptance Criteria

Develop performance criteria for new data being collected, 

acceptance criteria for data already collected.

Step 7. Develop the Detailed Plan for Obtaining Data

Select the most resource effective sampling and analysis plan -

that satisfies the performance or acceptance criteria.

Statistical

Hypothesis Testing

Estimation and other 

Step 1. State the Problem

Define the problem that motivates the 

Identify the planning team; examine budget, schedule.

Step 2. Identify the Goal of the Study

State how environmental data will be used in solving the

problem; identify study questions; define alternative outcomes.

Step 3. Identify Information Inputs

Step 4. Define the Boundaries of the Study

Specify the target population and characteristics of interest;

define spatial and temporal limits, scale of inference.

Step 5. Develop the Analytic Approach

Define the parameter of interest; specify the type of inference 

and develop logic for drawing conclusions from the findings.

Develop performance criteria for new data being collected, 

Select the most resourceStatistical Estimation and other 

analytical approaches

Step 1. State the Problem

Define the problem that motivates the study;

Identify the planning team; examine budget, s

Step 2. Identify the Goal of the Study

State how environmental data will be used in solving the

Step 3. Identify Information Inputs

Identify data and information needed to answer study questions.

Step 4. Define the Boundaries of the Study

Specify the target population and characteristics of interest;

define spatial and temporal limits, scale of inference.

Step 5. Develop the Analytic Approach

Define the parameter of interest

and develop logic for drawing conclusions from the findings.

Step 6. Specify Performance or Acceptance Criteria

Develop performance criteria for new data being collected, 

acceptance criteria for data already collected.

Step 7. Develop the Detailed Plan for Obtaining Data

Select the most resource effective sampling and analysis plan -

that satisfies the performance or acceptance criteria.

Statistical

Hypothesis Testing

Estimation and other 

Figure 1-1. The DQO Process 
EPA QA/G-9S February 2006 7

decision error rates, see Chapter 6 of Guidance on Data Quality Objectives EPA QA/G-4 (U.S. 

EPA 2000). 

If the project has been framed in terms of confidence intervals, then uncertainty is 

expressed as a combination of two interrelated terms: the width of the confidence interval 

(smaller intervals correspond to a smaller degree of uncertainty) or the confidence level that the 

true value of the parameter of interest lies within the interval (a higher confidence level 

represents a smaller degree of uncertainty). 

For the review of sampling design, recall that the key distinction in sampling design is 

between authoritative or judgmental sampling (in which sample numbers and locations are 

selected based on expert opinion) and probability sampling (in which sample numbers and 

locations are selected based on randomization and each member of the target population has a 

known probability of being included in the sample). Judgmental sampling should be considered 

only when the objectives of the investigation are not of a statistical nature (for example, when 

the objective of a study is to identify specific locations of leaks, or when the study is focused 

solely on the sampling locations themselves). Generally, conclusions drawn from authoritative 

samples apply only to the individual samples; aggregation may result in severe bias due to lack 

of representativeness and lead to highly erroneous conclusions. Judgmental sampling also 

precludes the use of the sample for any purpose other than the original one. If judgmental 

sample data are used, great care should be taken in interpreting any statistical statements 

concerning the conclusions to be drawn. Using some probabilistic statement with and 

judgmental sample is incorrect and should be avoided as it gives an illusion of correctness when 

there is none. The further the judgmental sample is from a truly random sample, the more 

questionable the conclusions. Guidance on Choosing a Sampling Design for Environmental 

Data Collection (EPA QA/G-5S) (U.S. EPA 2002) provides extensive information on sampling 

design issues and their implications for data interpretation. 

 The analyst should review the sampling design documentation with the data user's 

objectives in mind. Look for design features that support or contradict those objectives. For 

example, if the data user is interested in making a decision about the mean level of 

contamination in an effluent stream over time, then composite samples may be an appropriate 

sampling approach. On the other hand, if the data user is looking for hot spots of contamination 

at a hazardous waste site, compositing should be used with caution, to avoid "averaging away" 

hot spots. Also, look for potential problems in the implementation of the sampling design. For 

example, if simple random sampling has been used verify that each point in space (or time) had 

an equal probability of being selected for a simple random sampling design. Small deviations 

from a sampling plan may have minimal effect on the conclusions drawn from the data set. 

Significant or substantial deviations should be flagged and their potential effect carefully 

considered throughout the entire DQA. The most important point is to verify that the collected 

data are consistent with how the QA Project Plan, Sampling and Analysis Plan, or overall 

objectives of the study stated them to be. 
EPA QA/G-9S February 2006 8
EPA QA/G-9S February 2006 9

CHAPTER 2 

STEP 2: CONDUCT A PRELIMINARY DATA REVIEW

Step 2: Conduct a Preliminary Data Review

• Review quality assurance reports. 

 Look for problems or anomalies in the implementation of the sample collection and analysis 

procedures. 

 Examine QC data for information to verify assumptions underlying the Data Quality 

Objectives, the Sampling and Analysis Plan, and the QA Project Plans. 

• Calculate the statistical quantities. 

 Consider calculating appropriate percentiles (Section 2.2.1). 

 Select measures of central tendency (Section 2.2.2) and dispersion (Section 2.2.3). 

 If the data involve two variables, calculate the correlation coefficient (Section 2.2.4). 

• Display the data using graphical representations. 

 Select graphical representations (Section 2.3) that illuminate the structure of the data set and 

highlight assumptions underlying the Data Quality Objectives, the Sampling and Analysis 

Plan, and the QA Project Plans. 

 Use a variety of graphical representations that examine different features of the set. 

THE DATA QUALITY ASSESSMENT PROCESS

Review DQOs and Sampling Design

Conduct Preliminary Data Review 

Select the Statistical Method 

Verify the Assumptions 

Draw Conclusions from the Data 

CONDUCT PRELIMINARY DATA REVIEW

Purpose

Generate statistical quantities and graphical

representations that describe the data. Use this

information to learn about the structure of the data

and identify any patterns or relationships.

Activities

Review Quality Assurance Reports

Calculate Basic Statistical Quantities

Graph the Data

Tools

Statistical quantities

Graphical representations
EPA QA/G-9S February 2006 10

List of Boxes 

Page

Box 2-1: Directions for Calculating Percentiles with an Example.............................................. 13 

Box 2-2: Directions for Calculating the Measures of Central Tendency..................................... 14 

Box 2-3: Example Calculations of the Measures of Central Tendency....................................... 14 

Box 2-4: Directions for Calculating the Measures of Dispersion................................................ 15 

Box 2-5: Example Calculations of the Measures of Dispersion.................................................. 15 

Box 2-6: Directions for Calculating Pearson’s Correlation Coefficient with an Example.......... 17 

Box 2-7: Directions for Calculating Spearman’s Correlation Coefficient with an Example ...... 19 

Box 2-8: Directions for Estimating the Serial Correlation Coefficient with a Example ............. 19 

Box 2-9: Directions for Generating a Histogram......................................................................... 21 

Box 2-10: Example of Generating a Histogram .......................................................................... 21 

Box 2-11: Directions for Generating a Stem and Leaf Plot......................................................... 22 

Box 2-12: Example of Generating a Stem and Leaf Plot............................................................. 23 

Box 2-13: Directions for Generating a Box and Whiskers Plot................................................... 24 

Box 2-14: Example of a Box and Whiskers Plot ......................................................................... 24 

Box 2-15: Directions for Generating a Quantile Plot and a Ranked Data Plot............................ 26 

Box 2-16: Example of Generating a Quantile Plot ...................................................................... 26 

Box 2-17: Directions for Constructing a Normal Probability Plot .............................................. 28 

Box 2-18: Example of Normal Probability Plot .......................................................................... 28 

Box 2-19: Directions for Generating a Scatterplot and an Example ........................................... 29 

Box 2-20: Directions for Constructing an Empirical Q-Q Plot with an Example ....................... 32 

Box 2-21: Directions for Generating a Time Plot and an Example............................................. 34 

Box 2-22: Directions for Constructing a Correlogram ................................................................ 35 

Box 2-23: Example Calculations for Generating a Correlogram................................................. 36 

Box 2-24: Directions for Generating a Posting Plot and a Symbol Plot with an Example.......... 38 
EPA QA/G-9S February 2006 11

List of Figures 

Page

Figure 2-1. A Histogram of Concentration Frequencies. ............................................................ 20 

Figure 2-2. A Histogram of Concentration Densities.................................................................. 20 

Figure 2-3. Example of a Box-Plot.............................................................................................. 23 

Figure 2-4. Example of a Quantile Plot of Right-Skewed Data. ................................................. 25 

Figure 2-5. Example of a Scatterplot........................................................................................... 29 

Figure 2-6. Example of a Coded Scatterplot. .............................................................................. 30 

Figure 2-7. Example of a Parallel-Coordinate Plot. .................................................................... 30 

Figure 2-8. Example of a Scatterplot Matrix............................................................................... 31 

Figure 2-9. Example of an Empirical Q-Q Plot........................................................................... 31 

Figure 2-10. Example of a Time Plot. ......................................................................................... 33

Figure 2-11. Example of a Lag Plot............................................................................................. 34 

Figure 2-12. Example of a Correlogram...................................................................................... 35 

Figure 2-13. Example of a Four-Plot........................................................................................... 37 

Figure 2-14. Example of a Posting Plot....................................................................................... 38

Figure 2-15. Example of a Symbol Plot ...................................................................................... 39 

Figure 2-16. Example of a Bubble Plot ....................................................................................... 39 

Figure 2-17. Two Normal Curves, Common Mean, Different Variances................................... 40 

Figure 2-18. The Standard Normal Curve, Centered on Zero. .................................................... 40 

Figure 2-19. Three Different Lognormal Distributions............................................................... 41 
EPA QA/G-9S February 2006 12

CHAPTER 2 

STEP 2: CONDUCT A PRELIMINARY DATA REVIEW 

2.1 OVERVIEW AND ACTIVITIES 

In this step of DQA, the analyst conducts a preliminary evaluation of the data set, 

calculates some basic statistical quantities, and examines the data using graphical representations 

The first activity in conducting a preliminary data review is to review any relevant QA reports 

giving particular attention should be paid to information that can be used to check assumptions 

made in the DQO Process. Of great importance are apparent anomalies in recorded data, missing 

values, deviations from standard operating procedures, and the use of nonstandard data collection 

methodologies. 

Graphs can be used to identify patterns and trends, to quickly confirm or disprove 

hypotheses, to discover new phenomena, to identify potential problems, and to suggest corrective 

measures. Since no single graphical representation will provide a complete picture of the data 

set, the analyst should choose different graphical techniques to illuminate different features of 

the data. Section 2.3 provides descriptions and examples of common graphical representations. 

For a more extensive discussion of the overview and activities of this step, see Data 

Quality Assessment: A Reviewer’s Guide (EPA QA/G-9R) (U.S. EPA 2004). 

2.2 STATISTICAL QUANTITIES 

2.2.1 Measures of Relative Standing 

Sometimes the analyst is interested in knowing the relative position of one or several 

observations in relation to all of the observations. Percentiles or quantiles are measures of 

relative standing that are useful for summarizing data. A percentile is the data value that is 

greater than or equal to a given percentage of the data values. Stated in mathematical terms, the 

p

th percentile is a data value that is greater than or equal to p% of the data values and is less than 

or equal to (1-p)% of the data values. Therefore, if 'x' is the p

th percentile, then p% of the values 

in the data set are less than or equal to x, and (100-p)% of the values are greater than x. A 

sample percentile may fall between a pair of observations. For example, the 75th percentile of a 

data set of 10 observations is not uniquely defined. Therefore, there are several methods for 

computing sample percentiles, the most common of which is described in Box 2-1. 

Important percentiles usually reviewed are the quartiles of the data, the 25th, 50th, and 75th

percentiles. The 50th percentile is also called the sample median (Section 2.2.2), and the 25th and 

75th percentile are used to estimate the dispersion of a data set (Section 2.2.3). Also important 

for environmental data are the 90th, 95th, and 99th percentiles where a decision maker would like 

to be sure that 90%, 95%, or 99% of the contamination levels are below a fixed risk level. 
EPA QA/G-9S February 2006 13

Box 2-1: Directions for Calculating Percentiles with an Example 

Let X1, X2, ..., Xn represent the n data points. To compute the pth percentile, y(p), first rank the data from 

smallest to largest and label these points X(1), X(2), . . ., X(n). The pth percentile is: 

( ) ( ) 1 () ( ) +1 = − ⋅ + ⋅ i X i y p f X f

where r = (n-1)p + 1, i = floor(r), f = r – i, and X( ) () n+1 = X n . Note that floor(r) means calculate r and then 

discard all decimals. 

Example: The 90th and 95th percentile will be computed for the following 10 data points (ordered from 

smallest to largest) : 4, 4, 4, 5, 5, 6, 7, 7, 8, and 10 ppb. 

For the 95th percentile, r = (10-1)×0.95 + 1 = 9.55, i = floor(9.55) = 9, and f = 9.55 – 9 = 0.55. Therefore, 

the 95th percentile is y(0.95) = 0.45×8 + 0.55×10 = 9.1. 

For the 90th percentile, r = (10-1)×0.9 + 1 = 9.1, i = floor(9.1) = 9, and f = 9.1 – 9 = 0.1. Therefore, the 90th

percentile is y(0.9) = 0.9×8 + 0.1×10 = 8.2.

2.2.2 Measures of Central Tendency 

Measures of central tendency characterize the center of a data set. The three most 

common estimates are the mean, median, and the mode. Directions for calculating these 

quantities are contained in Box 2-2; examples are provided in Box 2-3. 

The most commonly used measure of the center of a data set is the sample mean, denoted 

by X . The sample mean can be thought of as the "center of gravity" of the data set. The sample 

mean is an arithmetic average for simple sampling designs; however, for complex sampling 

designs, such as stratification, the sample mean is a weighted arithmetic average. The sample 

mean is influenced by extreme values (large or small) and the treatment of non-detects (see 

Section 4.7). 

The sample median, is the second most popular measure of the center of the data. This 

value falls directly in the middle of the ordered data set. This means that 1⁄2 of the data are 

smaller than the sample median and 1⁄2 of the data are larger than the sample median. The 

median is another name for the 50th percentile (Section 2.2.1). The median is not influenced by 

extreme values and can easily be used if non-detects are present. 

Another method of measuring the center of the data is the sample mode. The sample 

mode is the value that occurs with the greatest frequency. Since the sample mode may not exist 

or be unique, it is the least commonly used measure of center. However, the mode is useful for 

qualitative data. 
EPA QA/G-9S February 2006 14

Box 2-2: Directions for Calculating the Measures of Central Tendency 

Let X1, X2, ..., Xn represent the n data points. 

Sample Mean: The sample mean, X , is the sum of the data points divided by the sample size, n: 

i

n

i

X

n

X = ∑=1

1

Sample Median: The sample median, ~

X , is the center of the ordered data set. To compute the sample 

median, sort the data from smallest to largest and label these points X( 1 ), X( 2 ), . . ., X( n ). Then, 

() ( ) [ ]

( ) ( ) ⎪

⎪

⎪

⎩

⎪⎪

⎪

⎨

⎧ +

=

+

+

if is odd

if is even

2

1

~

1 / 2

/ 2 / 2 1

X n

X X n

X

n

n n

 

Sample Mode: The sample mode is the value in the sample that occurs with the greatest frequency. The 

sample mode may not exist or be unique. Count the number of times each value occurs. The sample 

mode is the value that occurs most frequently. 

Box 2-3: Example Calculations of the Measures of Central Tendency 

The following is an example of computing the sample mean, median, and mode for the 10 data points: 

4, 4, 7, 7, 4, 10, 4, 3, 7, and 8. 

Sample mean:

5.8

10

58

10

4 4 7 7 4 10 4 3 7 8= = + + + + + + + + + X =

Sample median: The ordered data are: 3, 4, 4, 4, 4, 7, 7, 7, 8, and 10. Since n = 10 is even, the sample 

median is: 

median = [ ] X () ( ) + X = [ ] X () () + X = [ ] + = . / 4 7 5 5

2

1

2

1

2

1

10 2 10 / 2 +1 5 6

Sample mode: Computing the number of times each value occurs yields: 

4 appears 4 times; 5 appears 0 times; 6 appears 1 time; 7 appears 3 times; 8 appears 1 time; and 10 

appears 1 time. 

As the value of 4 occurs most often, it is the sample mode of this data set. 
EPA QA/G-9S February 2006 15

2.2.3 Measures of Dispersion 

Measures of central tendency are more meaningful if accompanied by a measure of the 

spread of values about the center. Measures of dispersion in a data set include the range, 

variance, sample standard deviation, coefficient of variation, and the interquartile range. 

Directions for computing these measures are given in Box 2-4; examples are given in Box 2-5. 

Box 2-4: Directions for Calculating the Measures of Dispersion 

Let X1, X2, ..., Xn represent the n data points. 

Sample Range: The sample range, R, is the difference between the largest and smallest values of the 

data set, i.e., R = max(Xi) - min(Xi). 

Sample Variance: To compute the sample variance, s2

, compute: 1

1

2

1 1

2

2

−

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

−

=

∑ ∑ = =

n

X

n

X

s

n

i

i

n

i

i

Sample Standard Deviation: The sample standard deviation, s, is the square root of the sample variance, 

i.e., 2 s = s

Coefficient of Variation: The coefficient of variation (CV) is the sample standard deviation divided by the 

sample mean (Section 2.2.2), i.e., CV = s X/ . The CV is often expressed as a percentage. 

Interquartile Range: The interquartile range (IQR) is the difference between the 75th and the 25th

percentiles, i.e., IQR = y(75) - y(25). 

Box 2-5: Example Calculations of the Measures of Dispersion 

The directions in Box 2-4 and the following 10 data points (in ppm): 4, 5, 6, 7, 4, 10, 4, 5, 7, and 8, are 

used to calculate the measures of dispersion. From Box 2-3, X = 6 ppm . 

Sample Range: R = max(Xi) - min(Xi) = 10 - 4 = 6 ppm 

Sample Variance: 

( ) ( ) 2

2 2 2 2 2

2 4 ppm 9

60

10

1 396

10 1

4 5 8

10

1 4 5 8

=

− ⋅

= −

+ + + − + + +

=

L L

s

Sample Standard Deviation: 4 2 ppm 2 s = s = =

Coefficient of Variation: CV = s / X = 2 ppm / 6 ppm = 0.33 = 33%

Interquartile Range: Using the directions in Section 2.2.1 to compute the 25th and 75th percentiles of the 

data are: y(25) = X(3) = 4.25 ppm and y(75) = X(8) = 7 ppm. The interquartile range (IQR) is the difference 

between these values: IQR = y(75) - y(25) = 7 – 4.25 = 2.75 ppm 

The easiest measure of dispersion to compute is the sample range. For small samples, the 

range is easy to interpret and may adequately represent the dispersion of the data. For large 
EPA QA/G-9S February 2006 16

samples, the range is not very informative because it only considers extreme values and is 

therefore greatly influenced by outliers. 

Generally speaking, the sample variance measures the average squared distance of data 

points from the sample mean. A large sample variance implies the data are not clustered close to 

the mean. A small sample variance (relative to the mean) implies most of the data are near the 

mean. The sample variance is affected by extreme values and by a large number of non-detects. 

The sample standard deviation is the square root of the sample variance and has the same unit of 

measure as the data. 

The coefficient of variation (CV) is a measure having no units that allows the comparison 

of dispersion across several sets of data. The CV (also known as the relative standard deviation) 

is often used in environmental applications because variability (when expressed as a standard 

deviation) is often proportional to the mean. 

When extreme values are present, the interquartile range may be more representative of 

the dispersion of the data than the standard deviation. This statistical quantity is the difference of 

the 75th and 25th percentiles and therefore, is not influenced by extreme values. 

2.2.4 Measures of Association 

Data sets often include measurements of several characteristics (variables) for each 

sampling point. There may be interest in understanding the relationship or level of association 

between two or more of these variables. One of the most common measures of association is the 

correlation coefficient. The correlation coefficient measures the relationship between two 

variables, such as a linear relationship between two sets of measurements. Note that the 

correlation coefficient does not imply cause and effect. The analyst may say the correlation 

between two variables is high and the relationship is strong, but may not say an increase or 

decrease in one variable causes the other variable to increase or decrease without further 

evidence and strong statistical controls. 

2.2.4.1 Pearson’s Correlation Coefficient 

The Pearson correlation coefficient measures the strength of the linear relationship 

between two variables. A linear association implies that as one variable increases, the other 

increases or decreases linearly. Values of the correlation coefficient close to +1 (positive 

correlation) imply that as one variable increases, the other increases nearly linearly. On the other 

hand, a correlation coefficient close to –1 implies that as one variable increases, the other 

decreases nearly linearly. Values close to 0 imply little linear correlation between the variables. 

When data are truly independent, the correlation between data points is zero (note, however, that 

a correlation of 0 does not necessarily imply independence). Directions and an example for 

calculating Pearson’s correlation coefficient are contained in Box 2-6. 
EPA QA/G-9S February 2006 17

Box 2-6: Directions for Calculating Pearson’s Correlation Coefficient with an Example 

Let X1, X2, ..., Xn represent one variable of the n data points and let Y1, Y2, ..., Yn represent a second 

variable of the n data points. The Pearson correlation coefficient, r, between X and Y is: 

⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

⋅ −

⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

−

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

−

=

∑ ∑ ∑ ∑

∑ ∑ ∑

= = = =

= = =

2

1 1

2

2

1 1

2

1 1 1

1 1

1

n

i

i

n

i

i

n

i

i

n

i

i

n

i

i

n

i

i

n

i

i i

Y

n

X Y

n

X

X Y

n

X Y

r

Example: Consider the following data set (in ppb): 

Sample No. 1 2 3 4 

Arsenic 8.0 6.0 2.0 1.0 

Lead 8.0 7.0 7.0 6.0 

17, 28, 105, 198, 8 8 6 7 2 7 1 6 126

4

1

4

1

2

4

1

2

4

1

4

1

∑ = ∑ = ∑ = ∑ = ∑ = ⋅ + ⋅ + ⋅ + ⋅ =

= = = = i =

i i

i

i

i

i

i

i

i

Xi Y X Y X Y

and 0 865

4

28 28 198

4

17 17 105

4

17 28 126

 = .

 - - 

 - 

r = 

⎟

⎠

⎞ ⎜

⎝

⎛ ⋅ ⎟

⎠

⎞ ⎜

⎝

⎛ ⋅

⋅

Since r is close to 1, there is a strong linear relationship between these two contaminants. 

The correlation coefficient does not detect nonlinear relationships so it should be used 

only in conjunction with a scatter plot (Section 2.3.7.2). A scatter plot can be used to determine 

if the correlation coefficient is meaningful or if some measure of nonlinear relationships should 

be used. The correlation coefficient can be significantly influenced by extreme values so a 

scatter plot should be used first to identify such values. 

 

An important property of the correlation coefficient is that it is unaffected by changes in 

location of the data (adding or subtracting a constant from all of the X measurements or all the Y

measurements) or by changes in scale of the data (multiplying or dividing all X or all Y values by 

a positive constant). Thus, linear transformations on the Xs and Ys do not affect the correlation 

of the measurements. This is reasonable since the correlation reflects the degree to which 

linearity between X and Y measurements occur and the degree of linearity is unaffected by 

changes in location or scale. For example, if a variable was temperature in Celsius, then the 

correlation would not change if Celsius was converted to Fahrenheit. 

On the other hand, if nonlinear transformations of the X and/or Y measurements are made, 

then the Pearson correlation between the transformed values will differ from the correlation of 

the original measurements. For example, if X and Y, respectively, represent PCB and dioxin 

concentrations in soil, and x = log(X) and y = log(Y), then the Pearson correlations between X and 
EPA QA/G-9S February 2006 18

Y, X and y, x and Y, and x and y, will all be different, in general, since the logarithmic 

transformation is a nonlinear transformation. 

Pearson’s correlation may be sensitive to the presence of one or two extreme values, 

especially when sample sizes are small. Such values may result in a high correlation, suggesting 

a strong linear trend, when only moderate trend is present. This may happen, for instance, if a 

single (X,Y) pair has very high values for both X and Y while the remaining data values are 

uncorrelated. Extreme values may also lead to low correlations between X and Y, thus tending to 

mask a strong linear trend. This may happen if all the (X,Y) pairs except one (or two) tend to 

cluster tightly about a straight line, and the exceptional point has a very large X value paired with 

a moderate or small Y value (or vice versa). As influences of extreme values can be important, it 

is wise to use a scatter plot (Section 2.3.7.2) in conjunction with a correlation coefficient. 

2.2.4.2 Spearman’s Rank Correlation 

An alternative to the Pearson correlation is Spearman’s rank correlation coefficient. It is 

calculated by first replacing each X value by its rank (i.e., 1 for the smallest X value, 2 for the 

second smallest X value, etc.) and each Y value by its rank. These pairs of ranks are then treated 

as the (X,Y) data and Spearman’s rank correlation is calculated using the same formulae as for 

Pearson’s correlation (Box 2-6). Directions and an example for calculating the Spearman’s Rank 

correlation coefficient are contained in Box 2-7. 

Since meaningful (i.e., monotonic increasing) transformations of the data will not alter 

the ranks of the respective variables (e.g., the ranks for log (X) will be the same for the ranks for 

X), Spearman’s correlation will not be altered by nonlinear increasing transformations of the Xs 

or the Ys. For instance, the Spearman correlation between PCB and dioxin concentrations (X and 

Y) in soil will be the same as the correlations between their logarithms (x and y). This desirable 

property, and the fact that Spearman’s correlation is less sensitive to extreme values, makes 

Spearman’s correlation a good alternative or complement to Pearson’s correlation coefficient. 

2.2.4.3 Serial Correlation 

For a sequence of data points taken serially in time, or one-by-one in a row, the serial 

correlation coefficient can be calculated by replacing the sequencing variable by the numbers 1 

through n and calculating Pearson’s correlation coefficient with x being the actual data values, 

and y being the numbers 1 through n. For example, for a sequence of data collected at a waste 

site along a straight transit line, the distances on the transit line of the data points are replaced by 

the numbers 1 through n, e.g., first 10-foot sample point = 1, the 20-foot sample point = 2, the 

30-foot sample point = 3, etc., for samples taken at 10-foot intervals. Directions for the Serial 

correlation coefficient, along with an example, are given in Box 2-8. 
EPA QA/G-9S February 2006 19

Box 2-7: Directions for Calculating Spearman’s Correlation Coefficient with an Example 

Let X1, X2, ..., Xn represent a set of ranks of the n data points of one data set and let Y1, Y2, ..., Yn

represent a set of ranks of a second variable of the n data points. The Spearman Rank correlation 

coefficient, r, between X and Y is computed by: 

⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

⋅ −

⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

−

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

−

=

∑ ∑ ∑ ∑

∑ ∑ ∑

= = = =

= = =

2

1 1

2

2

1 1

2

1 1 1

1 1

1

n

i

i

n

i

i

n

i

i

n

i

i

n

i

i

n

i

i

n

i

i i

Y

n

X Y

n

X

X Y

n

X Y

r

Example: Concentrations of arsenic and lead are taken at 4 sample locations. The following table gives 

the data set (in ppb) ranked according to the arsenic values. Ranks are given in parentheses. Note that 

any tied values are assigned average rank. 

 Sample No. 4 3 2 1 

 Arsenic 1.0 (1) 2.0 (2) 6.0 (3) 8.0 (4) 

 Lead 6.0 (1) 7.0 (2.5) 7.0 (2.5) 8.0 (4) 

Now, 10, 10, 30, 29.5, 1 1 2 2.5 3 2.5 4 4 29.5

4

1

4

1

2

4

1

2

4

1

4

1

∑ = ∑ = ∑ = ∑ = ∑ = ⋅ + ⋅ + ⋅ + ⋅ =

= = = = i =

i i

i

i

i

i

i

i

i

Xi Y X Y X Y

and 0.9487

4

10 10 29.5

4

10 10 30

4

10 10 29.5

 = 

 - - 

 - 

r = 

⎟

⎠

⎞ ⎜

⎝

⎛ ⋅ ⎟

⎠

⎞ ⎜

⎝

⎛ ⋅

⋅

Since r is close to 1, there is a strong linear relationship between these two contaminants when ranked. 

Box 2-8: Directions for Estimating the Serial Correlation Coefficient with a Example 

Directions: Let X1, X2, . . . , Xn represent the data values collected in sequence over equally spaced periods of 

time. Label the periods of time 1, 2, ..., n to match the data values. Use the directions in Box 2-6 to calculate the 

Pearson’s Correlation Coefficient between the data X and the time periods Y. 

Example: The following are hourly readings from a discharge monitor. 

Time 12:00 13:00 14:00 15:00 16:00 17:00 18:0

0 19:00 20:00 21:00 22:00 23:00 24:00 

Reading 6.5 6.6 6.7 6.4 6.3 6.4 6.2 6.2 6.3 6.6 6.8 6.9 7.0 

Time 

Periods 1 2 3 4 5 6 7 8 9 10 11 12 13 

Using Box 2-6, with the readings being the X values and the Time Periods being the Y values gives a serial 

correlation coefficient of 0.4318. 
EPA QA/G-9S February 2006 20

2.3 GRAPHICAL REPRESENTATIONS 

2.3.1 Histogram 

A histogram is a visual representation of 

the data collected into groups. This graphical 

technique provides a visual method of identifying 

the underlying distribution of the data. The data 

range is divided into several bins or classes and the 

data is sorted into the bins. A histogram is a bar 

graph conveying the bins and the frequency of data 

points in each bin. Other forms of the histogram 

use a normalization of the bin frequencies for the 

heights of the bars. The two most common 

normalizations are relative frequencies 

(frequencies divided by sample size) and densities 

(relative frequency divided by the bin width). 

Figure 2-1 is an example of a histogram using 

frequencies and Figure 2-2 is a histogram of 

densities. 

Histograms provide a visual method of 

accessing location, shape and spread of the data. 

Also, extreme values and multiple modes can be 

identified. The details of the data are lost, but an 

overall picture of the data is obtained. The stemand-leaf plot described in the next section offers the 

same insights into the data as a histogram, but the 

data values are retained. Therefore, stem-and-leaf 

plots can be more informative than histograms for 

smaller data sets. 

The visual impression of a histogram is 

sensitive to the choice of bins. A large number of 

bins will increase data detail while fewer bins will 

increase the smoothness of the histogram. A good 

starting point when choosing the number of bins is the square-root of the sample size. Another 

factor in choosing bins is the choice of endpoints. Using simple bin endpoints can improve the 

readability of the histogram. Simple bin endpoints include multiples of 5k units for some integer 

k. For example, 0 to <5, 5 to <10, etc. (Figure 2-1), or 1 to <1.5, 1.5 to <2, etc. Finally, when 

plotting a histogram for a continuous variable, e.g., concentration, it is necessary to decide on an 

endpoint convention; that is, what to do with data points that fall on the boundary of a bin. With 

discrete variables, (e.g., family size) the intervals can be centered in between the variables. For 

the family size data, the intervals can span between 1.5 and 2.5, 2.5 and 3.5, and so on. Then the 

whole numbers that relate to the family size can be centered within the box. Directions for 

generating a histogram are contained in Box 2-9 and an example is contained in Box 2-10. 

0 5 10 15 20 25 30 35 40

0

2

4

6

8

10

Concentration (ppm)

Number of Observations

Figure 2-1. A Histogram of 

Concentration Frequencies

0 5 10 15 20 25 30 35 40

0

2

4

6

8

Concentration (ppm)

Percentage of Observations (per ppm)

Figure 2-2. A Histogram of 

Concentration Densities
EPA QA/G-9S February 2006 21

Box 2-9: Directions for Generating a Histogram 

STEP 1: Partition the data range into several bins. Partitioning means the bins should be connected, but 

not overlapping. For example, 0 to <5, 5 to <10, etc. In almost all cases, the bin widths should 

be the same. When selecting bins, first choose a number of bins. A good rule of thumb is the 

square root of the sample size. Next, select bin endpoints that are simple (for example, 

multiples of 5k for some integer k) and will produce approximately the number of desired bins. 

STEP 2: Place each data point into the proper bin. This creates a summary of the data called a 

frequency table. If desired, compute one or both types of normalizations of the frequencies. 

The relative frequency is the frequency divided by the sample size. The density is the relative 

frequency divided by the bin width. 

STEP 3: Determine the horizontal axis based on the range of the frequencies or normalized frequencies. 

STEP 4: A histogram is a bar graph of the frequencies or normalized frequencies. For each bin, draw a 

bar using the bin endpoints on the x-axis as the width and the frequency or normalized 

frequency on the y-axis as the height. 

Box 2-10: Example of Generating a Histogram 

Consider the following 22 measurements of a contaminant concentration (in ppm): 17.7, 17.4, 22.8, 35.5, 

28.6, 17.2, 19.1, <4, 7.2, <4, 15.2, 14.7, 14.9, 10.9, 12.4, 12.4, 11.6, 14.7, 10.2, 5.2, 16.5, and 8.9. 

 

STEP 1: With 22 data points, a rough estimate of the number of bins is 22 = 4.69 or 5. Since the data 

ranges from 0 to 40, the suggested bins are 0 to <8, 8 to <16, etc. This is a little awkward and 

using multiples of 5 to create simpler bins leads to 0 to <5, 5 to <10, etc. This choice leads to 8 

bins which is close to suggested number and are the bins that will be used. 

STEP 2: Column 2 of the frequency table below shows the frequency or number of observations within 

each bin defined in Step 1. The third column shows the relative frequencies which is the 

frequency divided by the sample size. The final column of the table gives the densities or the 

relative frequencies divided by the bin widths. 

STEP 3: The horizontal axis for the 

data is from 0 to 40 ppm. 

The vertical axis for the 

histogram of frequencies is 

from 0 - 10 and the vertical 

axis for the histogram of 

densities is from 0% - 10%. 

STEP 4: The histogram of frequencies 

is shown in Figure 2-1 and 

the histogram of densities is 

shown in Figure 2-2. 

2.3.2 Stem-and-Leaf Plot 

The stem-and-leaf plot is used to show both the data values and visual information 

about the distribution of the data. Like a histogram, a stem-and-leaf plot is visual representation 

 Relative 

 Bin Frequency Frequency Density

 0 - 5 ppm 2 9.09 1.8 

 5 - 10 ppm 3 13.64 2.7 

10 - 15 ppm 8 36.36 7.3 

15 - 20 ppm 6 27.27 5.5 

20 - 25 ppm 1 4.55 0.9 

25 - 30 ppm 1 4.55 0.9 

30 - 35 ppm 0 0.00 0.0 

35 - 40 ppm 1 4.55 0.9 
EPA QA/G-9S February 2006 22

of the data collected into groups. However, data detail is retained in a stem-and-leaf. As with 

histograms, stem-and-leaf plots provide an understanding of the shape of the data (and the 

underlying distribution), i.e., location, shape, spread, number of modes and identification of 

potential outliers. Stem-and-leaf is also sensitive to bin selection. A stem-and-leaf plot can be 

more useful than a histogram in analyzing small data sets because it not only allows a 

visualization of the data distribution, but enables the data to be reconstructed. 

First, bins that divide the data range are selected in a similar manner as for a histogram; 

these are the stems. Since the number of data points is typically small the number of bins or 

stems should be approximately 4 to 8. Data points are then sorted into the proper stem. Each 

observation in the stem-and-leaf plot consists of two parts: the stem and the leaf. The stem is 

generally made up of the leading digit or digits of the data values while the leaf is made up of the 

trailing digit or digits. The stems are displayed on the vertical axis and the trailing digits of the 

data points make up the leaves. Changing the stem can be accomplished by increasing or 

decreasing the leading digits that are used, dividing the groupings of one stem (i.e., all numbers 

which start with the numeral 6 can be divided into smaller groupings), or multiplying the data by 

a constant factor (i.e., multiply the data by 10 or 100). Non-detects can be placed at the detection 

limit with the leaf indicating the observation was actually a nondetect. Directions for 

constructing a stem-and-leaf plot are given in Box 2-11 and an example is contained in 

Box 2-12. 

Box 2-11: Directions for Generating a Stem and Leaf Plot 

Let X1, X2, ..., Xn represent the n data points. To develop a stem-and-leaf plot, complete the following steps: 

STEP 1: Arrange the observations in ascending order. The ordered data is labeled (from smallest to largest) 

X(1), X(2), ..., X(n). 

STEP 2: Choose either one or more of the leading digits to be the stem values. For example, if data ranges 

from 0 to 30, then the best choice for stems would be the values of the tens column, 0, 1, 2, 3. 

 

STEP 3: List the stem values from smallest to largest along the vertical axis. Enter the trailing digits for each 

data point as the leaf. The leaves extend to the right of the stems. Continuing the example from 

above, the values of the ones column (and possible one decimal place) would be used as the leaf. 
EPA QA/G-9S February 2006 23

Box 2-12: Example of Generating a Stem and Leaf Plot 

Consider the following 22 samples of trifluorine (in ppm): 17.7, 17.4, 22.8, 35.5, 28.6, 17.2, 19.1, <4, 7.2, <4, 

15.2, 14.7, 14.9, 10.9, 12.4, 12.4, 11.6, 14.7, 10.2, 5.2, 16.5, and 8.9. 

 

STEP 1: Arrange the observations in ascending order: <4, <4, 5.2, 7.7, 8.9, 10.2, 10.9, 11.6, 12.4, 12.4, 

14.7, 14.7, 14.9, 15.2, 16.5, 17.2, 17.4, 17.7, 19.1, 22.8, 28.6, 35.5. 

STEP 2: The data ranges from 0 to 40 so the best choice for the stems is the tens column. Initially, this gives 

4 stems. We can divide the stems in half to give stems of 0 to <5, 5 to <10, etc. This gives 8 stems. 

STEP 3: List the stem values along the vertical axis from smallest to largest. Enter the leaf (the remaining 

digits) values in order from lowest to highest to the right of the stem. 

0 | <4 <4 

0 | 5.2 7.7 8.9 

1 | 0.2 0.9 1.6 2.4 2.4 4.7 4.7 4.9 

1 | 5.2 6.5 7.2 7.4 7.7 9.1 

2 | 2.8 

2 | 8.6 

3 | 

3 | 5.5 

Inspection of the stem-and-leaf indicates the data values are centered in the low teens with a slightly rightskewed distribution. Summary statistics can be calculated to strengthen these visual conclusions. 

2.3.3 Box-and-Whiskers Plot 

Box and Whisker plots, also known as box-plots, are useful in 

situations where a picture of the distribution is desired, but it is not 

necessary or feasible to portray all the details of the data. A box-plot 

(see Figure 2-3) displays several percentiles of the data set. It is a 

simple plot, yet provides insight into the location, shape, and spread 

of the data and underlying distribution. A simple box-plot contains 

only the 0th (minimum data value), 25th, 50th, 75th and 100th 

(maximum data value) percentiles. A more complex version includes 

identification of the mean with a plus-sign and potential outliers 

(identified by stars). Since the box-plot is compact (essentially onedimensional), several can be placed on to a single graph. This allows 

a simple method to compare the locations, spreads and shapes of 

several data sets or different groups within a single data set. In this 

situation, the width of the box can be proportional to the sample size 

of each data set. 

A box-plot divides the data into 4 sections, each containing 25% of the data. The length 

of the central box indicates the spread of the data (the central 50%) while the length of the 

whiskers shows the breadth of the tails of the distribution. The box-plot also demonstrates the 

shape of the data in the following manner. If the upper box and whisker are approximately the 

same length as the lower box and whisker, then the data are distributed symmetrically. If the 

upper box and whisker are longer than the lower box and whisker, then the data are rightFigure 2-3. Example of a 

Box-Plot
EPA QA/G-9S February 2006 24

skewed. If the upper box and whisker are shorter than the lower box and whisker, then the data 

are left-skewed. Directions for generating a box and whiskers plot are contained in Box 2-13, 

and an example is contained in Box 2-14. 

If the mean is added to the box-plot, then recall comparing the mean and median provides 

another method of identifying the shape of the data. If the mean is approximately equal to the 

median, then the data are distributed symmetrically. If the mean is greater than the median, then 

the data are right-skewed; if the mean is less than the median, then the data are left-skewed. 

Box 2-13: Directions for Generating a Box and Whiskers Plot 

STEP 1: Compute the 0th (minimum value), 25th, 50th (median), 75th and 100th (maximum value) percentiles. 

STEP 2: Plot these points either vertically or horizontally. If more than one box-plot is drawn, then the other 

axis can be used to identify the box-plots of the different groups or data sets. Draw a box around 

the 25th and 75th percentiles and add a line through the box at the 50th percentile. Optionally, 

make the width of the box proportional to the sample size. 

STEP 3: If desired, compute the mean and indicate this value on the box-plot with a plus-sign. Also, identify 

potential outliers if desired. A potential outlier is a value at a distance greater than 1.5×IQR from the 

closest end of the box. 

STEP 4: Draw the whiskers from each end of the box to the furthest data point that has not been identified as 

an outlier. Plot the potential outliers using asterisks. 

Box 2-14: Example of a Box and Whiskers Plot 

Consider the following 22 samples of trifluorine (in ppm) listed in order 

from smallest to largest: 4.0, 6.1, 9.8, 10.7, 10.8, 11.5, 11.6, 12.4, 

12.4, 14.6, 14.7, 14.7, 16.5, 17.0, 17.5, 20.6, 20.8, 25.7, 25.9, 26.5, 

32.0, and 35.5. 

STEP 1: The minimum value is 4.0, y(25) = 11.5, the median is 14.7, 

y(75) = 20.8, and the maximum value is 35.5. 

STEP 2: The simple box-plot is drawn using the summary statistics 

calculated in step 1 

STEP 3: The mean of the data is 16.9. The only potential outlier is 

35.5 because 35.5 – 20.8 = 14.7 which is greater than 

1.5×IQR = 13.95. 

STEP 4: The lower whisker extends from 11.5 to 4.0 and the upper 

whisker extends from 20.8 to 32.0 

Examining the box-plot, it is easy to see that the data are centered at 

15 ppm and 50% of the data lie between approximately 12 ppm and 

21 ppm. Also, since the upper box and whisker are longer than the 

lower box and whisker, the distribution of the data is right-skewed. 
EPA QA/G-9S February 2006 25

2.3.4 Quantile Plot and Ranked Data Plots 

The quantile plot and the ranked data plot are two further methods of visualizing the 

location, spread and shape of the data. Both plots are displays of the ranked data and differ only 

in the horizontal axis labels. No subjective decisions such as bin size are necessary and all of the 

data is plotted rather than summaries. 

A ranked data plot is a display of the data from smallest to largest at evenly spaced 

intervals. A quantile plot is a graph of the ranked data versus the fraction of data points it 

exceeds. Therefore, a quantile plot can be used to read the quantile information such as the 

median, quartiles, and the interquartile range. This additional information can aid in the 

interpretation of the shape of the data. 

Using either the quantile plot or the ranked data plot, the spread of the data may be 

visualized by examining the slope of the graph. The closer the general slope is to 0, the lower 

the variability of the data set. 

 

Also using either plot, the shape of the data maybe determined by inspecting the tails of 

the graph. If the left and right tails have approximately the same curvature, then the data are 

distributed symmetrically. If the 

curvature of the right-tail is greater 

than the curvature of the left-tail, then 

the data are right-skewed. This is the 

case of the data plotted in Figure 2-4. 

If the curvature of the left-tail is 

greater than the curvature of the righttail, then the data are left-skewed. 

Finally, the degree of curvature in the 

tails of either plot is proportional to 

the length of the tail of the data. In 

Figure 2-4, the plot rises slowly to a 

point, then the slope increases 

dramatically. This implies there is a 

large concentration of small data 

values and relatively few large data 

values. In this case, the left-tail of the 

data is very short and the right-tail is 

relatively long. 

Directions for developing a quantile plot are given in Box 2-15 and an example is given 

in Box 2-16. 

0 0.2 0.4 0.6 0.8 1

0

1

2

3

4

Fraction of Data (f-values)

Data Values

Upper

Quartile

Lower

Quartile

Interquartile Range

Median

Figure 2-4. Example of a Quantile Plot of RightSkewed Data 
EPA QA/G-9S February 2006 26

Box 2-15: Directions for Generating a Quantile Plot and a Ranked Data Plot 

Let X1, X2, ..., Xn represent the n data points. and X(1), X(2), ..., X(n) be the ordered data from smallest to 

largest . Compute the fractions fi = (i - 0.5)/n for i = 1,...,n. The quantile plot is a plot of the pairs (fi, X( i )), 

with straight lines connecting consecutive points. If desired, add vertical lines indication the quartiles, 

median, or other quantiles of interest. 

Alternatively, a ranked data plot can be made by simply plotting the ordered X values at equally spaced 

intervals along the horizontal axis.

Box 2-16: Example of Generating a Quantile Plot 

Consider the following 10 data points (in ppm): 4, 5, 6, 7, 4, 10, 4, 5, 7, and 8. The data ordered from 

smallest to largest, X(i), are shown in the second row of the table below and the third row displays the 

values fi for each i where fi = (i - 0.5)/n. 

i 1 2 3 4 5 6 7 8 9 10 

X( i ) 4 4 4 5 5 6 7 7 8 10 

fi 0.05 0.15 0.25 0.35 0.45 0.55 0.65 0.75 0.85 0.95 

The pairs (fi, X( i )) are then plotted to yield the following quantile plot: 

0 0.2 0.4 0.6 0.8 1

4

6

8

10

Fraction of Data (f-values)

Data (ppm)

Note that the graph curves upward. This indicates that the data are skewed to the right. 

2.3.5 Quantile-Quantile Plots and Probability Plots 

There are two types of quantile-quantile plots or q-q plots. One is an empirical q-q plot 

which involves plotting the quantiles of two data sets against each other. This is a technique to 

determine if the data sets were generated by the same underlying distribution and is discussed in 

Section 2.3.6.3. The other type of q-q plot involves graphing the quantiles of a data set against 

the quantiles of a specific probability distribution. This is a technique to determine if the data set 

was generated by the theoretical distribution. The following section will focus on the most 

common of these plots for environmental data, the normal probability plot (normal q-q plot). 

However, the discussion holds for other probability distributions. The normal probability plot is 
EPA QA/G-9S February 2006 27

a visual method to roughly determine how well the data set is modeled by a normal distribution. 

Formal tests are contained in Chapter 4, Section 2. Directions for developing a normal 

probability plot are given in Box 2-17 and an example is given in Box 2-18. A discussion of the 

normal distribution is contained in Section 2.4. 

A normal probability plot is the graph of the quantiles of a data set against the quantiles 

of the standard normal distribution. This can be accomplished by using a software package, 

plotting the sample quantiles against standard normal quantiles, or plotting the sample quantiles 

on normal probability paper. If the graph is approximately linear (the correlation coefficient 

being fairly high excluding outliers), then this is an indication that the data are normally 

distributed and a formal test should be performed. If the graph is not linear, then the departures 

from linearity give important information about how the data distribution deviates from a normal 

distribution. 

A nonlinear normal probability plot may be used to interpret the shape and tail behavior 

of the data. First, add the quartile line, the line through the first and third quartiles, to the plot. 

Next, examine the relationship of the tails of the normal probability plot to the quartile line. 

Relationship of the q-q plot to the 

quartile line 

Distribution of the data in relation 

to the normal distribution 

upper-tail lower-tail shape tail behavior 

above below symmetric heavy 

below above symmetric light 

above above right-skewed - 

below below left-skewed - 

A normal probability plot can also be used to identify potential outliers. A data value (or 

a few data values) much larger or much smaller will appear distant from the other values, which 

will be concentrated in the center of the graph. 

As a final note, using a simple natural log transformation of the data, the normal 

probability plot can be used to determine if the data are well modeled by a lognormal 

distribution. The lognormal is an important probability distribution when analyzing 

environmental data where normality cannot be assumed. 

2.3.6 Plots for Two or More Variables 

Data often consist of measurements of several characteristics (variables) for each sample 

point in the data set. For example, a data set may consist of measurements of lead, mercury, and 

chromium for each soil sample or may consist of daily concentration readings for several 

monitoring sites. In this case, graphs may be used to compare and contrast different variables. 
EPA QA/G-9S February 2006 28

Box 2-17: Directions for Constructing a Normal Probability Plot 

Let X1, X2, ..., Xn represent the n data points. Many software packages will produce normal probability 

plots and this is the recommended method. If one of these packages is unavailable, then use the 

following procedure. 

STEP 1: Compute the fractions fi = (i - 0.5)/n for i = 1,...,n. 

STEP 2: Compute the normal quantiles for the fi. These are ( ) i i Y f −1 = Φ , where Φ (⋅) −1 is the inverse 

of the standard normal cumulative density function. The Yi can be found using Table A-1: 

locate an fi in the body of the table, then the Yi is the found on the edges. For example, if fi

= 0.975, then Yi is 1.96. 

STEP 3: Plot the pairs (Yi, Xi), for i = 1, ...,n. 

STEP 4: If desired, then add the quartile line through the first and third quartiles. This acts as a guide 

to determine if the points follow a straight line, which may be fitted to the data. 

If the graph of these pairs approximately form a straight line, then the data may be well modeled by a 

normally distribution. Otherwise, use the table above to determine the shape and tail behavior. 

Box 2-18: Example of Normal Probability Plot 

Consider the following 15 data points: 5, 5, 6, 6, 8, 8, 9, 10, 10, 10, 10, 10, 12, 14, and 15. 

STEP 1: The computed fractions, fi, are: 0.0333, 0.1000, 0.1667, 0.2333,..., 0.9667. 

STEP 2: Working inside-out in Table A-1, the 

normal quantiles for the fi are: -1.83, -

1.28, -0.97, -0.73, -0.52, -0.34, -0.17, 

0.00, 0.17, 0.34, 0.52, 0.73, 0.97, 1.28, 

1.83. 

STEP 3: The data and normal quantile pairs are 

plotted in the graph below. 

STEP 4: Using Box 2-1, the first and third 

quartiles are 7 and 10, respectively. The 

first and third quartiles of the standard 

normal distribution are –0.67 and 0.67, 

respectively. The quartile line is the line 

through the points (-0.67, 7) and (0.67, 

10), and has been added to the plot. 

The plot looks approximately linear, but deviates 

from the quartile line in the upper-tail. However, no 

definite conclusion should be drawn from a plot of 

15 points. A formal test of normality (section 4.2) 

should be performed. 
EPA QA/G-9S February 2006 29

To compare and contrast several variables, collections of the single variable displays 

described in previous sections are useful. For example, the analyst may generate box-plots or 

histograms for each variable using the same axis for all of the variables. Separate plots for each 

variable may be overlaid on one graph, such as overlaying quantile plots for each variable. 

Another useful technique for comparing two variables is to place histograms (sometimes called 

bi-histograms) or stem-and-leaf plots back to back. Additional plots to display two or more 

variables are described in Sections 2.3.6.1 through 2.3.6.3. In many software packages, three 

dimensional scatter plots can be rotated and compared in order to find hidden relationships and 

anomalies. 

2.3.6.1 Scatterplot 

For data sets consisting of multiple 

observations per sampling point, a scatterplot is one 

of the most powerful graphical tools for analyzing 

the relationship between two or more variables. 

Scatterplots are easy to construct for two variables 

(Figure 2-5) and many software packages can 

construct 3-dimensional scatterplots. Directions for 

constructing a 2-dimensional scatterplot are given in 

Box 2-19 along with an example. 

Box 2-19: Directions for Generating a Scatterplot and an Example 

Let X1, X2, ..., Xn represent one variable of the n data points and let Y1, Y2, ..., Yn represent a second variable of 

the n data points. The paired data can be written as (Xi, Yi) for i = 1, ..., n. To construct a scatterplot, plot one 

variable along the horizontal axis and the other variable along the vertical axis. 

Example: PCE values are displayed on the vertical axis and Chromium VI values are displayed on the horizontal 

axis of Figure 2-5. 

Obs PCE 

(ppb) 

Chromium 

VI (ppb) Obs PCE (ppb) 

Chromium 

VI (ppb) Obs PCE (ppb) 

Chromium 

VI (ppb) 

1 14.49 3.76 9 2.23 0.77 17 4.14 2.36 

2 37.21 6.92 10 3.51 1.24 18 3.26 0.68 

3 10.78 1.05 11 6.42 3.48 19 5.22 0.65 

4 18.62 6.30 12 2.98 1.02 20 4.02 0.68 

5 7.44 1.43 13 3.04 1.15 21 6.30 1.93 

6 37.84 6.38 14 12.60 5.44 22 8.22 3.48 

7 13.59 5.07 15 3.56 2.49 23 1.32 2.73 

8 4.31 3.56 16 7.72 3.01 24 7.73 1.61 

 25 5.88 1.42

A scatterplot can clearly show the relationship between two variables if the data range is 

sufficiently large. Truly linear relationships can always be identified in scatter plots, but truly 

nonlinear relationships may appear linear (or some other form) if the data range is relatively 

small. Scatterplots of linearly correlated variables cluster about a straight line. As an example of 

a nonlinear relationship, consider two variables where one variable is approximately equal to the 

square of the other. With an adequate range in the data, a scatterplot of this data would display a 

02468

0

10

20

30

40

Chromium VI (ppb)

PCE (ppb)

Figure 2-5. Example of a Scatterplot 
EPA QA/G-9S February 2006 30

partial parabolic curve. Other important modeling relationships that may appear are exponential 

or logarithmic. Two additional uses of scatterplots are the identification of potential outliers for 

a single variable or for the paired variables and the identification of clustering in the data. 

2.3.6.2 Extensions of the Scatterplot 

It is easy to construct a 2-dimensional scatterplot by hand and many software packages 

can construct a useful 3-dimensional scatterplot. However, with more than 3 variables, it is 

difficult to construct and interpret a scatterplot. Therefore, several graphical representations 

have been developed that extend the idea of a scatterplot for data consisting of more than 2 

variables. 

The simplest of these graphical 

techniques is a coded scatterplot. All 

possible two-way combinations are given 

a code and the pairs of data are plotted on 

one scatterplot. The coded scatterplot 

does not provide information on threeway or higher interactions between the 

variables. If the data ranges for the 

variables are comparable, then a single 

set of axes may suffice, if greater data 

ranges, different scales will be required. 

As an example, consider a data set of 3 

variables, A, B, and C and assume the all 

of the data ranges are similar. An analyst 

may choose to display the pairs (Ai, Bi) 

using a small circle, the pairs (Ai, Ci) 

using a plus-sign, and the pairs (Bi, Ci) using a square on one scatterplot. The completed coded 

scatterplot is given in Figure 2-6. 

The parallel-coordinates 

method employs a scheme where 

coordinate axes are drawn in 

parallel (instead of perpendicular). 

 Figure 2-7 is an example of a 

parallel-coordinate plot. Consider 

a sample point X consisting of 

values X1 for measurement 1, X2

for measurement 2, through Xp for 

measurement p. A parallelcoordinate plot is constructed by 

placing an axis for each of the p

measurements in parallel and 

plotting X1 on axis 1, X2 on axis 2, 

and so on. The points are then 

0 10 20

0

10

20

30

40

(ppb)

(ppb)

Chromium vs. PCE

Atrazine vs. PCE

Atrazine vs. Chromium IV

Figure 2-6. Example of a Coded Scatterplot

Data Values for 

Variable 1

Data Values for 

Variable 2

012345678

0 10 20 30 40

Data Values for 

Variable 3

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Figure 2-7. Example of a Parallel-Coordinate Plot 
EPA QA/G-9S February 2006 31

joined with a broken line. This graph contains all of the information of a scatterplot in addition 

to information concerning 3-way and higher interactions. However, with p measurements, one 

must construct several parallelcoordinate plots in order to display 

all possible pairs. 

A scatterplot matrix is 

another useful method of extending 

scatterplots to higher dimensions. 

In this case, a scatterplot is created 

for all possible pairs of 

measurements which are then 

displayed in a matrix format. This 

method is easy to implement and 

provides a concise method of 

displaying the individual 

scatterplots. Interpretation proceeds 

as with a simple or coded 

scatterplot. As in those cases, this 

method does not provide 

information about 3-way or higher 

interactions. An example of a 

scatterplot matrix is contained in 

Figure 2-8. 

2.3.6.3 Empirical Quantile-Quantile Plot 

An empirical quantile-quantile (q-q) 

plot is a plot of the quantiles (Section 2.2.1) 

of two data sets against each other and is 

similar to the normal probability plot 

discussed in Section 2.3.5. This plot 

(Figure 2-9) is used to compare the 

distributions of two measurements; for 

example, the distributions of lead and iron 

concentrations in a drinking water well. This 

technique is used to determine if the data sets 

were generated from the same underlying 

distribution. If the graph is approximately 

linear, then the distributions are roughly the 

same. If the graph is not linear, then the 

departures from linearity give important information about how the two data distributions differ. 

 The interpretation is similar to that of a normal probability plot. Directions for constructing an 

empirical q-q plot with an example are given in Box 2-20. 

0 10 20 30 40

0

10

20

30

40

Chromium IV (ppb)

Chromium IV (ppb)

0 2 4 6 8 10 12 14

0

10

20

30

40

Atrazine (ppb)

Chromium VI (ppb)

0 10 20 30 40

0

2

4

6

8

10

12

14

Chromium IV (ppb)

Atrazine (ppb)

0 2 4 6 8 10 12 14

0

2

4

6

8

10

12

14

Atrazine (ppb)

Atrazine (ppb)

Figure 2-8. Example of a Scatterplot Matrix 

Figure 2-9. Example of an Empirical 

Q-Q Plot 
EPA QA/G-9S February 2006 32

Box 2-20: Directions for Constructing an Empirical Q-Q Plot with an Example 

Let X1, X2,..., Xn represent n data points of one variable and let Y1, Y2,..., Ym represent a second variable of 

m data points. Let X(1), X(2),..., X(n) and Y(1), Y(2),..., Y(m) be the ordered data sets. 

If n = m, then an empirical q-q plot of the two variables is simply a plot of the ordered values of the 

variables, i.e., a scatterplot of the pairs (X(1), Y(1)), (X(2), Y(2)), ..., (X(n), Y(n)). 

Suppose n > m. Then the empirical quantile-quantile plot will consist of m pairs of points with the ordered 

Y values plotted against the m evenly spaced quantiles of X. Using Box 2-1, compute the quantiles that 

correspond to the fractions q = (j −1) (m −1) j for j = 1,...,m. 

Example: Consider contaminant readings from two separate drinking water wells at the same site. 

well 1: 1.32, 3.26, 3.56, 4.02, 4.14, 5.22, 6.30, 7.72, 7.73, 8.22 

well 2: 0.65, 0.68, 0.68, 1.42, 1.61, 1.93, 2.36, 2.49, 2.73, 3.01, 3.48, 5.44. 

Since the sample sizes are not equal, we need to compute the 10 evenly space quantiles for well 2. 

q1 = 0, so the first quantile is y(q1) = 0.65. 

q2 = 1/9. So r = ( ) n −1 ⋅ q2 + 1= 11⋅1/ 9 + 1= 20 / 9 , i = floor(r ) = 2 , and 9 f = r − i = 2 / . 

Therefore, the second quantile is ( ) ( ) ( ) ( ) y q2 = 7 / 9 ⋅ X 2 + 2 / 9 ⋅ X( ) 3 = 0.68 . 

Continuing this process for j = 2,...,10 yields the following 10 quantiles: 

0.650, 0.680, 1.009, 1.547, 1.894, 2.374, 2.570, 2.886, 3.376, 5.440. 

These, paired with the well 1 data, are plotted in Figure 2-9. 

2.3.7 Plots for Temporal Data 

Data collected over specific time intervals (e.g., monthly, biweekly, or hourly) have a 

temporal component. For example, air monitoring measurements of a pollutant may be collected 

once a minute or once a day; water quality monitoring measurements of a contaminant level may 

be collected weekly or monthly. An analyst examining temporal data may be interested in the 

trends over time, correlation among time periods, or cyclical patterns. Some graphical 

techniques specific to temporal data are the time plot, lag plot, correlogram, and variogram. 

A data sequence collected at regular time intervals is called a time series. Time series 

data analysis is beyond the scope of this guidance. It is recommended that the interested reader 

consult a statistician. The graphical representations presented in this section are recommended 

for any data set that includes a temporal component regardless of the decision to perform a time 

series analysis. The graphical techniques described below will help identify temporal patterns 

that need to be accounted for in any analysis of the data. 

The analyst examining temporal environmental data may be interested in seasonal trends, 

directional trends, serial correlation, or stationarity. Seasonal trends are patterns in the data that 

repeat over time, i.e., the data rise and fall regularly over one or more time periods. Seasonal 

trends may be large scale, such as a yearly cycle where the data show the same pattern of rising 

and falling from year to year, or the trends may be small scale, such as a daily cycle. Directional 
EPA QA/G-9S February 2006 33

trends are positive or negative trends in the data which is of importance to environmental 

applications where contaminant levels may be increasing or decreasing. Serial correlation is a 

measure of the strength of the linear relationship of successive observations. If successive 

observations are related, statistical quantities calculated without accounting for the serial 

correlation may be biased. Finally, another item of interest for temporal data is stationarity 

(cyclical patterns). Stationary data look the same over all time periods. Directional trends or a 

change in the variability in the data imply non-stationarity. 

2.3.7.1 Time Plot 

A time plot (also known as a time series plot) is simply a plot of the data over time. This 

plot makes it easy to identify lack of randomness, changes in location, change in scale, smallscale trends, or large-scale trends over time. Small-scale trends are displayed as fluctuations 

over smaller time periods. For example, ozone levels over the course of one day typically rise 

until the afternoon, then decrease, and this process is repeated every day. Larger scale trends, 

such as seasonal fluctuations, appear as regular rises and drops in the graph. For example, ozone 

levels tend to be higher in the summer than in the winter so ozone data tend to show both a daily 

trend and a seasonal trend. A time plot can also show directional trends or changing variability 

over time. 

A time plot 

(Figure 2-10) is constructed 

by plotting the measurements 

on the vertical axis versus the 

actual time of observation or 

the order of observation on 

the horizontal axis. The 

points plotted may be 

connected by lines, but this 

may create an unfounded 

sense of continuity. It is 

important to use the actual 

time or number at which the 

observation was made. This 

can create discontinuities in the plot but are needed as the data that should have been collected 

now appear as “missing values” but do not disturb the integrity of the plot. Plotting the data at 

equally spaced intervals when in reality there were different time periods between observations 

is not advised. 

The scaling of the vertical axis of a time plot is of some importance. A wider scale tends 

to emphasize large-scale trends, whereas a narrower scale tends to emphasize small-scale trends. 

 Using the ozone example above, a wide scale would emphasize the seasonal component of the 

data, whereas a smaller scale would tend to emphasize the daily fluctuations. Directions for 

constructing a time plot are contained in Box 2-21 along with an example. 

0 5 10 15 20 25 30 35 40 45 50

0

5

10

15

20

Time 

Data Values

Figure 2-10. Example of a Time Plot 
EPA QA/G-9S February 2006 34

Box 2-21: Directions for Generating a Time Plot and an Example 

Let X1, X2, ..., Xn represent n data points listed in order by time, i.e., the subscript represents the ordered time 

interval. A plot of the pairs (i, Xi) is a time plot of the data. 

Example: Consider the following 50 daily observations listed in order by day: 

10.05, 11.22, 15.90, 11.15, 10.53, 13.33, 11.81, 14.78, 10.93, 10.31, 

 7.95, 10.11, 10.27, 14.25, 8.60, 9.18, 12.20, 9.52, 7.59, 10.33, 

12.13, 11.31, 10.13, 7.11, 6.72, 8.97, 10.11, 7.72, 9.57, 6.23, 

 7.25, 8.89, 9.14, 12.34, 9.99, 11.26, 5.57, 9.55, 8.91, 7.11, 

 6.04, 8.67, 5.62, 5.99, 5.78, 8.66, 5.80, 6.90, 7.70, 8.87. 

A time plot is constructed by plotting the pairs (i, Xi) where i represents the number of the day and Xi represents 

the concentration level. The plot is shown in Figure 2-10. Note the slight negative trend. 

2.3.7.2 Lag Plot 

A lag plot is another method to determine if the data set or time series is random. 

Nonrandom structure in the lag plot implies nonrandomness in the data. Examples of 

nonrandom structure are linear patterns or elliptical patterns. A linear pattern implies the data 

contain a directional trend while an elliptical pattern implies the data contain a seasonal 

component. 

If we have data points X1, X2,..., Xn, then a 

lag plot is a scatterplot of the points (Xt, Xt-k) for 

some integer lag k, the most common being lags 

1, 2, or 3. Figure 2-11 is a 1-lag plot for the data 

in the example from Box 2-21. Notice that there 

is a light linear structure suggestion a possible 

directional trend in the data. See Section 2.3.7.3 

for higher lags. 

2.3.7.3 Plot of the Autocorrelation Function 

(Correlogram) 

Serial correlation is a measure of the 

strength of the relationship between successive 

observations. If successive observations are 

related, then the data must be transformed or the 

relationship must be accounted for in the data 

analysis. The correlogram is a plot that is used to display serial correlations when the data are 

collected at equally spaced time intervals. The autocorrelation function is a summary of the 

serial correlations of data. The 1st sample autocorrelation coefficient, r1, is the correlation 

between points at lag 1 (points that are 1 time unit apart); r2 is the correlation between points at 

lag 2; etc. A correlogram (Figure 2-12) is a plot of the sample autocorrelation coefficients, rk, 

versus k. 

Figure 2-11. Example of a Lag Plot 
EPA QA/G-9S February 2006 35

For a large independent data sequence 

of n time points, the autocorrelations are 

approximately normally distributed with mean 

zero and variance 1/n. Therefore, to 

determine if the time points are independent, 

first plot the approximate 95% confidence 

lines ± 2 / n (shown as dashed lines in 

Figure 2-12) on the correlogram. If any of the 

autocorrelations lie outside the confidence 

lines, then there is evidence of serial 

correlation and we conclude that the time 

points are not independent. 

In examining Figure 2-12, there are 4 time points that lie outside the 95% confidence 

bounds. These are at lags 1, 2, 12 and 24. This demonstrates strong evidence that the sequence 

is serially correlated (most likely containing a positive trend along with an annual component). 

In application, the correlogram is only useful for data at equally spaced intervals and for 

irregular intervals a text on the geostatistical use of a variogram is recommended. Directions for 

constructing a correlogram are contained in Box 2-22; example calculations are contained in 

Box 2-23. For large sample sizes, a correlogram is tedious to construct by hand; therefore, 

statistical software should be used. 

Box 2-22: Directions for Constructing a Correlogram 

Let X1, X2, ..., Xn represent the data points ordered by time for equally spaced time points, i.e., X1 was collected 

at time 1, X2 was collected at time 2, and so on. To construct a correlogram, first compute the sample 

autocorrelation coefficients, rk. So for k = 0, 1, ..., compute 

k

k r

g

g =

0

 and ( ) X X ( ) X X

n

g = t t k

n

t k

k − − −

= +

∑

1

1

Once the rk have been computed, a correlogram is the graph (k, rk) for k = 0, 1, . . . , and so on. As an 

approximation, compute up to k = n/6. Also, note that r0 = 1. Finally, place horizontal lines at ± 2 / n . 

2.3.7.4 Multiple Observations in a Time Period 

Many times in environmental data collection, multiple samples are taken at each time 

point. For example, the data collection design may specify collecting 5 samples from a drinking 

well every Wednesday for three months. In this case, the time plot described in Section 2.3.7.1 

may be used to display the complete data set, display the mean weekly level, display a 

confidence interval for each mean, or display a confidence interval for each mean with the 

individual data values. A time plot of all the data will allow the analyst to determine if the 

variability for the different collection periods changes. A time plot of the means will allow the 

0 5 10 15 20 25 30

-0.5

-0.25

0

0.25

0.5

0.75

1

1.25

k

r(k)

Figure 2-12. Example of a Correlogram 
EPA QA/G-9S February 2006 36

analyst to determine if the means are possibly changing between the collection periods. In 

addition, each collection period may be treated as a distinct variable and the methods described 

in Section 2.3.6 may be applied. 

Box 2-23: Example Calculations for Generating a Correlogram 

A correlogram will be constructed using the following four hourly data points: hour 1: 4.5 ppm, hour 2: 3.5 ppm, 

hour 3: 2.5 ppm, and hour 4: 1.5 ppm. Only four data points are used so all computations may be shown for 

illustrative purposes. Therefore, the guideline of computing no more than n/6 autocorrelation coefficients will be 

ignored. The first step to constructing a correlogram is to compute the sample mean (Box 2-2), which is 3. 

Then, 

( )( ) ( ) [ ] ( )( )( )( ) 4.5 3 3.5 3 2.5 3 1.5 3 1.25

4

1

4

1

4

1 2 2 2 2 2

4

1

0

4

1

0 ∑ − − = ∑ − = − + − + − + − =

=

−

=

g = X X X X Xt X

t

t t

t

( )( ) [ ] ( )( ) ( )( ) ( )( ) 3.5 3 4.5 3 2.5 3 3.5 3 1.5 3 2.5 3 0.3125

4

1

4

1

1

4

2

1 − − − = − − + − − + − − =

=

g = ∑ Xt X Xt X

t

Similarly, g2 = -0.375 and g3 = -0.5625. 

So 0.25

1.25

0.3125

0

1

1 = = = g

g r , 0.3

1.25

0.375

0

2

2 − − = = g

g r and 0.45

1.25

0.5625

0

3

2 − − = = g

g r . 

Thus, the correlogram of these data is a plot of (0, 1) (1, 0.25), (2, -0.3) and (3, -0.45) with two confidence lines 

at ± 2 / n = ±1. This graph is shown below. 

01234

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

k

r(k)

-1.0

+1.0

(Hours)

2.3.7.5 Four-Plot 

A four-plot is a collection of 4 specific plots that provide different visual methods for 

illustrating a measurement process. There are 4 basic assumptions that underlie all measurement 

processes, there should be: random data, a fixed distribution, with a fixed location and variance. 

If all of these assumptions hold, then the measurement process or data set is considered to be 

stable. The fixed distribution we discuss here is a normal distribution, but any probability 

distribution of interest could be used. A four-plot consists of a time plot, a lag plot, a histogram 

and a normal probability plot. The data are random if the lag plot is structureless. The data are 
EPA QA/G-9S February 2006 37

from a normal distribution if the histogram is symmetric and bell-shaped and the normal 

probability plot is linear. If the time plot is flat and non-drifting, then the data have a fixed 

location. If the time plot has a constant spread, the data have a fixed variance. Figure 2-13 is a 

four-plot for the data contained in the example of Box 2-21. In this particular example, note that 

the data are not quite normal (deviations from the straight line on the plot), does not have a fixed 

location ( a downward trend in the time plot), and possibly has serial correlation present (the 

tendency of the lag plot to be increasing from left to right). 

Figure 2-13. Example of a Four-Plot 

2.3.8 Plots for Spatial Data 

The graphical representations of the preceding sections may be useful for exploring 

spatial data. However, an analyst examining spatial data may be interested in the location of 

extreme values, overall spatial trends, and the degree of continuity among neighboring locations. 

 Graphical representations for spatial data include postings, symbol plots, correlograms, h-scatter 

plots, and contour plots. 
EPA QA/G-9S February 2006 38

The graphical representations presented in this section are recommended for all spatial 

data regardless of whether or not geostatistical methods will be used to analyze the data. The 

graphical representations described below will help identify spatial patterns that need to be 

accounted for in the analysis of the data. 

2.3.8.1 Posting Plots 

A posting plot (Figure 2-14) is a map of 

data locations along with the corresponding 

data values. Data posting may reveal obvious 

errors in data location and identify data values 

that may be in error. The graph of the sampling 

locations gives the analyst an idea of where the 

data were collected (i.e., the sampling design), 

areas that may have been inaccessible, and 

areas of special interest to the decision maker 

which may have been heavily sampled. It is 

often useful to mark the highest and lowest 

values of the data to see if there are any obvious 

trends. If all of the highest concentrations fall 

in one region of the plot, the analyst may 

consider some method such as post-stratifying 

the data (stratification after the data are 

collected and analyzed) to account for this fact in the analysis. Directions for generating a 

posting plot are contained in Box 2-24. 

Box 2-24: Directions for Generating a Posting Plot and a Symbol Plot with an Example 

On a map of the site, plot the location of each sample. At each location, either indicate the value of the 

data point (a posting plot) or indicate the data value by a symbol (a symbol plot) or circle (a bubble plot). 

The circles on a bubble plot are equal to the square roots of the data values. 

Example: The spatial data displayed in the table below contains both a location (Northing and Easting) 

and a concentration level ([c]). The chosen symbols are floor (concentration/5) and range from ‘0’ to ‘7.’. 

The data values with corresponding symbols are: 

N E [c] Sym N E [c] Sym N E [c] Sym 

15 0 4.0 0 10 10 15.2 3 5 15 12.4 2 

15 5 11.6 2 10 15 35.5 7 5 20 22.8 4 

15 10 14.9 2 10 20 14.7 2 5 25 19.1 3 

15 15 17.4 3 10 25 16.5 3 0 10 10.2 2 

15 20 17.7 3 5 0 8.9 1 0 15 5.2 1 

15 25 12.4 2 5 5 14.7 2 0 20 4.9 0 

10 0 28.6 5 5 10 10.9 2 0 25 17.2 3 

10 5 7.7 1 

The posting plot of this data is displayed in Figure 2-14, the symbol plot is displayed in Figure 2-15, and the 

bubble plot is displayed in Figure 2-16. 

 4.0 11.6 14.9 17.4 17.7 12.4 

 28.6 7.7 15.2 35.5 14.7 16.5 

 8.9 14.7 10.9 12.4 22.8 19.1 

 10.2 5.2 4.9 17.2

Creek

Road

Figure 2-14. Example of a Posting Plot 
EPA QA/G-9S February 2006 39

2.3.8.2 Symbol Plots and Bubble Plots 

For large amounts of data, a posting plot may be visual unappealing. In this case, a 

symbol plot (Figure 2-15) or bubble plot (Figure 2-16) can be used. Rather than plotting the 

actual data values, symbols or bubbles that represent the data values are displayed. A symbol 

plot breaks the data range into bins and plots a bin label corresponding to the data value. A 

bubble plot consists of circles, with radii equal to the square roots of the data values. So unlike 

posting and symbol plots, a bubble plot gives a visual impression of the size of the data values. 

Directions for generating a symbol or bubble plot are contained in Box 2-24. 

 0 2 2 3 3 2 

 5 1 3 7 2 3 

 1 2 2 2 4 3 

 2 1 0 3

Creek

Road

Figure 2-15. Example of a Symbol Plot Figure 2-16. Example of a Bubble Plot 

2.3.8.3 Other Spatial Graphical Representations 

The three plots described in Sections 2.3.8.1 and 2.3.8.2 give information on the location 

of extreme values and spatial trends. The graphs below provide another item of interest to the 

data analyst, continuity of the spatial data. The graphical representations are not described in 

detail because they are used more for preliminary geostatistical analysis. These graphical 

representations can be difficult to develop and interpret. For more information on these 

representations, consult a statistician. 

An h-scatterplot is a plot of all possible pairs of data whose locations are separated by a 

fixed distance in a fixed direction (indexed by h). For example, an h-scatterplot could be based 

on all the pairs whose locations are 1 meter apart in a southerly direction. An h-scatterplot is 

similar in appearance to a scatterplot (Section 2.3.6.1). The shape of the spread of the data in an 

h-scatterplot indicates the degree of continuity among data values a certain distance apart in 

particular direction. If all the plotted values fall close to a fixed line, then the data values at 

locations separated by a fixed distance in a fixed location are very similar. As data values 

become less and less similar, the spread of the data around the fixed line increases outward. The 

data analyst may construct several h-scatterplots with different distances to evaluate the change 

in continuity in a fixed direction. 
EPA QA/G-9S February 2006 40

A spatial correlogram is a plot of the correlations of the h-scatterplots. As the 

h-scatterplot only displays the correlation between the pairs of data whose locations are 

separated by a fixed distance in a fixed direction, it is useful to have a graphical representation of 

how these correlations change for different separation distances in a fixed direction. The spatial 

correlogram is such a plot which allows the analyst to evaluate the change in continuity in a 

fixed direction as a function of the distance between two points. A spatial correlogram is similar 

in appearance to a temporal correlogram (Section 2.3.8.2). 

Contour plots are used to reveal overall spatial trends in the data by interpolating data 

values between sample locations. Most contour procedures depend on the density of the grid 

covering the sampling area (higher density grids usually provide more information than lower 

densities). A contour plot gives one of the best overall pictures of the important spatial features. 

 However, contouring often requires that the actual fluctuations in the data values are smoothed 

so that many spatial features of the data may not be visible. The contour map should be used 

with other graphical representations of the data and requires expert judgment to interpret. 

2.4 PROBABILITY DISTRIBUTIONS 

2.4.1 The Normal Distribution 

Data, especially measurements, often occur in 

natural patterns that can be considered to come from a 

distribution of values. In most instances, the data 

values will be grouped around some measure of central 

tendency such as the mean or median. The spread of 

the data is called the variance (the square root of this is 

called the standard deviation). A distribution with a 

large variance will be more spread out than one with a 

small variance (Figure 2-17). If a histogram of the data 

has a bell-shape (symmetric pattern about the mean 

with rapidly tapering tails), then the underlying 

distribution is often a normal distribution. 

If it is known or assumed that the 

underlying distribution is normal, then this is 

usually written as ‘distributed N(μ,σ

2

)’ where μ is 

the mean and σ

2

 is the variance. By subtracting μ

and dividing by σ, any normal distribution can be 

transformed to a standard normal distribution, 

N(0,1). A standard normal random variable is most 

often denoted by Z. A plot of the standard normal 

is given in Figure 2-18. It is frequently necessary 

to refer to the percentiles of a standard normal and 

Figure 2-17. Two Normal Curves, 

Common Mean, Different 

Variances 

Figure 2-18. The Standard Normal 

Curve, Centered on Zero 
EPA QA/G-9S February 2006 41

in this guidance document, the subscript to a quoted Z-value will denote the percentile (or area 

under the curve, cumulative from the left), see Figure 2-18 (showing the area, 0.8413 

corresponding to a Z-value of 1.00, and written as Z0.8413 =1.00). Although common practice to 

use this subscript notation, some text tables and software programs use a different nomenclature; 

the user is advised to verify the meaning of any statistic encountered. 

2.4.2 The t-Distribution 

The standard normal curve is used when exact information on the mean and variance are 

available, but when only estimates from a sample are available, a different type of sampling 

distribution applies. When only information from a random sample on sample mean and sample 

variance is known for decision making purposes, a Student’s-t distribution is appropriate. It 

resembles a standard normal but is lower in the center and fatter in the tails. The degree of 

fatness in the tails is a function of the degrees of freedom available, which in turn is related to 

sample size. As the sample size increases, the estimates of the mean and variance improve. As a 

result, the Student’s-t distribution more closely resembles the standard normal distribution. 

2.4.3 The Lognormal Distribution 

Another commonly used distribution in 

environmental work is the lognormal distribution. 

The lognormal distribution is bounded on the left by 

0, has a fatter right-tail than the normal distribution, 

and has a right-skewed shape. These characteristics 

are shown in Figure 2-19. The lognormal and 

normal distributions are related by a simple 

transformation: if X is distributed lognormally, then 

Y = ln(X) is distributed normally. However, logtransforming lognormal data to perform statistical 

procedures requiring normal data is a practice that 

should be done with care 

2.4.4 Central Limit Theorem 

In many testing and estimation situations in environmental work, the focus of the 

investigation centers on the mean of a population. It is rare that true normality of the 

observations can be assumed. In most cases, the normally-based statistical tests are not overly 

affected by the lack of normality since tests are very robust and perform tolerably well unless 

gross non-normality is present. In addition, many tests become increasingly tolerant of 

deviations from normality as the number of observations increase. In simple terms, the 

underlying distribution of the sample mean more closely resembles a normal distribution as the 

number of observations increases. This occurs no matter what the underlying distribution of an 

individual observation. This phenomenon is called the Central Limit Theorem and is the basis of 

many statistical procedures. 

Figure 2-19. Three Different Lognormal 

Distributions 
EPA QA/G-9S February 2006 42
EPA QA/G-9S February 2006 43

CHAPTER 3 

STEP 3: SELECT THE STATISTICAL METHOD 

Step 3: Select the Statistical Method

• Select the statistical method based on the data user's objectives and the results of the preliminary 

data review. 

 If the problem involves comparing study results to a fixed threshold, such as a regulatory 

standard, consider the methods in Section 3.2. 

 If the problem involves comparing two populations, such as comparing data from two 

different locations or processes, then consider the hypothesis tests in Section 3.3. 

• Identify the assumptions underlying the statistical method. 

 List the key underlying assumptions of the statistical procedure, such as distributional form, 

dispersion, independence, etc. 

 Note any sensitive assumptions where relatively small deviations could jeopardize the 

validity of the results. 

THE DATA QUALITY ASSESSMENT PROCESS

Review DQOs and Sampling Design

Conduct Preliminary Data Review

Select the Statistical Method 

Verify the Assumptions 

Draw Conclusions from the Data 

SELECT THE STATISTICAL METHOD

Purpose

Select an appropriate procedure for analyzing

data based on the preliminary data review. 

 

Activities

Select Statistical Method

Identify Assumptions Underlying Test 

Tools

Hypothesis tests for a single population 

Hypothesis tests for comparing two populations

Confidence Intervals for a single population

Confidence Intervals for comparing two populations
EPA QA/G-9S February 2006 44

List of Boxes 

Page

Box 3-1: Directions for the One-Sample t-Test........................................................................... 49 

Box 3-2: Directions for Computing a One-Sample t Confidence Interval or Limit .................... 49 

Box 3-3: A One-Sample t-Test Example..................................................................................... 50 

Box 3-4: An Example of a One-Sample t Upper Confidence Limit for a Population Mean....... 50 

Box 3-5: Directions for Computing a One-Sample Tolerance Interval or Limit......................... 51 

Box 3-6: An Example of a One-Sample Upper Tolerance Limit ................................................ 51 

Box 3-7: Directions for a One-Sample t-Test for a Stratified Random Sample .......................... 54 

Box 3-8: An Example of a One-Sample t-Test for a Stratified Random Sample........................ 55 

Box 3-9: Directions for the Chen Test......................................................................................... 56 

Box 3-10: Example of the Chen Test........................................................................................... 57 

Box 3-11: Directions for Computing Confidence Limits for the Population 

Mean of a Lognormal Distribution Using Land’s Method ......................................... 58 

Box 3-12: An Example Using Land’s Method ............................................................................ 58 

Box 3-13: Directions for the One-Sample Test for Proportions.................................................. 59 

Box 3-14: Directions for Computing a Confidence Interval for a Population Proportion........... 59 

Box 3-15: An Example of the One-Sample Test for Proportions................................................ 60 

Box 3-16: Directions for the Sign Test (One-Sample) ................................................................ 62 

Box 3-17: An Example of the Sign Test (One-Sample) .............................................................. 63 

Box 3-18: Directions for the Wilcoxon Signed Rank Test (One-Sample) .................................. 64 

Box 3-19: An Example of the Wilcoxon Signed Rank Test (One-Sample) ................................ 65 

Box 3-20: Directions for the Two-Sample t-Test (Equal Variances) .......................................... 67 

Box 3-21: Directions for a Two-Sample t Confidence Interval (Equal Variances)..................... 68 

Box 3-22: An Example of a Two-Sample t-Test (Equal Variances) ........................................... 68 

Box 3-23: Directions for the Two-Sample t-Test (Unequal Variances)...................................... 69 

Box 3-24: Directions for a Two-Sample t Confidence Interval (Unequal Variances) ................ 70 

Box 3-25: An Example of the Two-Sample t-Test (Unequal Variances).................................... 70 

Box 3-26: Directions for a Two-Sample Test for Proportions .................................................... 72 

Box 3-27: Directions for Computing a Confidence Interval for the Difference Between 

Population Proportions................................................................................................ 73 

Box 3-28: An Example of a Two-Sample Test for Proportions .................................................. 73 

Box 3-29: Directions for the Paired t-Test................................................................................... 74 

Box 3-30: Directions for Computing the Paired t Confidence Interval....................................... 74 

Box 3-31: An Example of the Paired t-Test................................................................................. 75 

Box 3-32: Directions for the Wilcoxon Rank Sum Test.............................................................. 77 

Box 3-33: An Example of the Wilcoxon Rank Sum Test............................................................ 78 

Box 3-34: A Large Sample Example of the Wilcoxon Rank Sum Test ...................................... 79 

Box 3-35: Directions for the Quantile Test.................................................................................. 80

Box 3-36: A Example of the Quantile Test ................................................................................. 80 

Box 3-37: Directions for the Slippage Test ................................................................................. 81 

Box 3-38: A Example of the Slippage Test ................................................................................. 82 

Box 3-39: Directions for the Sign Test (Paired Samples)............................................................ 83 
EPA QA/G-9S February 2006 45

Page

Box 3-40: An Example of the Sign Test (Paired Samples) ......................................................... 84 

Box 3-41: Directions for the Wilcoxon Signed Rank Test (Paired Samples).............................. 85 

Box 3-42: An Example of the Wilcoxon Signed Rank Test (Paired Samples) ........................... 86 

Box 3-43: A Large Sample Example of the Wilcoxon Signed Rank Test (Paired Samples) ...... 87 

Box 3-44: Directions for Dunnett’s Test ..................................................................................... 88

Box 3-45: An Example of Dunnett’s Test ................................................................................... 89 

Box 3-46: Directions of the Fligner-Wolfe Test.......................................................................... 91 

Box 3-47: An Example of the Fligner-Wolfe Test ...................................................................... 92 
EPA QA/G-9S February 2006 46

CHAPTER 3 

STEP 3: SELECT THE STATISTICAL METHOD 

3.1 OVERVIEW AND ACTIVITIES 

This chapter provides an overview of issues associated with selecting an appropriate 

statistical method that will be used to draw conclusions from the data. There are two important 

outputs from this step: (1) the chosen method, and (2) the assumptions underlying the method. If 

a particular statistical procedure has been specified either in the DQO Process, the QA Project 

Plan, or the particular program or study, the analyst should use the results of the preliminary data 

review to determine if it is appropriate for the data collected. If a particular procedure has not 

been specified, then the analyst should select one based upon the data user's objectives and the 

preliminary data review. 

One division in the methods of this section is between parametric and nonparametric 

hypothesis tests. Parametric tests typically concern the population mean or quantile, use the 

actual data values, and assume data values follow a specific probability distribution. 

Nonparametric tests typically concern the population mean or median, use data ranks, and don’t 

assume a specific probability distribution. Parametric tests will have more power than a 

nonparametric counterpart if the assumptions are met. However, the distributional assumptions 

are often strict or undesirable for the parametric tests and deviations can lead to misleading 

results. 

3.2 METHODS FOR A SINGLE POPULATION 

The methods of this section concern comparing a single population parameter to a 

regulatory value (i.e. a fixed number) or the estimation of the population parameter. If the 

regulatory or action-value was estimated, then a one-sample method is not appropriate and a 

two-sample test should be selected. 

An example of a one-sample test would be to determine if 95% of all companies emitting 

sulfur dioxide into the air are below a fixed discharge level. For this example, the population 

parameter is a proportion and the threshold value is 95% (0.95). Comparing the mean 

contaminant concentration of a hazardous site to the mean concentration of a background area 

would be a considered a two-sample test. 

The hypothesis tests discussed in this section may be used to determine if there is 

evidence that θ < θ0, θ > θ0, or θ ≠θ0 where θ represents the population mean, median, 

proportion, or quantile, and θ0 represents the threshold value. There are also 

confidence/tolerance interval procedures to estimate θ. Section 3.2.1 discusses parametric 

hypothesis tests and confidence/tolerance intervals for a population mean or a population 

proportion. Section 3.2.2 discusses nonparametric hypothesis tests for the population median or 

population mean. 
EPA QA/G-9S 47 

Decision Tree for Selecting the Specific Method

Test Population Parameter D

t-Test and CI Mean 

Stratified t-Test Mean 

Chen Test Mean R

Land’s CI Method Mean 

Test for a Proportion and CI Proportion 

Sign Test Median 

Wilcoxon Signed Ranks Test Median /Mean 

t-Test and CI (equal variances) Diff in Means 

t-Test and CI (unequal variances) Diff in Means 

Test for Proportions and CI Diff in Props 

Paired t-Test Diff in Means 

Wilcoxon Rank Sum Test Diff in Means S

Quantile Test Right-Tail 

Slippage Test Right-Tail 

Sign Test Median 

Wilcoxon Signed Ranks Test Median /Mean 

Dunnett’s Test Mean 

Kruskal-Wallis Test Mean 

Paire

d

k-Sample 

Non

parametric

One-Sample 

Nonparametric 

Paire

d

Inde

pendent Non

parametric

Inde

pendent

Parametric 

Parametric

Two-Sample 

Parametric
EPA QA/G-9S 48 February 2006 

3.2.1 Parametric Methods 

These methods rely on the knowing the specific distribution of the population or of the 

statistic of interest. 

3.2.1.1 The One-Sample t-test and Confidence Interval or Limit 

Purpose: Test for a difference between a population mean and a fixed threshold or to estimate a 

population mean. 

Data: A simple or systematic random sample, x1,...,xn, from the population of interest. 

Assumptions: The data are independent and come from an approximately normal distribution or 

the sample size is large (n ≥ 30). 

Limitations and Robustness: One-sample t methods are robust against the population distribution 

deviating moderately from normality. However, they are not robust against outliers and have 

difficulty dealing with non-detects. The nonparametric methods of section 3.2.2 are an 

alternative. The substitution or adjustment procedures of chapter 4 can be used for non-detects. 

Directions for the one-sample t-test are contained in Box 3-1, with an example in Box 3–3. 

Directions for the one-sample confidence interval are contained in Box 3-2, with an example in 

Box 3-4. 

3.2.1.2 The One-Sample Tolerance Interval or Limit 

Purpose: A tolerance interval specifies a region that contains a certain proportion of the 

population with a certain confidence. For example, “the 99% tolerance interval for 90% of the 

population is 7.5 to 9.9”, is interpreted as, “I can be 99% certain that the interval 7.5 to 9.9 

captures 90% of the population.” 

Data: A simple or systematic random sample, x1,...,xn, from the population of interest. The 

sample may or may not contain compositing. 

Assumptions: The data are independent and approximately normally distributed. 

Limitations and Robustness: Tolerance intervals are robust against the population distribution 

deviating moderately from normality. However, they are not robust against outliers. 

Directions for the one-sample tolerance interval are contained in Box 3-5, with an example in 

Box 3-6. 
EPA QA/G-9S 49 February 2006 

Box 3-1: Directions for the One-Sample t-Test 

COMPUTATIONS: Compute the mean, X , and standard deviation, s, of the data set. 

STEP 1. Null Hypothesis: C H : 0 μ =

STEP 2. Alternative Hypothesis: i) C HA : μ > (upper-tail test) 

 ii) C HA : μ < (lower-tail test) 

 iii) C HA : μ ≠ (two-tail test) 

STEP 3. Test Statistic: 

n

s

X t − C

0 =

STEP 4. a) Critical Value: Use Table A-2 to find: 

 i) tn−1,1−α

 ii) −tn−1,1−α

 iii) n−1,1−α 2 t

STEP 4. b) p-value: Use Table A-2 to find: 

 i) ( ) 1 0 t t P n− >

 ii) ( ) 1 0 t t P n− <

 iii) ( ) 2 P 1 0 t t ⋅ n− > , where 0t is the absolute value of t0. 

STEP 5. a) Conclusion: i) If t0 > tn−1,1−α , then reject the null hypothesis that the true population 

mean is equal to the threshold C. 

 ii) If t0 < −tn−1,1−α , then reject the null hypothesis. 

 iii) If 0 > n−1,1−α / 2 t t , then reject the null hypothesis. 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis that the true population mean is 

 equal to the threshold C. 

STEP 6. If the null hypothesis was not rejected, there is only one false acceptance error rate (β at μ1), and if 

( )

( ) C 2

2

1

2

1

2

1 1

2 α β α

μ

− − ′ − ′ + −

+

≥

s z z z

n , then the sample size was probably large enough to achieve the DQOs. 

The value of α′ is α for a one-sided test and 2 α / for a two-sided test. 

Box 3-2: Directions for Computing a One-Sample t Confidence Interval or Limit 

COMPUTATIONS: Compute the sample mean, X , and sample standard deviation, s, of the data set. 

A 100(1 - α)% confidence interval for μ is 

n

s X tn ± ⋅ −1,1−α 2 , where Table A-2 is used to find n−1,1−α 2 t . 

A 100(1 - α)% upper confidence limit (UCL) for μ is 

n

s X tn + ⋅ −1, 1−α . 

A 100(1 - α)% lower confidence limit (LCL) for μ is 

n

s X tn − ⋅ −1, 1−α . 
EPA QA/G-9S 50 February 2006 

Box 3-3: A One-Sample t-Test Example 

Consider the following 9 random data values (in ppm): 

82.39, 103.46, 104.93, 105.52, 98.37, 113.23, 86.62, 91.72, 108.21. 

This data will be used to test: H0: μ ≤ 95 ppm vs. HA: μ > 95 ppm. The decision maker has specified a 5% false 

rejection error rate (α) at 95 ppm (C), and a 20% false acceptance error rate (β) at 105 ppm (μ1). 

COMPUTATIONS: The mean is X = 99.38 ppm and the standard deviation is s = 10.41 ppm. 

STEP 1. Null Hypothesis: H0: μ ≤ 95 

STEP 2. Alternative Hypothesis: HA: μ > 95 (upper-tail test) 

STEP 3. Test Statistic: 126

9

10.41

99 38 95

0 = . . = 

n

s

X C t = − −

STEP 4. a) Critical Value: Using Table A-2, 86 1. tn−1,1−α = t8, 0.95 =

STEP 4. b) p-value Using Table A-2, 0.10 < p-value < 0.15. Using statistical software, 

p - value = P(tn−1 > t0 ) = P(t8 > 1.26) = 0.1216 . 

STEP 5. a) Conclusion: Since 1.26 < 1.86, we fail to reject the null hypothesis that the true population 

 mean is at most 95 ppm. 

STEP 5. b) Conclusion: Since p-value = 0.1216 > 0.05 = significance level, we fail to reject the 

 null hypothesis that the true population mean is at most 95 ppm. 

STEP 6. Since the null hypothesis was not rejected and there is only one false acceptance error rate, it is possible 

to use the sample size formula to determine if the error rate has been satisfied. Since, 

( )

( )

( )

( ) 8 049

2

1645

95 105

10.41 1645 0 842

C 2

2

2

2 2 2

1

2

1

2

1 1

2

 = . . + - 

. + . = 

s z + z z

n α β α

μ

− − ′ − ′ + − ≥ , 

the false acceptance error rate has probably been satisfied. 

Box 3-4: An Example of a One-Sample t Upper Confidence Limit for a Population Mean 

The effluent from a discharge point in a plating manufacturing plant was sampled 7 times over the course of 4 

days for the presence of Arsenic with the following results (in ppm): 8.1, 7.9, 7.9, 8.2, 8.2, 8.0, 7.9. A 95% upper 

confidence limit for the population mean will be computed. 

COMPUTATIONS: The sample mean is X = 8.03 ppm and the sample standard deviation is s = 0.138 ppm. 

A 95% upper confidence limit for μ is: 

7

0.138 + 1, 1 ⋅ . .8.03 ±1.943⋅ − − i e

n

s X tn α or 8.131 ppm. 
EPA QA/G-9S 51 February 2006 

Box 3-5: Directions for Computing a One-Sample Tolerance Interval or Limit 

COMPUTATIONS: Compute the sample mean, X , and sample standard deviation, s, of the data set. 

A 100(1 – α)% tolerance interval for (1 – p)% of the population is X ± k2 ⋅ s , where 2

1,

2

2 1 / 2

1

χ − α

− ⋅

− = ⋅

n

p

n

n k z . 

Table A-1 is used to find z1−p / 2 and Table A-9 is used to find 2 χ n−1, α . 

A 100(1 – α)% upper tolerance limit for (1 – p)% of the population is X + k1 ⋅ s and 

a 100(1 – α)% lower tolerance limit for (1 – p)% of the population is X − k ⋅ s 1 , where 

a

z z ab

k p + p − = − −

2

1 1

1 , 2 ( ) 1

1

2

1

⋅ − = − −n

z

a α , and 

n

z b z p

2

2 1

1 −α = − − . 

Box 3-6: An Example of a One-Sample Upper Tolerance Limit 

The effluent from a discharge point in a plating manufacturing plant was sampled 7 times over the course of 4 

days for the presence of Arsenic with the following results (in ppm): 8.1, 7.9, 7.9, 8.2, 8.2, 8.0, 7.9. A 95% upper 

tolerance limit for 90% of the population will be computed. 

COMPUTATIONS: The sample mean is X = 8.03 ppm and the sample standard deviation is s = 0.138 ppm. 

Also, 0.7731

2 6

1.65 1

2

= ⋅ a = − , 1.2495

7

1.65 1.28

2

2 b = − = , and 2.716

0.7731

1.28 1.28 0.7731 1.2495 2

1 = + − ⋅ k = . 

A 95% upper tolerance limit for 90% of the population is: X + k1 ⋅s i.e. 8.03 + 2.716 ⋅ 0.138 or 8.405 

ppm. 

So we are 95% confident that at least 90% of the population is less than 8.405 ppm. 

3.2.1.3 Stratified Random Sampling 

This section provides a brief introduction of stratified random sampling and directions for 

a one-sample t-test with stratified data. For a more in-depth discussion of stratified random 

sampling, see Guidance on Choosing a Sampling Design for Environmental Data Collection, 

EPA QA/G-5S (U.S. EPA 2002). 

A stratified random sample occurs when a population is split into subpopulations (called 

strata) and simple random samples are taken from the subpopulations. Reasons for 

implementing stratified random sampling include administrative convenience and a gain in the 

precision in the estimates if a heterogeneous population can be split into homogeneous 

subpopulations. 
EPA QA/G-9S 52 February 2006 

Suppose the cost of taking a sample is C = c0 + Σchnh, where c0 is the overhead cost, ch is 

the cost of sampling in stratum h, and nh is the sample size taken from stratum h. Then the 

variance of the sample mean is minimized for given cost C and the cost is minimized for a 

specified variance if nh is proportional to h h h W σ / c , where Wh is the stratum weight (the size 

of the subpopulation in relation to the entire population) and σh is the population standard 

deviation for stratum h. Therefore, a larger sample is taken in a given stratum when the stratum 

is large, the stratum variance is large, or sampling is cheaper in the stratum. If the cost of 

sampling is equal across all strata, then the variance of the sample mean is minimized for a fixed 

total sample size n when 

∑ = ⋅

h h

h h

h W

W

n n

σ

σ . 

This type of sample size selection is called Neyman allocation. In practice, the unknown σh 

is replaced by the observed sample standard deviation. 

Purpose: Test for a difference between a population mean and a fixed threshold or to estimate 

the population mean. 

Data: A stratified random sample, h hnh x , , x 1 K , for h = 1,...,L, where the nh are not necessarily 

equal. Also, let the total sample size be L n = n +L+ n 1 . 

Assumptions: The data in each stratum comes from an approximately normal distribution. 

Limitations and Robustness: Defining the strata requires previous knowledge of the population. 

Further, determining samples for each stratum requires knowledge or assumptions about the 

variance in each stratum. Finally, if strata are chosen poorly, then observations may appear to be 

outliers when they are actually observations from a different stratum. 

Directions for a one-sample t-test for a stratified sample are contained in Box 3-7, with an 

example in Box 3-8. 

3.2.1.4 The Chen Test 

Purpose: Test for a difference between a population mean and a fixed threshold when the 

underlying distribution is right-skewed, i.e., the right-tail of the distribution is longer than the 

left-tail. The standard one-sample t-test works best when the underlying distribution is normal or 

at least symmetric. Often times though, environmental data has an underlying distribution that is 

right-skewed. Therefore, the Chen test may be more appropriate when most of the data values 

are relatively small, but there are also a few relatively large values. 

Data: A simple or systematic random sample, x1,...,xn, from the population of interest. 
EPA QA/G-9S 53 February 2006 

Assumptions: A simple or systematic random sample, x1,...,xn, from the population of interest 

which is believed to be right-skewed. This can be verified by inspection of a histogram of the 

data or a sample skewness that is greater than one. 
EPA QA/G-9S 54 February 2006 

Box 3-7: Directions for a One-Sample t-Test for a Stratified Random Sample 

COMPUTATIONS: Calculate the stratum weights, ∑ =

L

h W h =Vh Vh 1 , where Vh is the surface area of stratum h

multiplied by the depth of sampling in stratum h. For each stratum, calculate the sample stratum mean and the 

sample stratum standard deviation, 

h

h i

n

X hi

n

X

h

=

=

∑1

1

 and ( ) X X . n

s = hi h

n

i h

h

h 2

1

2

1

1 − − ∑=

Compute the overall sample mean and the sample variance of the mean, ∑=

=

L

h

Xst WhXh

1

 and ∑==

L

h h

h n st n

W s

s

1

2

2 . 

Finally, calculate the approximate degrees of freedom, 

∑ ( ) = −

= L

h h h

h h

st

n n

W s

s df

1 2

4 4

4

1

 (round up to the next integer). 

STEP 1. Null Hypothesis: C H : 0 μ =

STEP 2. Alternative Hypothesis: i) C HA : μ > (upper-tail test) 

 ii) C HA : μ < (lower-tail test) 

 iii) C HA : μ ≠ (two-tail test) 

STEP 3. Test Statistic: 

st

st

s

X t C

0

− =

STEP 4. a) Critical Value: Use Table A-2 t find: 

 i) tdf ,1−α

 ii) −tdf ,1−α

 iii) df ,1−α 2 t

STEP 4. b) p-value: Use Table A-2 to find: 

 i) ( ) P 0 t t df >

 ii) ( ) P 0 t t df <

 iii) ( ) 2 P 0 t t ⋅ df >

STEP 5. a) Conclusion: i) If t0 > tdf ,1−α , then reject the null hypothesis that the true population mean 

is equal to the threshold C. 

 ii) If t0 < −tdf ,1−α , then reject the null hypothesis. 

 iii) If 0 > df , 1−α / 2 t t , then reject the null hypothesis. 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis that the true population mean is 

 equal to the threshold C. 

STEP 6. If the null hypothesis was not rejected, there is only one false acceptance error rate (β at μ1), and if 

( )

( ) C 2

2

1

2

1

2

1 1

2 α β α

μ

− − ′ − ′ + −

+

≥

s z z z

n

st , then the sample size was probably large enough to achieve the DQOs. 

The value of α′ is α for a one-sided test and 2 α / for a two-sided test. 
EPA QA/G-9S 55 February 2006 

Box 3-8: An Example of a One-Sample t-Test for a Stratified Random Sample 

Consider a stratified random sample consisting of two strata where stratum 1 comprises 25% of the total site 

surface area and stratum 2 comprises the other 75%. Suppose 40 samples were collected from stratum 1, and 

60 samples were collected from stratum 2. This information will be used to test the null hypothesis that the 

overall site mean is 40 ppm versus the lower-tail alternative. The decision maker has specified a 1% false 

rejection decision limit (α) at 40 ppm and a 20% false acceptance decision error limit (β) at 35 ppm (μ1). 

COMPUTATIONS: The stratum weights are W1 = 0.25, W2 = 0.75. For stratum 1, the sample mean is 31 ppm 

and the sample standard deviation is 18.2 ppm. For stratum 2, the sample mean is 35 ppm, and the sample 

standard deviation is 20.5 ppm. 

The sample overall mean concentration is 0.25 31 0.75 35 34

1

=∑ = ⋅ + ⋅ =

=

L

h

Xst WhXh and the sample variance 

of the mean is 

4.46

60

0.75 20.5

40

0.25 18.2 2 2 2 2

1

2 2

2 = ⋅ + ⋅ =∑ =

=

L

h h

h h

st n

W s

s . 

The approximate degrees of freedom is: 

( )

73.7 74

60 59

0.75 20.5

40 39

0.25 18.2

4.46

1 2

4 4

2

4 4

2

1 2

4 4

4

= →

⋅

⋅ + ⋅⋅ =

−

=

∑ =

L

h h h

h h

st

n n

W s

s df . 

STEP 1. Null Hypothesis: 0 H0 : μ = 4

STEP 2. Alternative Hypothesis: 0 HA : μ < 4 (lower-tail test) 

STEP 3. Test Statistic: 2.841

4.46

C 34 40

0 = − − = − =

st

st

s

X t

STEP 4. a) Critical Value: Using Table A-2, 378 2. −tdf ,1−α = −t74, 0.99 = − . 

STEP 4. b) p-value: Using Table A-2, p-value < 0.005. (the exact value is 

 P( ) tdf < t0 = P(t74 < −2.841) = 0.0029 ). 

STEP 5. a) Conclusion: Since test statistic = -2.841 < -2.378 = critical value, we reject the null 

 hypothesis that the true population mean is equal to 40. 

STEP 5. b) Conclusion: Since p-value = 0.0029 < 0.01 = significance level, we reject the null 

 hypothesis that the true population mean is equal to 40. 

Limitations and Robustness: Chen’s test is a generalization of the one-sample t-test. Like the ttest, this Chen’s test can have some difficulties in dealing with non-detects, especially if there are 

a large number of them. For a moderate amount of non-detects, a substitution method (e.g.,1⁄2 of 

the detection limit) will suffice if the threshold level C is much larger than the detection limit, 

otherwise refer to the methods discussed in Chapter 4. 
EPA QA/G-9S 56 February 2006 

Directions for the Chen test are contained in Box 3-9, with an example in Box 3-10. 

Box 3-9: Directions for the Chen Test 

COMPUTATIONS: Visually check the assumption of right-skewness by inspecting a histogram. If at most 15% of 

the data points are below the detection limit (DL) and C is much larger than the DL, then replace values below the 

DL with DL/2. Compute the mean, X , and the standard deviation, s of the data set. Then compute 

( )( ) b ( ) n

n ns

X X i

i

n

= − − −

= 1 2 3 ∑ 3

1

 (sample skewness), a

b

n = 6

, and t X C

s

n

0 = − . 

NOTE: The sample skewness should be greater than 1 to indicate the underlying distribution is right-skewed. 

STEP 1. Null Hypothesis: C H : 0 μ =

STEP 2. Alternative Hypothesis: i) C HA : μ > (upper-tail test) 

 ii) C HA : μ < (lower-tail test) 

 iii) C HA : μ ≠ (two-tail test) 

STEP 3. Test Statistic: T t a t at t =+ + + + 0 0 ( ) ( ) 2 2 0 03 12 4 2

STEP 4. a) Critical Value: Use Table A-1to find: 

 i) z1−α

 ii) zα

 iii) z1−α 2

STEP 4. b) p-value: Use Table A-1 to find: 

 i) P(Z > T )

 ii) P(Z < T )

 iii) 2⋅P(Z > T )

STEP 5. a) Conclusion: i) If T > z1−α , then reject the null hypothesis that the true population mean is 

equal to the threshold value C. 

 ii) If T < zα , then reject the null hypothesis. 

 iii) If > 1−α / 2 T z , then reject the null hypothesis. 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis that the true population mean is 

equal to the threshold value C. 
EPA QA/G-9S 57 February 2006 

Box 3-10: Example of the Chen Test 

Consider the following sample of a contaminant concentration measurements (in ppm): 

2.0, 2.0, 5.0, 5.2, 5.9, 6.6, 7.4, 7.4, 9.7, 9.7, 10.2, 11.5, 

12.4, 12.7, 14.1, 15.2, 17.7, 18.9, 22.8, 28.6, 30.5, 35.5. 

We want to test the null hypothesis that the mean μ is less than 10 ppm versus the alternative that it exceeds 10 

ppm. A significance level of 0.05 is to be used. 

COMPUTATIONS: A histogram of the 22 data points indicates a right-skewed distribution. It is found that 

X = 13.227 ppm and s = 9.173 ppm. Also, 

( ) 13.227 1.078

21 20 9.173

22 22

1

3

3 − = ⋅ ⋅ = ∑

i =

b Xi , 0.0383

6 22

1.078

a = = , and 1.650

22

9.173

13.227 10

0 = − = − =

n

s

X C t . 

A skewness value of 1.078 indicates the data are right-skewed. 

STEP 1. Null Hypothesis: H0: μ ≤ 10 

STEP 2. Alternative Hypothesis: HA: μ > 10 (upper-tail test) 

STEP 3. Test Statistic: (1 2 ) 4 ( 2 ) 1.96 3

0 0

2 2 T = t0 + a ⋅ + t0 + a ⋅ t + t = . 

STEP 4. a) Critical Value: Using Table A-1, z1−α = z0.95 = 1.645. 

STEP 4. b) p-value: Using Table A-1, P(Z > 1.96) = 1− 0.9750 = 0.0250 . 

STEP 5. a) Conclusion: Since the test statistic = 1.96 > 1.645 = the critical value, we reject 

 the null hypothesis. 

STEP 5. b) Conclusion: Since the p-value = 0.0250 < 0.05 = the significance level, we reject 

 the null hypothesis. 

3.2.1.5 Land’s Method for Lognormally Distributed Data 

Purpose: Estimate the mean of a lognormally distributed population using confidence limits. 

Another alternative is to log-transform the data to make it approximately normal, then use the t

confidence interval described in Box 3-2, and finally transforming the confidence bounds back to 

the original scale. This method produces a biased interval estimate and should be avoided, 

unless a median is being estimated in which case this method produces an unbiased estimate. 

Data: A simple or systematic random sample, x1,...,xn, from the population of interest. The 

sample may or may not contain compositing. 

Assumptions: The underlying distribution of the data is approximately lognormal. 

Limitations and Robustness: Land’s method is extremely sensitive to outliers since the mean and 

standard deviation are not robust against outliers unless a preventative limit on variance is used. 
EPA QA/G-9S 58 February 2006 

Directions for Land’s Method are contained in Box 3-11, with an example in Box 3-12. 

Box 3-11: Directions for Computing Confidence Limits for the Population 

Mean of a Lognormal Distribution Using Land’s Method 

COMPUTATIONS: Transform the data: yi = ln xi, i = 1,...,n. Next, compute the sample mean, y , and sample 

variance, 2

y s , of the transformed data. The values for H1−α and Hα come from Table A-17. 

An upper one-sided 100(1 - α)% confidence limit for the population mean is ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

− + + −

2 1

exp

1

2

n

s s H

y y y α . 

A lower one-sided 100α% confidence limit for population mean is ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

− + +

2 1

exp

2

n

s s H

y y y α . 

Box 3-12: An Example Using Land’s Method 

A random sample of 15 concentrations from a monitoring process (assumed to be lognormal) is reported: 

8.12, 7.32, 4.82, 6.52, 7.80, 11.89, 12.94, 7.51, 18.14, 4.09, 5.70, 15.57, 6.68, 8.15, 5.56. 

Compute an upper one-sided 95% confidence limit for the population mean of the process. 

COMPUTATIONS: The log-transformed data set is: 

2.09, 1.99, 1.57, 1.87, 2.05, 2.48, 2.56, 2.02, 2.90, 1.41, 1.74, 2.75, 1.90, 2.10, 1.72 

The sample mean and sample standard deviation of the transformed data are y = 2.0767 and 4272 = 0. sy . 

The value of H0.95 is found by interpolation in Table A-17. 

An upper one-sided 95% confidence limit for the population mean is 

10.97

14

0.4272 1.9939

2

0.4272

exp 2.0767

2 1

exp

2 0.95

2

=⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ ⋅ = + + ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

− + +

n

s s H

y y y

3.2.1.6 The One-Sample Proportion Test and Confidence Interval 

Purpose: Test for a difference between a population proportion, P, and a fixed threshold (P0) or 

to estimate the population proportion. If P0 = 0.5, this test is equivalent to the Sign test. 

Data: A simple or systematic random sample, x1,...,xn, from the population of interest. 

Assumptions: The data constitutes an independent random sample from the population. 

Limitations and Robustness: Both nP0 and n(1-P0) must be at least 5 to apply the normal 

approximation. Otherwise, exact tests must be used and a statistician should be consulted. 
EPA QA/G-9S 59 February 2006 

Directions for the one-sample test for proportions are contained in Box 3-13, with an example in 

Box 3-15. Directions for computing a confidence interval for a proportion are contained in 

Box 3-14. 

Box 3-13: Directions for the One-Sample Test for Proportions 

COMPUTATIONS: Compute p, the sample proportion of data values that fit the desired characteristic. 

STEP 1. Null Hypothesis: 0 0 H : P = P

STEP 2. Alternative Hypothesis: i) A 0 H : P > P (upper-tail test) 

 ii) A 0 H : P < P (lower-tail test) 

 iii) A 0 H : P ≠ P (two-tail test) 

STEP 3. Test Statistic: 

P ( P )/n

p c P z

0 0

0

0 1−

+ − = , where 

⎩

⎨

⎧

<

− > =

0

0

0.5 / if

0.5 / if

n p P

n p P

c

STEP 4. a) Critical Value: Use Table A-1 to find: 

 i) z1−α

 ii) zα

 iii) z1−α 2

STEP 4. b) p-value: Use Table A-1 to find: 

 i) ( ) P Z > z0

 ii) ( ) P Z < z0

 iii) ( ) 2 P Z > z0 ⋅

STEP 5. a) Conclusion: i) If z0 > z1−α , then reject the null hypothesis that the true proportion is 

equal to the threshold value P0. 

 ii) If z0 < zα , then reject H0. 

 iii) If 0 > 1−α / 2 z z , then reject H0. 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis that the true proportion is 

equal to the threshold value P0. 

STEP 6. If the null hypothesis was not rejected, there is only one false acceptance error rate (β at P1), and if 

⎥

⎥

⎦

⎤

⎢

⎢

⎣

⎡

−

− − ≥ ′

P P

z P ( P )+ z P ( P )

n - -

1 0

1 0 0 1 1 1

2

α 1 β 1

, then the sample size was large enough to achieve the 

DQOs. The value of α′ is α for a one-sided test and 2 α / for a two-sided test. 

Box 3-14: Directions for Computing a Confidence Interval for a Population Proportion 

COMPUTATIONS: Compute p, the sample proportion of data values that fit the desired characteristic. 

A 100(1 - α)% confidence interval for P is ( )

n

p p p z − ± ⋅ −

1

1 α 2 , where Table A-1 is used to find z1−α 2 . 
EPA QA/G-9S 60 February 2006 

Box 3-15: An Example of the One-Sample Test for Proportions 

Consider 85 concentration samples of which 11 have are greater than the clean-up standard. Test the null 

hypothesis that the population proportion of concentrations greater than the clean-up standard is 0.2 versus the 

lower-tailed alternative. The decision maker has specified a 5% false rejection rate (α) for P0 = 0.2, and a false 

acceptance rate (β) of 20% for P1 = 0.15. 

COMPUTATIONS: Since both nP0 = 85⋅0.2 = 17 and n(1- P0) = 85⋅ (1-0.2) = 68 are at least 5, the normal 

approximation can be applied. From the data, the sample proportion is p = 11/85 = 0.1294 

STEP 1. Null Hypothesis: H0: P ≥ 0.20 

STEP 2. Alternative Hypothesis: HA: P < 0.20 

STEP 3. Test Statistic: 149

0 2 1 0 2 85

01294 0 5 85 0 2

1

0 5

0 0

0

0 . . ( . )/

. . / . = 

P ( P )/n

p . /n P z = − −+ −

−

+ − =

STEP 4. a) Critical Value: Using Table A-1, zα = z0.05 = -1.645 . 

STEP 4. b) p-value: Using Tables A-1, P(Z < −1.49) = 1− 0.9319 = 0.0681. 

STEP 5. a) Conclusion: Since the test statistic = -1.49 > -1.645, we fail to reject the null 

 hypothesis that the true population proportion is at least 0.20. 

STEP 5. b) Conclusion: Since p-value = 0.0681 > 0.05 = significance level, we fail to reject the 

 null hypothesis that the true population proportion is at least 0.20. 

STEP 6. Since the null hypothesis was not rejected and there is only one false acceptance error rate, it is 

possible to use the sample size formula to determine if the error rate has been satisfied. Since, 

( ) ( ) ( ) ( ) 365.53

015 0.2

1 1 164 0 2 1 0 2 0.84 015 1 015

2

1 0

1 0 0 1 1 1

2

 = .

. . . . .

P P

z P P + z P P

n - -

⎥

⎥

⎦

⎤

⎢

⎢

⎣

⎡

−

− + − = ⎥

⎥

⎦

⎤

⎢

⎢

⎣

⎡

−

− − < α′ β , 

the sample size was not large enough to achieve the DQOs. So the null hypothesis was not rejected, 

but the false acceptance error rate was not satisfied. Therefore, there is insufficient evidence that the 

proportion is less than 0.20, but this conclusion is uncertain because the sample size was too small. 

3.2.2 Nonparametric Methods 

These methods rely on the relative rankings of data values. Knowledge of the precise 

form of the population distribution is not necessary. 

3.2.2.1 The Sign Test 

Purpose: Test for a difference between the population median, and a fixed threshold. 

Data: A simple or systematic random sample, x1,...,xn, from the population of interest. The 

sample may or may not contain compositing. 
EPA QA/G-9S 61 February 2006 

Assumptions: The Sign test can be used for any underlying population distribution. 

Limitations and Robustness: The Sign test has less power than the one-sample t-test or the 

Wilcoxon Signed Rank test. However, the Sign test makes no distributional assumptions like the 

other two tests and it can handle non-detects if the detection limit is below the threshold. 

Directions for conducting a sign test are contained in Box 3-16, with an example in Box 3-17. 

3.2.2.2 The Wilcoxon Signed Rank Test 

Purpose: Test for a difference between the true location (mean or median) of a population and a 

fixed threshold. If the underlying population distribution is approximately normal, then the onesample t-test will have more power (chance of rejecting the null hypothesis when it is false) than 

the Wilcoxon Signed Rank test. For symmetric distributions, the Wilcoxon Signed Rank test 

will have more power than the Sign test. If the sample size is small and the data is not 

approximately symmetric nor normally distributed, then the Sign test should be used. 

Data: A simple or systematic random sample, x1,...,xn, from the population of interest. 

Assumptions: The data set comes from an approximately symmetric distribution. 

Limitations and Robustness For large sample sizes (n > 50), the one-sample t-test is more robust 

against violations of its assumptions than the Wilcoxon Signed Rank test. The Wilcoxon signed 

rank test may produce misleading results if there are many tied data values. If possible, 

measurements should be recorded with sufficient accuracy so that a large number of tied values 

do not occur. Estimated concentrations should be reported for data below the detection limit, 

even if these estimates are negative, as their relative magnitude is of importance. If this is not 

possible, substitute the value DL/2 for each value below the detection limit providing all the data 

have the same detection limit. When different detection limits are present, all data could be 

censored at the highest detection limit but this will substantially weaken the test. A statistician 

should be consulted on the potential use of Gehan ranking. 

Directions for the Wilcoxon signed rank test are contained in Box 3-18, with an example in 

Box 3-19. 
EPA QA/G-9S 62 February 2006 

Box 3-16: Directions for the Sign Test (One-Sample) 

COMPUTATIONS: Let C be the threshold. Compute the deviations C di = xi − . If any of the deviations are 

zero delete them and correspondingly reduce the sample size. Finally, compute B, the number of times di > 0. 

STEP 1: Null Hypothesis: H0: median = C 

STEP 2: Alternative Hypothesis: i) HA: median > C (upper-tail test) 

 ii) HA: median < C (lower-tail test) 

 iii) HA: median ≠ C (two-tail test) 

STEP 3: Test Statistic: If n ≤ 20, then the test statistic is B. 

 If n > 20, then the test statistic is 

/ 4

/ 2

0

n

B n

z − = . 

STEP 4 a): Critical Value: If n ≤ 20, then use Table A-18 to find 

 i) Bupper(n, 2α) 

 ii) Blower(n, 2α) - 1 

 iii) Blower(n, α) – 1 and Bupper (n, α) 

 If n > 20, then use Table A-1 to find 

 i) z1−α

 ii) zα

 iii) z1−α / 2

STEP 4 b): p-value: If n ≤ 20, then let ( ) n i

n

p i

2

1

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ = and the p-value is 

 i) ∑ ( ) =

n

i B p i

 ii) ∑ ( ) =

B

i p i0

 iii) () () ⎭

⎬

⎫

⎩

⎨

⎧ ⋅ ∑ = ∑ =

n

i B

B

i

2 min p i , p i

0

 If n > 20, then use Table A-1 to find 

 i) ( ) P Z > z0

 ii) ( ) P Z < z0

 iii) ( ) 2 P 0 ⋅ Z > z

STEP 5 a): Conclusion: If n ≤ 20, then 

 i) If B ≥ Bupper(n, 2α), then reject the null hypothesis that the true 

population median is equal to the threshold value C. 

 ii) If B ≤ Blower(n, 2α) - 1, then reject the null hypothesis. 

 iii) If B ≥ Bupper(n, α) or B ≤ Blower(n, α) - 1, then reject the null hypothesis. 

 If n > 20, then 

 i) If z0 > z1−α , then reject the null hypothesis that the true population 

median is equal to the threshold value C. 

 ii) If z0 < zα , then reject the null hypothesis. 

 iii) If |z0| > z1−α / 2 , then reject the null hypothesis. 

STEP 5 b): Conclusion: If p-value < α, then reject the null hypothesis that the true population median is 

equal to the threshold value C. 
EPA QA/G-9S 63 February 2006 

Box 3-17: An Example of the Sign Test (One-Sample) 

The following 10 data points (in ppb) will be used to test the null hypothesis that the population median is at 

least 1000 ppb versus the lower-tail alternative. The decision maker has specified a 10% false rejection error 

rate (α) at 1000 ppb (C), and a 20% false acceptance error rate (β) at 900 ppb ( 1

~μ ). 

COMPUTATIONS: The table below displays the data values and the deviations: 

xi 974 1044 1093 897 879 1161 839 824 796 <750 (DL) 

di -26 44 93 -103 -121 161 -161 -176 -204 -625 

Therefore, B = number of di > 0 = 3. 

STEP 1: Null Hypothesis: H0: median ≥ 1000 ppb 

STEP 2: Alternative Hypothesis: HA: median < 1000 ppb 

STEP 3: Test Statistic: Since n ≤ 20, the test statistic is B = 3. 

STEP 4 a): Critical Value: Since n ≤ 20, Table A-18 is used to find Blower(n, 2α) - 1 = 2. 

STEP 4 b): p-value: Since n ≤ 20, p-value = 0.1719

2

10 1 3

0

10 = ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

∑

i = i . 

STEP 5 a): Conclusion: Since test statistic = 3 > 2 = critical value, we fail to reject the null 

hypothesis that the true median is at least 1000 ppb. 

STEP 5 b): Conclusion: Since p-value = 0.1719 > 0.10 = significance level, we fail to reject the 

null hypothesis that the true median is at least 1000 ppb. 

3.3 COMPARING TWO POPULATIONS 

The two-sample methods of this section concern the comparison of two population 

parameters (means, medians, or proportions). During Step 1: Review DQOs and Sampling 

Design, the population parameters were specified. In environmental applications, the two 

populations to be compared may be a potentially contaminated area with a background area or 

concentration levels from an upgradient and a downgradient well. Another example is 

comparing site concentration levels before and after clean-up to test for significant improvement. 

The populations of interest can be one of two types, independent or paired. With 

independent populations, measurements are taken separately from the two groups of sampling 

units; for example, concentrations from a contaminated site and a background site. The 

observations from paired populations are correlated. Here, two measurements are taken upon 

one set of sampling units at separate instances; for example, measurements before and after 

clean-up or two labs making separate measurements on a single set of samples. Parametric and 

nonparametric methods are described for both independent and paired populations. 
EPA QA/G-9S 64 February 2006 

Box 3-18: Directions for the Wilcoxon Signed Rank Test (One-Sample) 

COMPUTATIONS: Let C be the threshold. Compute the deviations di = xi - C. If any of the deviations are 

zero, then delete them and correspondingly reduce the sample size. Rank the absolute deviations, |di|, from 

smallest to largest. If there are tied observations, then assign the average rank. Let Ri be the signed rank of 

|di|, where the sign of Ri is determined by the sign of di. 

STEP 1. Null Hypothesis: H0: location = C 

STEP 2. Alternative Hypothesis: i) HA: location > C (upper-tail test) 

 ii) HA: location < C (lower-tail test) 

 iii) HA: location ≠ C (two-tail test) 

STEP 3. Test Statistic: If n ≤ 20, then 

{ } ∑>

+ =

i: Ri 0

T Ri , the sum of the positive signed ranks. 

 If n > 20, then ( )

( ) +

+ − + =

T

T n n

z

var

1 / 4

0 , 

 where ( ) ( )( ) ∑ ( ) =

+ − − + + =

g

j

j j t t n n n T

1

2 1

48

1

24

1 2 1

var , g is the number of 

 tied di groups and tj is the size of group j. 

STEP 4. a) Critical Value: If n ≤ 20, then use Table A-7 to find: If n > 20, then use Table A-1 to find: 

 i) n(n+1)/2 - wα i) z1-α

 ii) wα ii) zα

 iii) wα/2 iii) z1- α /2

 

STEP 4. b) p-value: If n ≤ 20, then use Table A-7 to find: If n > 20, then use Table A-1 to find: 

 i) ( ( ) ) + P W ≤ n n + 1 / 2 −T i) ( ) P 0 Z > z

 ii) ( ) + P W ≤ T ii) ( ) P 0 Z < z

 iii) { ( ( ) ) ( ) } + + 2 ⋅ min P W ≤ n n + 1 / 2 −T , P W ≤ T iii) 2 P(Z>| 0 z |)

 STEP 5. a) Conclusion: If n ≤ 20, then: If n > 20, then: 

i) If + T ≥ n(n+1)/2 - wα, then reject the null i) If z0 > z1-α, then reject the null 

 ii) If + T ≤ wα, then reject the null hypothesis. ii) If z0 < zα, then reject the null 

iii) If + T ≥ n(n+1)/2 - wα/2 or + T ≤ wα, then reject the null iii) If |z0| > z1-α/2, then reject the null 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis that the true population 

location is equal to the threshold C. 

STEP 6. If the null hypothesis was not rejected and n>20, then the sample size necessary to achieve the 

DQOs assuming only one false acceptance error rate (β at μ1) has been specified is 

( )( )

( ) ⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧

+ −

+

≥ ⋅ − − ′ − ′ +

C 2

var

1.16

2

1

2

1

2

1 α 1 β α

μ

T z z z

n . 

If this is true then the false acceptance error rate has probably been satisfied (the value of α′ is α for 

a one-sided test and 2 α / for a two-sided test). 
EPA QA/G-9S 65 February 2006 

Box 3-19: An Example of the Wilcoxon Signed Rank Test (One-Sample) 

The following 10 data points (in ppb) will be used to test the null hypothesis that the population median is at 

least 1000 ppb versus the lower-tail alternative. The decision maker has specified a 10% false rejection error 

rate (α) at 1000 ppb (C), and a 20% false acceptance error rate (β) at 900 ppb). 

974, 1044, 1093, 897, 879, 1161, 839, 824, 796, <750 (detection limit) 

COMPUTATIONS: For this example, the only option is to assign the value 375 ppb (DL/2) to the nondetect. 

The table below displays the remaining computations. 

xi 974 1044 1093 897 879 1161 839 824 796 375 

di -26 44 93 -103 -121 161 -161 -176 -204 -625 

|di| 26 44 93 103 121 161 161 176 204 625 

rank 1 2 3 4 5 6.5 6.5 8 9 10 

Ri -1 2 3 -4 -5 6.5 -6.5 -8 -9 -10 

STEP 1. Null Hypothesis: H0: median ≥ 1000 ppb 

STEP 2. Alternative Hypothesis: HA: median < 1000 ppb 

STEP 3. Test Statistic: Since n ≤ 20, compute 

{ }

2 3 6.5 11.5

: 0

= ∑ = + + =

>

+

Ri i

T Ri

STEP 4. a) Critical Value: Since n ≤ 20, Table A-7 is used to find w0.10 = 14. 

STEP 4. b) p-value: Since n ≤ 20, Table A-7 is used to find the p-value is between 0.05 and 

0.075. Using software, the p-value is found to be 0.0527. 

STEP 5. a) Conclusion: Since test statistic = + T = 11.5 ≤ 14 = critical value, we reject the null 

hypothesis that the true population median is at least 1000. 

STEP 5. b) Conclusion: Since p-value = 0.0527 < 0.10 = significance level, we reject the null 

hypothesis that the true population median is at least 1000. 

NOTE: The Sign test failed to reject the null hypothesis for this example. Recall that the Wilcoxon Signed 

Rank test has more power than the Sign test if the distribution of the population is symmetric. 

The hypothesis tests in this section may be used to determine if there is evidence that 

θ1 – θ2 < δ0, θ1 – θ2 > δ0, or θ1 – θ2 ≠ δ0, where θ1 and θ2 represent population means, medians, or 

proportions and δ0 represents the threshold value. Also, there are confidence interval procedures 

to estimate θ1 – θ2 . Section 3.3.1.1 covers parametric methods for comparing two independent 

populations, while Section 3.3.1.2 covers parametric methods for comparing two paired 

populations. Section 3.3.2 describes nonparametric counterparts to the parametric methods. 

3.3.1 Parametric Methods 

These methods rely on the knowing the specific distribution of the populations or of the 

statistics of interest. 
EPA QA/G-9S 66 February 2006 

3.3.1.1 Independent Samples 

3.3.1.1.1 The Two-Sample t-Test and Confidence Interval (Equal Variances) 

Purpose: Test for a difference or estimate the difference between two population means when it 

can be assumed the population variances are approximately equal. 

Data: A simple or systematic random sample x1, x2, . . . , xm from one population, and an 

independent simple or systematic random sample y1, y2, . . . , yn from the second population. 

Assumptions: The two populations are independent. If not, then it is possible that a paired 

method could be used. Both are approximately normally distributed or the sample sizes are large 

(m and n both at least 30). If this is not the case, then a nonparametric procedure is an 

alternative. Finally, the variances of both populations are approximately equal. If the population 

variances are not equal (tests are available in Section 4.5), then use the methods of the next 

section. 

Limitations and Robustness: The two-sample t-test with equal variances is robust to moderate 

violations of the assumption of normality, but not to large inequalities of variances. An 

alternative is the parametric methods for unequal variances described in the next section. The ttest is not robust against outliers because sample means and standard deviations are sensitive to 

outliers. 

Directions for the two-sample t-test with equal variances are contained in Box 3-20, with an 

example in Box 3-22. Directions for a two-sample confidence interval with equal variances are 

contained in Box 3-21. 

3.3.1.1.2 The Two-Sample t-Test and Confidence Interval (Unequal Variances) 

Purpose: Test for a difference or estimate the difference between two population means when it 

is suspected the population variances are not equal. 

Data: A simple or systematic random sample x1, x2, . . . , xm from the one population, and an 

independent simple or systematic random sample y1, y2, . . . , yn from the second population. 

Assumptions: The two populations are independent. If not, then it is possible that a paired 

method could be used. Both are approximately normally distributed or the sample sizes are large 

(m and n both at least 30). If this is not the case, then a nonparametric procedure is an 

alternative. 

Limitations and Robustness: The two-sample t-test with unequal variances is robust to moderate 

violations of the assumption of normality. The t-test is also not robust to outliers because sample 

means and standard deviations are sensitive to outliers. 
EPA QA/G-9S 67 February 2006 

Directions for the two-sample t-test with unequal variances are contained in Box 3-23, with an 

example in Box 3-25. Directions for a two-sample confidence interval with unequal variances 

are contained in Box 3-24. 

Box 3-20: Directions for the Two-Sample t-Test (Equal Variances) 

COMPUTATIONS: Calculate the sample means, X and Y , and the sample variances, sX

2 and sY2 of the two 

populations. Also compute the pooled standard deviation estimate, ( ) ()

2

1 1 2 2

+ −

− ⋅ + − ⋅ = m n

m s n s

s

X y

p

STEP 1. Null Hypothesis: 0 0 H : μ X − μY = δ

STEP 2. Alternative Hypothesis: i) A 0 H : μ X − μY > δ (upper-tail test) 

 ii) A 0 H : μ X − μY < δ (lower-tail test) 

 iii) A 0 H : μ X − μY ≠ δ (two-tail test) 

STEP 3. Test Statistic: ( )

m n

s

X Y t

p

1 1

0

0

+

− − = δ

STEP 4. a) Critical Value: Use Table A-2 to find: 

 i) tm+n−2,1−α

 ii) −tm+n−2,1−α

 iii) m+n−2,1−α 2 t

STEP 4. b) p-value: Use Table A-2 to find: 

 i) ( ) P 2 0 t t m+n− >

 ii) ( ) P 2 0 t t m+n− <

 iii) ( ) 2 P 2 0 t t ⋅ m+n− >

STEP 5. a) Conclusion: i) If t0 > tm+n−2,1−α , then reject the null hypothesis that the true difference 

between population means is equal to the threshold δ0. 

 ii) If t0 < −tm+n−2,1−α , then reject the null hypothesis. 

 iii) If 0 > m+n−2,1−α / 2 t t , then reject the null hypothesis. 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis that the true difference 

between population means is equal to the threshold δ0. 

STEP 6. If the null hypothesis was not rejected, there is only one false acceptance error rate (β at δ1), and both m

and n are at least ( )

( ) 4

2 2

1

2

1 0

2

1 1

2 α β α

δ δ

− − ′ − ′ + −

sp z + z z , then the sample sizes were probably large enough to 

achieve the DQOs. The value of α′ is α for a one-sided test and 2 α / for a two-sided test. 
EPA QA/G-9S 68 February 2006 

Box 3-21: Directions for a Two-Sample t Confidence Interval (Equal Variances) 

COMPUTATIONS: Calculate the sample means, X and Y , and the sample variances, sX

2 and sY2 of the two 

populations. Also compute the pooled standard deviation estimate, ( ) ()

2

1 1 2 2

+ −

− ⋅ + − ⋅ = m n

m s n s

s X y

p . 

A 100(1 - α)% confidence interval for μ X − μY is ( ) m n

X Y t s m n p

1 1 − ± + −2, 1−α 2 ⋅ + , where Table A-2 is used to 

find m+n−2, 1−α 2 t . 

Box 3-22: An Example of a Two-Sample t-Test (Equal Variances) 

At a hazardous waste site, an area cleaned using an in-situ methodology (area 1) was compared with a similar, but 

relatively uncontaminated reference area (area 2). If the in-situ methodology worked, then the average contaminant 

levels at the two sites should be approximately equal. If the methodology did not work, then area 1 should have a 

higher average than the reference area. Suppose 7 random samples were taken from area 1, and 8 were taken 

from area 2. Methods described in Section 4.5 were used to determine that the variances were essentially equal. 

The false rejection error rate was set at 5% and the false acceptance error rate was set at 20% (β) if the difference 

between the areas is 2.5 ppm (δ1). 

COMPUTATIONS: The sample means and sample variances are X1 = 7.8 , X2 = 6.6 , 2.1 2 s1 = , and 2.2 2 s2 = . 

The pooled standard deviation is: ( ) ( ) 14676

7 8 2

7 1 21 8 1 2 2 . . + . s p = + −− ⋅ − ⋅ =

STEP 1. Null Hypothesis: 0 H : 0 μ1 − μ2 =

STEP 2. Alternative Hypothesis: 0 H : A μ1 − μ2 >

STEP 3. Test Statistic: ( ) 15799

8

1

7

1 14676

7 8 6 6 0

1 1

1 2 0

0 .

.

. . = 

m n

s

X X t

p

=

+

− −

+

− − = δ

STEP 4. a) Critical Value: Using Table A-2, tm+n−2,1−α = t13, 0.95 = 1.771

STEP 4. b) p-value: Using Table A-2, 0.05 < p-value < 0.10. (The exact 

 p - value = P( tm+n−2 > t0 ) = P( t13 > 1.5799 ) = 0.0691 ) 

STEP 5. a) Conclusion: Since test statistic = 1.5799 < 1.771 = critical value, we fail to reject the null 

hypothesis of no difference between population means. 

STEP 5. b) Conclusion: Since p-value = 0.0691 > 0.05 = significance level, we fail to reject the null 

hypothesis of no difference between population means. 

STEP 6. Since the null hypothesis was not rejected and there is only one false acceptance error rate, we can 

compute the sample sizes necessary to achieve the DQOs. Each sample size must be at least 

( )

( ) 0.25 1.645 4.94

2.5 0

2 1.4676 1.645 0.842 2

2

2 2

+ ⋅ = −

⋅ + . 

Since m and n are both greater than 4.94, the false acceptance error rate has probably been satisfied. 
EPA QA/G-9S 69 February 2006 

Box 3-23: Directions for the Two-Sample t-Test (Unequal Variances) 

COMPUTATIONS: Calculate the sample means, X and Y , and the sample variances, sX

2 and sY2 of the two 

populations. Also, compute the degrees of freedom (Satterthwaite’s approximation) for the test: 

( ) 1 ( ) 1 2

2

2

2

2 2 2

− + −

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+

=

n n

s

m m

s

n

s

m

s

df

X Y

X Y

, rounded down to the next integer. 

STEP 1. Null Hypothesis: 0 0 H : μ X − μY = δ

STEP 2. Alternative Hypothesis: i) A 0 H : μ X − μY > δ (upper-tail test) 

 ii) A 0 H : μ X − μY < δ (lower-tail test) 

 iii) A 0 H : μ X − μY ≠ δ (two-tail test) 

STEP 3. Test Statistic: ( )

n

s

m

s

X Y t

X Y

2 2

0

0

+

− − = δ

STEP 4. a) Critical Value: Use Table A-2 to find: 

 i) tdf ,1−α

 ii) −tdf,1−α

 iii) df ,1−α 2 t

STEP 4. b) p-value: Use Table A-2 to find: 

 i) ( ) P 0 t t df >

 ii) ( ) P 0 t t df <

 iii) ( ) 2 P 0 t t ⋅ df >

STEP 5. a) Conclusion: i) If t0 > tdf ,1−α , then reject the null hypothesis that the difference 

between the population means is equal to δ0. 

 ii) If t0 < −tdf ,1−α , then reject H0. 

 iii) If 0 > df ,1−α / 2 t t , then reject H0. 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis that the difference between 

the population means is equal to δ0. 

STEP 6. If the null hypothesis was not rejected, then there is no simple method to determine if the sample sizes 

were large enough to achieve the DQOs. 
EPA QA/G-9S 70 February 2006 

Box 3-24: Directions for a Two-Sample t Confidence Interval (Unequal Variances) 

COMPUTATIONS: Calculate the sample means, X and Y , and the sample variances, sX

2 and sY2 of the two 

populations. Also, compute the degrees of freedom (Satterthwaite’s approximation) for the confidence interval: 

( ) 1 ( ) 1 2

4

2

4

2 2 2

− + −

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+

=

n n

s

m m

s

n

s

m

s

df

X Y

X Y

, rounded down to the next integer. 

A 100(1 - α)% confidence interval for μ X − μY is ( ) n

s

m

s X Y t X Y

df

2 2

− ± , 1−α 2 ⋅ + , where Table A-2 is used to find 

m+n−2, 1−α 2 t . 

Box 3-25: An Example of the Two-Sample t-Test (Unequal Variances) 

At a hazardous waste site, an area cleaned using a new methodology (area 1) was compared with a similar area 

cleaned with the standard technology (area 2). If the new methodology worked, then the two sites should be 

approximately equal in average contaminant levels. If the new methodology did not work, then area 1 should 

have a higher average than the reference area. Suppose 7 random samples were taken from area 1 and 8 were 

taken from area 2. As the contaminant concentrations in the two areas are supposedly equal, we will test no 

difference in population means versus the upper-tail alternative. Using Section 4.5, it was determined that the 

variances of the two populations were not equal, therefore using Satterthwaite's method is appropriate. The false 

rejection error rate was set at 5% and the false acceptance error rate was set at 20% (β) if the difference between 

the areas is 2.5 ppm. 

COMPUTATIONS: The sample means and sample variances are X1 = 9.2 , X2 = 6.1 , 1.3 2 s1 = , and 5.7 2 s2 = . 

 Satterthwaite’s approximation of the degrees of freedom for the test is: 

( ) ( ) ( ) ( )

10.3 10

8 8 1

5.7

7 7 1

1.3

8

5.7

7

1.3

1 1 2

2

2

2

2

2

4

2

4

2 2 2

= →

− + −

⎟

⎠

⎞ ⎜

⎝

⎛ +

=

− + −

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+

=

n n

s

m m

s

n

s

m

s

df

X Y

X Y

STEP 1. Null Hypothesis: 0 H : 0 μ1 − μ2 =

STEP 2. Alternative Hypothesis: 0 H : A μ1 − μ2 >

STEP 3. Test Statistic: ( ) 3.271

8

5.7

7

1.3

9.2 61 0

2

2

2

1

1 2 0

0 =

+

− −

+

− − = . = 

n

s

m

s

X X t δ

STEP 4. a) Critical Value: Using Table A-2, 812 1. tdf,1−α = t10, 0.95 =

STEP 4. b) p-value: Using Table A-2, p-value < 0.005. (The exact 

 p - value = P(tdf > t0 ) = P(t10 > 3.271) = 0.0042 ) 

STEP 5. a) Conclusion: Since test statistic = 3.271 > 1.812 = critical value, we reject the null 

hypothesis that there is no difference between the population means. 

STEP 5. b) Conclusion: Since p-value = 0.0042 < 0.05 = significance level, we reject the null 

hypothesis that there is no difference between the population means. 
EPA QA/G-9S 71 February 2006 

3.3.1.1.3 Two-Sample Test for Population Proportions 

Purpose: Test for a difference between two population proportions, P1 and P2. A simple or 

systematic random sample, x1,...,xn, from one population and an independent simple or 

systematic random sample, y1, y2, . . . , yn, from second population of interest. 

Assumptions: The data constitutes independent simple or systematic random samples from the 

populations. 

Limitations and Robustness: To ensure the normal approximation is appropriate, compute mp1, 

m(1-p1), np2, and n(1-p2), where m and n are the sample sizes and p1 and p2 are the sample 

proportions. If all of the products are at least 5, then the normal approximation may be used. 

Otherwise, seek the assistance from a statistician as exact tests must be used. Since data 

positioning is used rather than actual data values, the procedures are robust to outliers. 

Directions for the two-sample test for proportions are contained in Box 3-26, with an example in 

Box 3-28. Directions for a two-sample confidence interval for the difference between 

proportions are contained in Box 3-27. 

3.3.1.2 Paired Samples 

Observations from paired populations are correlated. The general set up for this teat 

involves taken two measurements upon one group of sampling units at separate instances; for 

example, measurements before and after clean-up, or two labs making separate measurements on 

a single set of objects. 

3.3.1.2.1 The Paired t-Test and Confidence Interval 

Purpose: Test for or estimate the difference between two paired population means. 

Data: Two paired data sets x1,...,xn and y1,...,yn. 

Assumptions: The two data sets come from approximately normal distributions or n ≥ 30. 

Limitations and Robustness: Since there is really only one sample, the limitations for the paired 

t-test are the same as those for the one-sample t-test. These methods are robust against the 

population distribution deviating moderately from normality. However, they are not robust 

against outliers. In either case, the nonparametric methods of section 3.3.2.2 offer an alternative. 

 Finally, these methods have difficulty dealing with non-detects. The substitution or adjustment 

procedures of chapter 4 are an alternative, but it is best to use a nonparametric method. 

Directions for the paired t-test are given in Box 3-29, with an example in Box 3-31. Directions 

for a paired t confidence interval are given in Box 3-30. 
EPA QA/G-9S 72 February 2006 

Box 3-26: Directions for a Two-Sample Test for Proportions 

COMPUTATIONS: Let pi, i = 1,2, denote the sample proportion of data values from population i that fit the 

characteristic of interest. Also, calculate the pooled sample proportion p, which is the total number of data 

values that fit the characteristic divided by the total sample size, m + n . 

To ensure the normal approximation is appropriate, compute mp1, m(1-p1), np2, n(1-p2). If all of these values 

are at least 5, then the normal approximation may be used. Otherwise, seek assistance from a statistician. 

STEP 1. Null Hypothesis: 0 1 2 0 H : P − P = δ

STEP 2. Alternative Hypothesis: i) A 1 2 0 H : P − P > δ (upper-tail test) 

 ii) A 1 2 0 H : P − P < δ (lower-tail test) 

 iii) A 1 2 0 H : P − P ≠ δ (two-tail test) 

STEP 3. Test Statistic: ( )

( )

.

1 1 1

1 2 0

0

⎟

⎠

⎞ ⎜

⎝

⎛ − ⋅ +

− − =

m n

p p

p p z

δ

STEP 4. a) Critical Value: Use Table A-1 to find: 

 i) z1−α

 ii) zα

 iii) z1−α 2

STEP 4. b) p-value: Use Table A-1 to find: 

 i) ( ) P Z > z0

 ii) ( ) P Z < z0

 iii) ( ) 2 P 0 ⋅ Z > z

STEP 5. a) Conclusion: i) If z0 > z1−α , then reject the null hypothesis that the difference in 

population proportions is equal to δ0. 

 ii) If z0 < zα , then reject the null hypothesis. 

 iii) If 0 > 1−α / 2 z z , then reject the null hypothesis. 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis that the difference in 

population proportions is equal to δ0. 

STEP 6. If the null hypothesis was not rejected, there is only one false acceptance error rate (β at P1-P2), and 

both m and n are at least 

( )

( )2

2 1

2 2 1 1 1

P P

z z p( p) - -

−

α′ + β ⋅ − where 

2

1 2 p p p

+ = , 

then the sample sizes were probably large enough to achieve the DQOs. The value of α′ is α for a 

one-sided test and 2 α / for a two-sided test. If only one of m or n meets the sample size criterion, 

then use statistical software to calculate the power of the test, assuming that the true values for the 

proportions P1 and P2 are those obtained in the sample. If the estimated power is below 1-β, the false 

acceptance error rate has not been satisfied. 
EPA QA/G-9S 73 February 2006 

Box 3-27: Directions for Computing a Confidence Interval for the 

Difference Between Population Proportions 

COMPUTATIONS: Let pi, i = 1,2, denote the sample proportion of data values from population i that fit the 

characteristic of interest. Also, calculate the pooled sample proportion p, which is the total number of data values 

that fit the characteristic divided by the total sample size, m + n . 

To ensure the normal approximation is appropriate, compute mp1, m(1-p1), np2, n(1-p2). If all of these values are 

at least 5, then the normal approximation may be used. Otherwise, seek the assistance from a statistician as 

exact tests must be used. 

A 100(1 - α)% confidence interval for the difference between population proportions, P1 – P2, is 

( ) ( ) ⎟

⎠

⎞ ⎜

⎝

⎛ − ± − ⋅ − +

m n

p p z p p

1 1

1 2 1 α 2 1 , where Table A-1 is used to find z1−α 2

Box 3-28: An Example of a Two-Sample Test for Proportions 

At a hazardous waste site, investigators must determine whether an area suspected to be contaminated with 

dioxin needs to be remediated. The possibly contaminated area (area 1) will be compared to a reference area 

(area 2) to see if dioxin levels in area 1 are greater than dioxin levels in the reference area. An inexpensive 

surrogate probe was used to determine if each individual sample is either "contaminated," i.e., over the health 

standard of 1 ppb, or "clean." The decision maker is willing to accept a false rejection decision error rate of 10% 

(α) and a false-negative decision error rate of 5% (β) when the difference in proportions between areas exceeds 

0.1. A team collected 92 readings from area 1 (of which 12 were contaminated) and 80 from area 2, the 

reference area, (of which 10 were contaminated). 

COMPUTATIONS: The sample proportion for area 1 is p1 = 12/92 = 0.130, the sample proportion for area 2 is 

p2 = 10/80 = 0.125, and the pooled sample proportion is p = (12 + 10) / (92 + 80 ) = 0.128. Since mp1 = 12, 

m(1-p1) = 80, np2 = 10, n(1-p2) = 70 are all at least 5, the normal approximation is appropriate. 

STEP 1. Null Hypothesis: 0 H : 0 P1 − P2 =

STEP 2. Alternative Hypothesis: 0 H : A P1 − P2 > (upper-tail test) 

STEP 3. Test Statistic: 

( ) ( )

010.

80

1

92

1 0.128 1 0.128

0.13 0.125

1 1 1

1 2

0 .

m n

p p

p p z =

⎟

⎠

⎞ ⎜

⎝

⎛ − +

− =

⎟

⎠

⎞ ⎜

⎝

⎛ − +

− =

STEP 4. a) Critical Value: Using Table A-1, 282 1. z1−α = z0.90 =

STEP 4. b) p-value: Using Table A-1, P(Z > z0 ) = P(Z > 0.10) = 1− 0.5398 = 0.4602

STEP 5. a) Conclusion: Since test statistic = 0.10 < 1.282 = critical value, we fail to reject the null 

hypothesis of no difference between population proportions. 

STEP 5. b) Conclusion: Since p-value = 0.4602 > 0.10 = significance level, we fail to reject the 

null hypothesis of no difference between population proportions. 

STEP 6. Since the null hypothesis was not rejected and there is only one false acceptance error rate (β = 0.05 

at a difference of P1 - P2 = 0.1) has been specified, it is possible to calculate the sample sizes that 

achieve the DQOs. So m and n must be of size at least 

( ) ( ) 190.6

0.1

2 1.282 1.645 0.1275 1 0.1275

2

2

= + − , 

since 0.1275

2

0.13 0.125 = + P = . As both m and n are less than 190.6, the false acceptance error 

rate has not been satisfied. Therefore, the null hypothesis was not rejected, but the samples sizes 

were not large enough to ensure adequate power in the test. 
EPA QA/G-9S 74 February 2006 

Box 3-29: Directions for the Paired t-Test 

COMPUTATIONS: Compute the differences di = yi - xi for i = 1,...,n and the sample mean, D , and sample 

standard deviation, sD, of the differences. 

STEP 1. Null Hypothesis: 0 0 H : μ D = δ

STEP 2. Alternative Hypothesis: i) A 0 H : μD > δ (upper-tail test) 

 ii) A 0 H : μ D < δ (lower-tail test) 

 iii) A 0 H : μD ≠ δ (two-tail test) 

STEP 3. Test Statistic: 

n

s

D t

D

0

0

− δ =

STEP 4. a) Critical Value: Use Table A-2 to find: 

 i) tn−1,1−α

 ii) −tn−1,1−α

 iii) n−1,1−α 2 t

STEP 4. b) p-value: Use Table A-2 to find: 

 i) ( ) P 1 0 t t n− >

 ii) ( ) P 1 0 t t n− <

 iii) ( ) 2 P 1 0 t t ⋅ n− >

STEP 5. a) Conclusion: i) If t0 > tn−1,1−α , then reject the null hypothesis that there is no 

difference between the population means. 

 ii) If t0 < −tn−1,1−α , then reject the null hypothesis. 

 iii) If 0 > n−1, 1−α / 2 t t , then reject the null hypothesis. 

STEP 5. b) Conclusion: If p-value < α, then reject that the difference between the population 

means is δ0. 

STEP 6. If the null hypothesis was not rejected, there is only one false acceptance error rate (β at μ1), and 

( )

( ) C 2

2

1

2

1

2

1 1

2 α β α

μ

− − ′ − ′ + −

+

≥

s z z z

n

D , then the sample size was probably large enough to achieve the DQOs. 

The value of α′ is α for a one-sided test and 2 α / for a two-sided test. 

Box 3-30: Directions for Computing the Paired t Confidence Interval 

COMPUTATIONS: Compute the differences di = yi - xi for i = 1,...,n and the sample mean, D , and sample 

standard deviation, sD, of the differences. 

A 100(1 - α)% confidence interval for μD is 

n

s D t D n ± ⋅ −1,1−α 2 , where Table A-2 is used to find n−1,1−α 2 t .
EPA QA/G-9S 75 February 2006 

Box 3-31: An Example of the Paired t-Test 

Consider the following 9 pairs of data points (in ppb): 

xi 178 52 161 245 164 184 157 308 130 

yi 92 67 62 206 106 126 108 314 126 

This data will be used to test the null hypothesis that there is no difference in means versus the alternative that 

the mean of the first population is greater than the mean of the second population. The decision maker has 

specified a 5% false rejection error rate and a 10% false acceptance error rate at a difference of 50 ppb. 

COMPUTATIONS: The table below displays the computations. 

di -86 15 -99 -39 -58 -58 -49 6 -4 

The sample mean of the differences is -41.33 and the sample standard deviation is 39.92. 

STEP 1. Null Hypothesis: 0 H : 0 μD =

STEP 2. Alternative Hypothesis: 0 HA : μD < (lower-tail test) 

STEP 3. Test Statistic: 3.11

9

39.92

0 41.33 0

0 = − − − = − =

n

s

D t

D

δ . 

STEP 4. a) Critical Value: Using Table A-2, 860 1. −tn−1,1−α = −t8, 0.95 = − . 

STEP 4. b) p-value: Using Table A-2, 0.005 < p-value < 0.01. (The exact 

 p - value = P(tn−1 < t0 ) = P(t8 < −3.11) = 0.0072 ). 

STEP 5. a) Conclusion: Since test statistic = 3.11 < 1.860 = critical value, we reject the null 

hypothesis of no difference between population means. 

STEP 5. b) Conclusion: Since p-value = 0.0072 < 0.05 = significance level, we reject the null 

hypothesis of no difference between population means. 

3.3.2 Nonparametric Methods 

These methods rely on the relative rankings of data values. Knowledge of the precise 

form of the population distributions is not necessary. 

3.3.2.1 Independent Samples 

3.3.2.1.1 The Wilcoxon Rank Sum Test 

Purpose: Test for a difference between two population means. The Wilcoxon Rank Sum test 

applied with the Quantile test, provides a powerful combination for detecting true differences 

between two population distributions. 
EPA QA/G-9S 76 February 2006 

Data: A random sample x1, x2, . . . , xm from one population, and an independent random 

sample y1, y2, . . . , yn from the second population. 

Assumptions: The validity of the random sampling and independence assumptions should be 

verified by review of the procedures used to select the sampling points. The two underlying 

distributions are assumed to have approximately the same shape (variance) and that the only 

difference between them is a shift in location. A qualitative test of this assumption can be done 

by comparing histograms. 

Limitations and Robustness: The Wilcoxon signed rank test may produce misleading results if 

there are many tied data values. When many ties are present, their relative ranks are the same, 

and this has the effect of diluting the statistical power of the Wilcoxon test. If possible, results 

should be recorded with sufficient accuracy so that a large number of tied values do not occur. 

Estimated concentrations should be reported for data below the detection limit, even if these 

estimates are negative, as their relative magnitude to the rest of the data is of importance. If this 

is not possible, substitute the value DL/2 for each value below the detection limit providing all 

the data have the same detection limit. When different detection limits are present, all data could 

be censored at the highest detection limit but this will substantially weaken the test. A 

statistician should be consulted on the potential use of Gehan ranking. 

Directions for the Wilcoxon Rank Sum test are given in Box 3-32, with an example in Box 3-33. 

 Directions for the large sample approximation for the Wilcoxon Ranked Sum test are given in 

the example in Box 3-34. 

3.3.2.1.2 The Quantile Test 

Purpose: Test for a shift to the right in the right-tail of population 1 versus population 2 This 

may be regarded as being equivalent to detecting if the values in the right-tail of population 1 

distribution are generally larger than the values in the right-tail of the population 2 distribution. 

Data: A simple or systematic random sample, x1, x2, . . ., xn, from the site population and an 

independent simple or systematic random sample, y1, y2, . . ., ym, from the background 

population. 

Assumptions: The validity of the random sampling and independence assumptions is assured by 

using proper randomization procedures, which can be verified by reviewing the procedures used 

to select the sampling points. 

Limitations and Robustness: Since the Quantile test focuses on the right-tail, large outliers will 

bias results. Also, the Quantile test says nothing about the center of the two distributions. 

Therefore, this test should be used in combination with a location test like the t-test (if the data 

are normally distributed) or the Wilcoxon Rank Sum test. 
EPA QA/G-9S 77 February 2006 

Box 3-32: Directions for the Wilcoxon Rank Sum Test 

COMPUTATIONS: Rank the pooled data from smallest to largest assigning average rank to ties. Sum the ranks 

of the first population and denote this by R1. Then compute 

( )

2

1

0 1

+ = − m m W R

STEP 1. Null Hypothesis: 0 H : 0 μ X − μY = (no difference between population means) 

STEP 2. Alternative Hypothesis: i) 0 HA : μ X − μY > (upper-tail test) 

 ii) 0 HA : μ X − μY < (lower-tail test) 

 iii) 0 HA : μ X − μY ≠ (two-tail test) 

STEP 3. Test Statistic: If m and n are at most 20, then the test statistic is W0. 

 If m and n are greater than 20, then compute ( ) 0

0

0

var

/ 2

W

W mn

z − = , 

 where ( ) ( )

( )( ) ( ) ⎪

⎭

⎪

⎬

⎫

⎪

⎩

⎪

⎨

⎧

− + + − − + + = ∑=

g

j

j j t t

m n m n

mn m n mn W

1

2

0 1

12 12 1

1

var , 

 g is the number of tied groups, and tj is the number of ties in the j

th group. 

STEP 4. a) Critical Value: If m and n< =20, use Table A-8 to find: If m and n >20 use Table A-1 

to find: 

 i) mn - wα i) z1-α

 ii) wα ii) zα

 iii) wα/2 iii) z1-α/2

STEP 4. b) p-value: If m and n <=20, use Table A-8 to find: If m and n > 20 use Table A-1 to find: 

 i) ( ) P Wrs < mn −W0 i) ( ) P Z > z0

 ii) ( ) P Wrs < W0 ii) ( ) P Z < z0

 iii) 2 ⋅ min{P(Wrs < mn −W0 ), P(Wrs < W0 )} iii) ( ) 2 P Z > z0 ⋅

STEP 5. a) Conclusion: If m and n are at most 20, then 

 i) If W0 ≥ mn - wα, then reject the null hypothesis. 

 ii) If W0 ≤ wα, then reject the null hypothesis. 

 iii) If W0 ≥ mn - wα/2 or W0 ≤ wα/2, then reject the null hypothesis. 

 If m and n are greater than 20, then 

 i) If z0 > z1-α, then reject the null hypothesis of no difference 

between population means. 

 ii) If z0 < zα, then reject the null hypothesis. 

 iii) If 0 > 1−α / 2 z z , then reject the null hypothesis. 

STEP 5. b) Conclusion: If p-value < significance level, then reject the null hypothesis 

STEP 6. If the null hypothesis was not rejected, then the sample sizes necessary to achieve the DQOs should be 

computed. If the sample sizes are large, only one false acceptance error rate (β at δ1) has been 

specified, then the false acceptance error rate has probably been satisfied if both m and n are at least 

( ) ( )

⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧

+ ⋅ ⋅ + ⋅ − − ′ − ′

4

2 var

1.16

2

1

2

1

2

0 1 α 1 β α

δ

W z z z . 

NOTE: The value of α′ is α for a one-sided test and 2 α / for a two-sided test. The large sample normal 

approximation is adequate as long as min(m, n) > 10. 
EPA QA/G-9S 78 February 2006 

Box 3-33: An Example of the Wilcoxon Rank Sum Test 

At a hazardous waste site, an area cleaned (area 1) was compared with a relatively uncontaminated reference area 

(area 2). If the methodology worked, then the two sites should be approximately equal in average contaminant 

levels. If the methodology did not work, then area 1 should have a higher average than the reference area. The false 

rejection error rate was set at 10% (α) and the false acceptance error rate was set at 20% (β) if the difference 

between the areas is 2.5 ppb. Seven random samples were taken from area 1 and 8 samples were taken from area 

2: 

Area 1: 17, 23, 26, 5, 13, 13, 12 

Area 2: 16, 20, 5, 4, 8, 10, 7, 3 

COMPUTATIONS: The ordered pooled data and their ranks are (Area 1 denoted by *): 

Pooled 

data 3 4 5* 5 7 8 10 12* 13* 13* 16 17* 20 23* 26* 

Rank 1 2 3.5* 3.5 5 6 7 8* 9.5* 9.5* 11 12* 13 14* 15* 

The sum of the ranks of area 1 is R1 = 3.5 + 8 + 9.5 + 9.5 + 12 + 14 + 15 = 71.5 and W0 = 71.5 - 7(7 + 1)/2 = 43.5 

STEP 1. Null Hypothesis: 0 H : 0 μ X − μY =

STEP 2. Alternative Hypothesis: 0 HA : μ X − μY >

STEP 3. Test Statistic: Since m and n are less than 20, the test statistic is W0. 

STEP 4. a) Critical Value: Using Table A-8, mn - w0.10 = 56 – 16 = 40 

STEP 4. b) p-value: Using Table A-8, mn - w0.05 < W0 < mn - w0.025, which implies 

 0.05 < p-value < 0.025. Using statistical software, p-value = 0.0410. 

STEP 5. a) Conclusion: Since Test Statistic = 43.5 > 40 = Critical Value, we reject the null hypothesis of no 

difference between population means. 

STEP 5. b) Conclusion: Since p-value = 0.0410 < 0.10 (significance level), we reject the null hypothesis 

Directions for the Quantile test are contained in Box 3-35, with an example in Box 3-36. 

3.3.2.1.3 The Slippage Test 

Purpose: Test for a shift to the right in the extreme right-tail of population 1 (site) versus 

population 2 (background). This is equivalent to asking if a set of the largest values of the site 

distribution are larger than the maximum value of the background distribution. 

Data: A simple or systematic random sample, x1, x2, . . ., xm, from the site population and an 

independent simple or systematic random sample, y1, y2, . . ., yn, from the background 

population. 
EPA QA/G-9S 79 February 2006 

Box 3-34: A Large Sample Example of the Wilcoxon Rank Sum Test 

Arsenic concentrations (in ppm) from a site are to be compared to a reference area. The null hypothesis is that the 

means of the two areas are equal versus the upper-tail alternative. The false rejection error rate was set at 5% (α) and 

the false acceptance error rate was set at 20% (β) if the difference between the areas is 2.5 ppm. 

Site concentrations (m = 22): 11.2, 11.3, 12.2, 13.2, 14.2, 15.9, 16.3, 17.1, 18.6, 19.2, 21.5, 

22.3, 22.4, 22.7, 22.8, 23.3, 24.1, 25.8, 30.2, 30.7, 31.4, 37.1 

background concentrations (n = 21): 6.1, 8.5, 11.1, 11.3, 12.6, 12.8, 13.6, 15.0, 15.2, 15.3, 16.1, 

16.2, 17.0, 17.1, 17.6, 19.2, 19.2, 19.6, 21.1, 22.2, 25.0 

COMPUTATIONS: The ordered pooled data and their ranks are (site concentrations are denoted by *): 

Value Rank Value Rank Value Rank Value Rank 

6.1 1 14.2* 12* 17.6 23 22.7* 34* 

8.5 2 15.0 13 18.6* 24* 22.8* 35* 

11.1 3 15.2 14 19.2 26 23.3* 36* 

11.2* 4* 15.3 15 19.2 26 24.1* 37* 

11.3 5.5 15.9* 16* 19.2* 26* 25.0 38 

11.3* 5.5* 16.1 17 19.6 28 25.8* 39* 

12.2* 7* 16.2 18 21.1 29 30.2* 40* 

12.6 8 16.3* 19* 21.5* 30* 30.7* 41* 

12.8 9 17.0 20 22.2 31 31.4* 42* 

13.2* 10* 17.1 21.5 22.3* 32* 37.1* 43* 

13.6 11 17.1* 21.5* 22.4* 33* 

The sum of the site ranks is 

R1 = 4 + 5.5 + 7 + 10 + 12 + 16 + 19 + 21.5 + 24 + 26 + 30 + 32 + 33 + 34 + 35 + 36 + 37 + 39 + 40 + 41 + 42 + 43 = 587 

and W0 = 587 – 22(22 + 1)/2 = 334. 

STEP 1. Null Hypothesis: 0 H : 0 μ X − μY =

STEP 2. Alternative Hypothesis: 0 HA : μ X − μY >

STEP 3. Test Statistic: Since m and n are greater than 20, the test statistic is ( ) 0

0

0

var

/ 2

W

W mn

z − = . To 

 compute var(W0), we need to determine the number of tied groups and the number 

 of values in each group. There are 3 tied groups, so g = 3. The number of tied 

 values in the groups are 2 (at 11.3), 2 (at 17.1), and 3 (at 19.2). Therefore, 

( ) ( )

( )( ) [2 ( 2 1) 2 (2 1)( ) 3 3 1 ] 1693.23

12 22 21 22 21 1

22 21

12

22 21 22 21 1

var 2 2 2

0 =

⎭

⎬

⎫

⎩

⎨

⎧ ⋅ − + ⋅ − + ⋅ − ⋅ + + −

⋅ − ⋅ ⋅ + + W = . 

 Finally, 2.50

1693.23

334 22 21/ 2

0 = − ⋅ z = . 

STEP 4. a) Critical Value: Using Table A-1, z0.95 = 1.645. 

STEP 4. b) p-value: Using Table A-1, P(Z > 2.50) = 1− 0.9938 = 0.0062 . 

STEP 5. a) Conclusion: Since Test Statistic = 2.50 > 1.645 = Critical Value, we reject the null hypothesis of no 

difference between population means. 

STEP 5. b) Conclusion: Since p-value = 0.0062 < 0.05 = significance level, we reject the null hypothesis of no 

difference between population means. 
EPA QA/G-9S 80 February 2006 

Box 3-35: Directions for the Quantile Test 

COMPUTATIONS: Select a quantile, 0.5 ≤ b1 < 1. Rank the pooled data from smallest to largest. Find the 

number of pooled data points larger than the th b1 quantile, 

c = m + n − floor{ } ( ) m + n −1 ⋅ b1 −1, where floor means to calculate the value and discard all decimals. 

STEP 1. Null Hypothesis: H0: The right-tails of the two population distributions are the same. 

STEP 2. Alternative Hypothesis: HA: The right-tail of the distribution of Population 1 (site) is shifted to the 

 right, i.e., the values in the right-tail of the site distribution are larger 

 than the values in the right-tail of the background distribution. 

STEP 3. Test Statistic: s = number of site samples greater than the th b1 quantile. 

STEP 4. a) Critical Value: If m and n are both at most 20, then use Table A-19 to find qα. 

STEP 4. b) p-value: If m and n are both at least 15, then use Table A-1 to find 

⎟

⎠

⎞ ⎜

⎝

⎛ − − − < σ

0.5 μ 1 P s Z , where 

m n

c

n

+

μ = ⋅ and 

1

1

+ − ⎟⋅

⎠

⎞ ⎜

⎝

⎛

+ ⋅ − + = ⋅ m n

m

m n

c

m n

c σ n

STEP 5. a) Conclusion: If s ≥ qα, then reject the null hypothesis that the right tails of the two 

 population distributions are the same. 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis that the right tails of the two 

population distributions are the same. 

Box 3-36: A Example of the Quantile Test 

At a hazardous waste site a new, cheaper, in-site methodology was compared against an existing methodology 

by remediating separate areas of the site using each method. It will be assumed that the new methodology 

works as well as the old and we will test for evidence that the new method leaves higher concentrations. A 

Quantile Test with a significance level of 0.05 will be used to make this determination based on 12 samples from 

the area remediated using the new methodology and 7 samples from the area remediated using the standard 

methodology. 

New Methodology: 7, 18, 2, 4, 6, 11, 5, 9, 10, 2, 3, 3 

Standard Methodology: 17, 8, 20, 4, 6, 5, 4 

COMPUTATIONS: The 0.75 quantile is selected. The ranked pooled data set is: 

2*, 2*, 3*, 3*, 4, 4, 4*, 5*, 5, 6, 6*, 7*, 8, 9*, 10*, 11*, 17, 18*, 20, 

where * denoted samples from the new methodology portion of the site. The number of values larger than the 

0.75 quantile is: 

c = m + n − floor{ } ( ) m + n −1 ⋅ b1 −1= 7 + 12 − floor{(7 + 12 −1)⋅ 0.75}−1= 5

STEP 1. Null Hypothesis: H0: The right-tails of the two population distributions are the same. 

STEP 2. Alternative Hypothesis: HA: The right-tail of the distribution of Population 1 (site) is shifted to the 

 right, i.e., the values in the right-tail of the site distribution are larger 

 than the values in the right-tail of the background distribution. 

STEP 3. Test Statistic: s = number of site samples greater than the 0.75 quantile = 3 

STEP 4. a) Critical Value: Since m and n are both less than 20, we use Table A-19 to find q0.10 = 5 

STEP 4. b) p-value: Since m and n are both less than 15, we can’t use the normal 

 approximation for the p-value. 

STEP 5. a) Conclusion: Since s = 3 < 5 = q0.10, we fail to reject the null hypothesis that the right-tails 

of the two-population distributions are the same. 
EPA QA/G-9S 81 February 2006 

Assumptions: The validity of the random sampling and independence assumptions is assured by 

using proper randomization procedures, which can be verified by reviewing the procedures used 

to select the sampling points. 

Limitations and Robustness: Since the Slippage test focuses on the right-tail, large outliers will 

bias results. Also, the Slippage test says nothing about the center of the two distributions. 

Therefore, this test should be used in combination with a location test like the t-test (if the data 

are normally distributed) or the Wilcoxon Rank Sum test. 

Directions for the slippage test are contained in Box 3-37, with an example in Box 3-38. 

Box 3-37: Directions for the Slippage Test 

COMPUTATIONS: Let x1,...,xm be the site data and y1,...,yn be the background data. Order the samples 

separately. 

STEP 1. Null Hypothesis: H0: The right-tails of the two population distributions are the same. 

STEP 2. Alternative Hypothesis: HA: The extreme right-tail of the distribution of Population 1 (site) is 

 shifted to the right, i.e., the largest values of the site distribution are 

 larger than the largest values of the background distribution. 

STEP 3. Test Statistic: s = number of site values greater than the maximum background value. 

STEP 4. a) Critical Value: Use Table A-14 to find Sα. Note that α can be 0.01, 0.05, or 0.10. 

STEP 4. b) p-value: ( )

( )

( ) ( )

( ) ∑ ∑ ( )

−

= = −

+ − − ⋅ +

⋅ = − −

+ − − ⋅ +

⋅ 1

0

!

1 !

!

! 1

!

1 !

!

! s

i

m

i s m i

m n i

m n

n m

m i

m n i

m n

n m

 Where “n!’ = nx(n-1)x(n-2)x(n-3)x...3x2x1 

STEP 5. a) Conclusion: If s ≥ Sα, then reject the null hypothesis that the right-tails of the two 

population distributions are the same. 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis that the right-tails of the two 

population distributions are the same. 

3.3.2.2 Paired Samples 

Recall that the observations from paired populations are correlated. The general setting 

involves taken two measurements upon one group of sampling units at separate instances; for 

example, measurements before and after clean-up or two labs making separate measurements on 

a single set of objects. 

3.3.2.2.1 The Sign Test 

Purpose: Test for a difference between the medians of two paired populations. This test is very 

similar to the one sample version presented in section 3.2.2.1. 
EPA QA/G-9S 82 February 2006 

Box 3-38: A Example of the Slippage Test 

At a hazardous waste site a new, cheaper, in-site methodology was compared against an existing methodology 

by remediating separate areas of the site using each method. It will be assumed that the new methodology 

works as well as the old and we will test for evidence that the new method leaves higher concentrations. The 

Slippage Test with a significance level of 0.05 will be used to make this determination based on 12 samples 

from the area remediated using the new methodology and 7 samples from the area remediated using the 

standard methodology. 

New Methodology: 7, 18, 2, 4, 6, 11, 5, 9, 10, 2, 3, 3 

Standard Methodology: 17, 8, 20, 4, 6, 5, 4 

COMPUTATIONS: The ordered samples are: 

New Methodology: 2, 2, 3, 3, 4, 5, 6, 7, 9, 10, 11, 18 

Standard Methodology: 4, 4, 5, 6, 8, 17, 20 

STEP 1. Null Hypothesis: H0: The right-tails of the two population distributions are the same. 

STEP 2. Alternative Hypothesis: HA: The extreme right-tail of the distribution of Population 1 (site) is 

 shifted to the right, i.e., the largest values of the site distribution are 

 larger than the largest values of the background distribution. 

STEP 3. Test Statistic: s = number of site values greater than the maximum background value = 0. 

STEP 4. a) Critical Value: Using Table A-14, S0.05 = 6. 

STEP 4. b) p-value: ( )

( )

( ) 1 0 1

!

1!

!

! 1

0 1

0

= − = −

+ − − ⋅ +

⋅ − ∑

−

i = m i

m n i

m n

n m

STEP 5. a) Conclusion: Since test statistic = 0 < 6 = S0.05, we fail to reject the null hypothesis that 

the right-tails of the two population distributions are the same. 

STEP 5. b) Conclusion: Since p-value = 1 > 0.05 = significance level, we fail to reject the null 

hypothesis that the right-tails of the two population distributions are the 

same. 

Data: Two paired data sets x1,...,xn and y1,...,yn selected randomly or systematically. 

Assumptions: The Sign test can be used no matter what the underlying distributions may be. 

Limitations and Robustness: The Sign test has less power than the two-sample t-test or the 

Wilcoxon Signed Rank Test. However, the Sign test makes no distributional assumptions like 

the other two tests and it can handle non-detects (if the detection limit is below the threshold). 

Directions for the paired populations sign test are contained in Box 3-39, with an example in 

Box 3-40. 

3.3.2.2.2 The Wilcoxon Signed Rank Test 

Purpose: Test for a difference between the locations (means or medians) of two paired 

populations. This test is very similar to the one sample version presented in Section 3.2.2.2. 
EPA QA/G-9S 83 February 2006 

Box 3-39: Directions for the Sign Test (Paired Samples) 

COMPUTATIONS: Compute the deviations di = xi - yi. If any of the deviations are zero delete them and 

correspondingly reduce the sample size. Finally, compute B = number of differences greater than zero. 

STEP 1. Null Hypothesis: H0: median 1 = median 2 

STEP 2. Alternative Hypothesis: i) HA: median 1 > median 2 (upper-tail test) 

 ii) HA: median1 < median 2 (lower-tail test) 

 iii) HA: median 1 ≠ median 2 (two-tail test) 

STEP 3. Test Statistic: If n ≤ 20, then the test statistic is B. 

 If n > 20, then the test statistic is 

/ 4

/ 2

0

n

B n

z − = . 

STEP 4. a) Critical Value: If n ≤ 20, then use Table A-18 to find: If n > 20, then use Table A-1 to find: 

 i) Bupper(n, 2α) i) z1-α

 ii) Blower(n, 2α) – 1 ii) zα

 iii) Blower(n, α) and Blower (n, α) iii) z1- α /2

STEP 4. b) p-value: If n ≤ 20, then let ( ) n i

n

p i

2

1

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ = where 

( )! !

!

n i i

n

i

n

− =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

 i) ∑ ( ) =

n

i B

p i

 ii) ∑ ( ) =

B

i o

p i

 iii) () ()⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧

⋅ ∑ ∑ = =

n

i B

B

i

2 min p i , p i

0

 If n > 20, then use Table A-1 to find: 

 i) ( ) P Z > z0

 ii) ( ) P Z < z0

 iii) ( ) 2 P 0 ⋅ Z > z

STEP 5. a) Conclusion: If n ≤ 20, then 

 i) If B ≥ Bupper(n, 2α), then reject the null hypothesis that the 

two population medians are equal. 

 ii) If B ≤ Blower(n, 2α) - 1, then reject the null hypothesis. 

 iii) If B ≥ Bupper(n, α) or B ≤ Blower(n, α) - 1, then reject the null 

hypothesis. 

 If n > 20, then 

 i) If z0 > z1-α, then reject the null hypothesis that the two 

population medians are equal. 

 ii) If z0 < zα, then reject the null hypothesis. 

 iii) If |z0| > z1-α/2, then reject the null hypothesis. 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis that the two population 

medians are equal. 
EPA QA/G-9S 84 February 2006 

Box 3-40: An Example of the Sign Test (Paired Samples) 

Consider the following 9 pairs of data points (in ppb): 

xi 178 52 161 245 164 184 157 308 130 

yi 92 67 62 206 106 126 108 314 126 

This data will be used to test the null hypothesis that there is no difference in medians versus the alternative 

that the median of the first population is greater than the median of the second population. The decision maker 

has specified a 5% false rejection error rate and a 10% false acceptance error rate at a difference of 50 ppb. 

COMPUTATIONS: The table below displays the computations. 

di 86 -15 99 39 58 58 49 -6 4 

Therefore, B = the number of differences greater than zero = 7. 

STEP 1. Null Hypothesis: H0: 1 2 ~ ~ μ = μ

STEP 2. Alternative Hypothesis: HA: 1 2 ~ ~ μ > μ

STEP 3. Test Statistic: Since n ≤ 20, the test statistic is B = 7. 

STEP 4. a) Critical Value: Since n ≤ 20, Table A-18 is used to find Bupper(n, 2α) = 8. 

STEP 4. b) p-value: Since n ≤ 20, p-value = 0.0898

2

9 1 9

7

9 = ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

∑

i = i . 

STEP 5. a) Conclusion: Since test statistic = 7 < 8 = critical value, we fail to reject H0. 

STEP 5. b) Conclusion: Since p-value = 0.0898 > 0.05 = significance level, we fail to reject H0. 

Data: Two paired data sets x1,...,xn and y1,...,yn selected randomly or systematically. 

Assumptions: The data sets come from a approximately symmetric distributions. 

Limitations and Robustness: For large sample sizes (n > 50), the paired t-test is more robust to 

violations of its assumptions than the Wilcoxon signed rank test. For small sample sizes, if the 

data are not approximately symmetric or normally distributed, the sign test should be used. 

The Wilcoxon signed rank test may produce misleading results if there are many tied data values. 

 Ties have the effect of diluting the statistical power of the Wilcoxon test. If possible, results 

should be recorded with sufficient accuracy so that a large number of tied values do not occur. 

Estimated concentrations should be reported for data below the detection limit, even if these 

estimates are negative, as their relative magnitude to the rest of the data is of importance. If this 

is not possible, substitute the value DL/2 for each value below the detection limit providing all 

the data have the same detection limit. When different detection limits are present, all data could 

be censored at the highest detection limit but this will substantially weaken the test. A 

statistician should be consulted on the potential use of Gehan ranking. 
EPA QA/G-9S 85 February 2006 

Directions for the paired sample Wilcoxon signed rank test are contained in Box 3-41, 

with an example in Box 3-42. Directions for the large sample version of the test are contained in 

the example in Box 3-43. 

Box 3-41: Directions for the Wilcoxon Signed Rank Test (Paired Samples) 

COMPUTATIONS: Compute the deviations di = xi - yi. If any of the deviations are zero delete them and 

correspondingly reduce the sample size. Rank the absolute deviations, |di|, from smallest to largest. If there 

are tied observations, then assign the average rank. Let Ri be the signed rank of |di|, where the sign of Ri is 

determined by the sign of di. 

STEP 1. Null Hypothesis: H0: location1 = location2 

STEP 2. Alternative Hypothesis: i) HA: location1 > location2 (upper-tail test) 

 ii) HA: location1 < location2 (lower-tail test) 

 iii) HA: location1 ≠ location2 (two-tail test) 

STEP 3. Test Statistic: If n ≤ 20, then 

{ } ∑>

+ =

i: Ri 0

T Ri , the sum of the positive signed ranks. 

 If n > 20, then ( )

( ) +

+ − + =

T

T n n

z

var

1 / 4

0 , 

 where ( ) ( )( ) ∑ ( ) =

+ − − + + =

g

j

j j t t n n n T

1

2 1

48

1

24

1 2 1

var , g is the number of 

 tied di groups and tj is the size of group j. 

STEP 4. a) Critical Value: If n ≤ 20, use Table A-7 to find: If n > 20, use Table A-1 to find: 

 i) n(n+1)/2 - wα i) z1-α

 ii) wα ii) zα

 iii) wα/2 iii) z1-α/2

STEP 4. b) p-value: If n ≤ 20,use Table A-7 to find: If n > 20, use Table A-1 to find: 

 i) ( ( ) ) + P W ≤ n n + 1 / 2 −T i) ( ) P Z > z0

 ii) ( ) + P ii) W ≤ T ( ) P Z < z0

 iii) { ( ( ) ) ( )} + + 2 iii) ⋅min P W ≤ n n + 1 / 2 −T ,P W ≤ T ( ) 2 P 0 ⋅ Z > z

STEP 5. a) Conclusion: If n ≤ 20, If n > 20, 

 i) If + T ≥ n(n+1)/2 - wα, then reject the null i) If z0 > z1-α, then reject the null 

 ii) If + T ≤ wα, then reject the null hypothesis. ii) If z0 < zα, then reject the null 

 iii) If + T ≥ n(n+1)/2 - wα/2 or + T ≤ wα, then reject the null iii) If |z0| > z1-α/2, then reject the null 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis that the two population 

locations are equal. 

STEP 6. If the null hypothesis was not rejected, then the sample sizes necessary to achieve the DQOs 

should be computed. If the sample size is large, only one false acceptance error rate (β at δ1) has 

been specified, and 

( )( )

⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧

+

+

≥ ⋅ − − −

+

2

var

1.16

2

1

2

1

2

1 α 1 β α

δ

T z z z

n , 

then the false acceptance error rate has probably been satisfied. The value of α′ is α for a onesided test and 2 α / for a two-sided test. 
EPA QA/G-9S 86 February 2006 

Box 3-42: An Example of the Wilcoxon Signed Rank Test (Paired Samples) 

Consider the following 9 pairs of data points (in ppb): 

xi 178 52 161 245 164 184 157 308 130 

yi 92 67 62 206 106 126 108 314 126 

This data will be used to test the null hypothesis that there is no difference in medians versus the alternative 

that the median of the first population is greater than the median of the second population. The decision maker 

has specified a 5% false rejection error rate and a 10% false acceptance error rate at a difference of 50 ppb. 

COMPUTATIONS: The table below displays the computations. 

di 86 -15 99 39 58 58 49 -6 4 

|di| 86 15 99 39 58 58 49 6 4 

rank 8 3 9 4 6.5 6.5 5 2 1 

Ri 8 -3 9 4 6.5 6.5 5 -2 1 

STEP 1. Null Hypothesis: H0: median1 = median 2 

STEP 2. Alternative Hypothesis: HA: median 1< median 2 (lower-tail test) 

STEP 3. Test Statistic: Since n ≤ 20, compute 

{ }

1 4 5 6.5 6.5 8 9 40

: 0

= ∑ = + + + + + + =

>

+

Ri i

T Ri

STEP 4. a) Critical Value: Since n ≤ 20, Table A-7 is used to find n(n+1)/2 - w0.05 = 45 – 8 = 37. 

STEP 4. b) p-value: Since n ≤ 20, Table A-7 is used to find the approximate p-value of 

 0.025. Using software, the p-value is found to be 0.0195. 

STEP 5. a) Conclusion: Since test statistic = 40 ≥ 37 = critical value, we reject the null 

 hypothesis of equal population medians. 

STEP 5. b) Conclusion: Since p-value = 0.0195 < 0.05 = significance level, we reject the null 

 hypothesis of equal population medians. 

NOTE: The Sign test failed to reject the null hypothesis for this example. The Wilcoxon Signed Rank test has 

more power than the Sign test if the distribution of the population is symmetric. 

3.4 COMPARING SEVERAL POPULATIONS SIMULTANEOUSLY 

This section describes procedures for comparing several population means 

simultaneously. The comparison is made between several treatment populations versus a single 

control population. For example, we can simultaneously test for a difference between the 

concentrations at several sites versus a single background area. 

The methods in section compare the several populations while controlling the overall 

significance level. If individual two-sample t-tests are performed at significance level α, then 

the overall significance level is higher than α. Possible much higher if there are many 

experimental groups. For example, comparing two experimental groups to a control using two 

two-sample t-tests at significance level 0.05 results in an overall significance level of 1-(1-

0.05)(1-0.05) = 0.0975. The tests in this section are more powerful in detecting differences 

between the experimental groups and the control than other multiple comparison methods that 

compare all possible pairs of population means, e.g., the ANOVA F-test and the Kruskal-Wallis 

test. 
EPA QA/G-9S 87 February 2006 

Box 3-43: A Large Sample Example of the Wilcoxon Signed Rank Test (Paired Samples) 

A hazardous waste site has recently gone through remediation. To determine if the remediation method was 

effective, 24 paired samples (before and after clean-up) will be compared. The following data will be used to 

test the null hypothesis that there is no difference in medians versus the alternative that the median 

concentration before is greater than the median concentration after clean-up. The decision maker has specified 

a 5% false rejection error rate and a 10% false acceptance error rate at a difference of 30 ppb. 

before: 331, 351, 259, 323, 305, 336, 196, 233, 336, 349, 352, 341, 172, 253, 275, 285, 212, 349, 301, 343, 368, 332, 374, 311 

after: 246, 270, 229, 326, 295, 238, 278, 302, 331, 264, 267, 249, 288, 272, 270, 313, 337, 284, 271, 253, 295, 271, 289, 281 

COMPUTATIONS: The table below displays the computation of the signed ranks. 

di |di| rank Ri di |di| rank Ri di |di| rank Ri

85 85 17.5 17.5 5 5 2.5 2.5 -125 125 24 -24 

81 81 14 14 85 85 17.5 17.5 65 65 11 11 

30 30 8 8 85 85 17.5 17.5 30 30 8 8 

-3 3 1 -1 93 93 21 21 90 90 20 20 

10 10 4 4 -117 117 23 -23 73 73 13 13 

98 98 22 22 -19 19 5 -5 60 60 10 10 

-82 82 15 -15 5 5 2.5 2.5 85 85 17.5 17.5 

-69 69 12 -12 -28 28 6 -6 30 30 8 8 

STEP 1. Null Hypothesis: H0: median 1 = median 2 

STEP 2. Alternative Hypothesis: HA: median 1 > median 2 (upper-tail test) 

STEP 3. Test Statistic: Since n > 20, compute ( )

( ) +

+ − + =

T

T n n

z

var

1 / 4

0 . First, 

 

{ }

17.5 14 8 214

: 0

= ∑ = + + + =

>

+ L

Ri i

T Ri . To compute ( ) + var T , we need 

 to identify the number of tied groups and the number of values in each 

 of the tied groups. There are 3 tied groups so g = 3. The number of 

 values in the tied groups are 2 (at 2.5), 3 (at 30), and 4 (at 85). 

 Therefore, 

( ) ( )( ) {2 (2 1) 3 (3 1) 4 (4 1)} 1223.125

48

1

24

24 24 1 2 24 1

var 2 2 2 − ⋅ − + ⋅ − + ⋅ − = ⋅ + ⋅ + = + T .

 Finally, ( ) 1.83

1223.125

214 24 24 1 / 4

0 = − ⋅ +

z =

STEP 4. a) Critical Value: Since n > 20, Table A-1 is used to find z0.95 = 1.645. 

STEP 4. b) p-value: Since n > 20, Table A-1 is used to find P(Z > z0 ) = 1− 0.9664 = 0.0336 . 

STEP 5. a) Conclusion: Since test statistic = 1.83 > 1.645 = critical value, we reject the null 

 hypothesis of equal population medians. 

STEP 5. b) Conclusion: Since p-value = 0.0336 < 0.05 = significance level, we reject the null 

 hypothesis of equal population medians. 

The Dunnett test of Section 3.4.1.1 is a parametric test that compares several treatment 

means to a control mean. Section 3.4.2.1 presents a nonparametric alternative to the Dunnett 

test. 
EPA QA/G-9S 88 February 2006 

3.4.1 Parametric Methods 

These methods rely on the knowing the specific distribution of the populations. 

3.4.1.1 Dunnett’s Test 

Purpose: Test simultaneously for a difference between several population means and the mean 

of a control population. A typical application would involve comparing different potentially 

cleaned areas of a hazardous waste site to an uncontaminated reference area. 

Data: A set of k-1 independent experimental random samples, i i in x , , x 1 K , i = 1,..., k-1 and an 

independent random sample from the control population, c c cn x , , x 1 K . 

Assumptions: The Dunnett test is similar to the two-sample t-test so the populations need to be 

approximately normal or the sample sizes need to be large (≥30). If this is not the case, then the 

nonparametric Fligner-Wolfe test can be used. However, that test simple detects a difference 

between all population means. Both tests assume that the k populations have equal variances. 

Limitations and Robustness: The Dunnett critical values (Table A-14) are for the case of equal 

number of samples in the control and experimental groups, but are approximately correct 

provided the number of samples from the investigated groups are more than half but less than 

double the size of the control group. Also, Table A-14 is for one-tailed tests only. 

Directions for Dunnett’s test are contained in Box 3-44, with an example in Box 3-45. 

Box 3-44: Directions for Dunnett’s Test 

COMPUTATIONS: Calculate the sample mean, Xi , and the sample variance, si

2 for each of the k-1 populations 

along with Xc and 2 sc . Also compute ( ) () ∑

−

=

= − + −

1

1

2 2 SSE 1 1

k

i

nc sc ni si

STEP 1. Null Hypothesis: H0: μI = μc, for i = 1, ...,k-1 

STEP 2. Alternative Hypotheses: i) HA: At least one μI > μc, for i = 1, ...,k-1. 

 ii) HA: At least one μI < μc, for i = 1, ...,k-1. 

STEP 3. Test Statistic: For each of the k-1 experimental populations, compute: 

 

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+

− =

i c

i c

i

N-k n n

X X

t

SSE 1 1

, where N is the total sample size. 

STEP 4. a) Critical Value: Use Table A-15 to determine the critical value TD(α) where the degrees of 

 freedom are N-k. 

STEP 4. b) p-value: Too complex for this guidance. 

STEP 5. a) Conclusion: For each i, reject the corresponding null hypothesis if ti > TD(α). 

STEP 5. b) Conclusion: Use the critical value approach. 
EPA QA/G-9S 89 February 2006 

Box 3-45: An Example of Dunnett’s Test 

At a hazardous work site, 6 designated areas previously identified as ‘problems’ have been cleaned. In order for 

these areas to be admitted to the overall work site abatement program these areas should be sown to be the same 

as the reference area. The means of these areas will be compared to mean of a reference area located on the site 

using Dunnett’s test. The null hypothesis is no difference between the means of the ‘problem’ areas and the mean 

of the reference area and the alternative hypotheses are that the ‘problem’ means are greater than the reference 

area mean. The significance level is 0.05. Summary statistics for the data are given in the table below. 

Reference IAK3 ZBF6 3BG5 4GH2 5FF3 6GW4

ni 7 6 5 6 7 8 7 

Xi 10.3 11.4 12.2 10.2 11.4 11.9 12.1 

si

2 2.5 2.6 3.3 3.0 3.2 2.6 2.8 

nc/ni 1.16 1.4 1.16 1 0.875 1 

ti 1.18 1.93 0.11 1.22 1.84 2.00 

Since all of the sample size ratios fall between 0.5 and 2.0, Dunnett’s test may be used. 

COMPUTATIONS: The sample means and sample variances are displayed in rows 2 and 3 above. The ti row is 

the collection of test statistics that are generated in Step 3 below. The final computation is 

MSE = () () () 7 − 1 ⋅ 2.5 + 6 − 1 ⋅ 2.6 +L+ 7 −1 ⋅ 2.8 = 110.4 . 

STEP 1. Null Hypothesis: H0: μI = μc, for i = 1, ...,6 

STEP 2. Alternative Hypotheses: HA: μI > μc, for i = 1, ...,6 

STEP 3. Test Statistic: For each of the 6 ‘problem’ areas, ti was computed. For 

 example, 1.18

7

1

6

1

46 7

110.4

11.4 10.3

1 =

⎟

⎠

⎞ ⎜

⎝

⎛

⋅ + −

− t = . The results are displayed in 

 the 5th row of the table. 

STEP 4. a) Critical Value: Using Table A-15 of Appendix A with N-k = 46-7 = 39 degrees of 

 freedom, the critical value TD(0.95) = 2.37. 

STEP 5. a) Conclusion: Since ti < 2.37 = TD(0.95) for all i, we fail to reject the null hypothesis. 

 We conclude that none of the ‘problem’ areas have contamination 

 levels significantly higher than the reference area. Therefore, these 

 areas may be admitted to the work site abatement program. 

3.4.2 Nonparametric Methods 

These methods rely on the relative rankings of data values. Knowledge of the precise 

form of the population distributions is not necessary. 

3.4.2.1 The Fligner-Wolfe Test 

Purpose: Test simultaneously for a difference between several population locations (means or 

medians) and the location of a control population. A typical application would involve 
EPA QA/G-9S 90 February 2006 

comparing different potentially cleaned areas of a hazardous waste site to an uncontaminated 

reference area. This test is similar to the Wilcoxon Rank Sum test. 

Data: A set of k-1 independent experimental simple or systematic random samples, i i in x , , x 1 K , i

= 1,..., k-1 and an independent simple or systematic random sample from the control population, 

c c cn x , , x 1 K . Let ∑ −

= = 1

1

* k

i i N n and N = N*

 + nc. 

Assumptions: All data sets come from distributions with similar shapes. 

Limitations and Robustness: The alternative hypothesis is quite restrictive as it demands all of 

the experimental groups have population means that are either at least as large as the control 

mean (or at most the control mean). They are not appropriate tests when some experimental 

group means may be higher than the control mean and some might be lower. 

Directions for the Fligner-Wolfe test are contained in Box 3-46, with an example in Box 3-47. 
EPA QA/G-9S 91 February 2006 

Box 3-46: Directions of the Fligner-Wolfe Test 

COMPUTATIONS: Rank the N pooled data points from smallest to largest assigning average rank to ties. Let rij

denote the rank of data point Xij. Compute 

( )

2

1 1 * *

1 1

0

+ =∑∑ −

−

= =

N N FW r

k

j

n

i

ij

i

. 

Note that the first term of FW is the sum of the ranks from all experimental groups. 

STEP 1. Null Hypothesis: 1 H : location location , for 1, , 0 i = c i = K k −

STEP 2. Alternative Hypothesis: i) 1 HA : locationi ≥ locationc , for i = 1,K,k − (at least one strict inequality) 

 ii) 1 H : location location , for 1, , 0 i ≤ c i = K k − (at least one strict inequality) 

STEP 3. Test Statistic: If min(nc, N*

) ≤ 20, then the test statistic is FW0. 

 If min(nc, N*

) > 20, then the test statistic is ( ) 0

* 0

0

var

/ 2

FW

FW n N

z − c = , where 

 ( ) ( )

( )∑ ( ) =

− − − + =

g

j

j j c c t t

N N

n N N n N FW

1

2 * *

0 1

12 12 1

1

var , g is the number of tied 

 groups and tj is the number of tied values in the j th group. 

STEP 4. a) Critical Value: If min(nc, N*

) ≤ 20, then use Table A-8 to find wα. 

 If min(nc, N*

) > 20, then use Table A-1 to find: 

 i) z1-α

 ii) zα

STEP 4. b) p-value: If min(nc, N*

) ≤ 20, then use Table A-8 to find: 

 i) ( ) 0 * P Wrs < ncN − FW

 ii) ( ) P Wrs < FW0

 If min(nc, N*

) > 20, then use Table A-1to find: 

 i) ( ) P Z > z0

 ii) ( ) P Z < z0

STEP 5. a) Conclusion: If min(nc, N*

) ≤ 20, then 

 i) If FW0 ≥ nc N*

- wα, then reject the null hypothesis of no difference 

between population locations. 

 ii) If FW0 ≤ wα, then reject the null hypothesis. 

 If min(nc, N*

) > 20, then 

 i) If z0 > z1-α, then reject the null hypothesis of no difference 

between population locations. 

 ii) If z0 < zα, then reject the null hypothesis. 

STEP 5. b) Conclusion: If p-value < significance level, then reject the null hypothesis of no difference 

between population locations. 

NOTE: The large sample normal approximation is adequate as long as min(nc, N*

) > 10. 
EPA QA/G-9S 92 February 2006 

Box 3-47: An Example of the Fligner-Wolfe Test 

4 contaminated ponds are to be combined with a fifth (reference) pond before further work can commence. If the 

ponds are approximately equal, the proposed remediation method will be acceptable. The assumption of 

normality cannot be made but all ponds were produced by the same waste process, they should exhibit similar 

characteristics. The significance level is 0.05. Data values with ranks in parentheses are given in the table 

below. 

Reference 

Pond 

North East South West

10 (1) 15 (3) 20 (8) 12 (2) 19 (5.5) 

19 (5.5) 25 (14) 26 (15) 16 (4) 20 (8) 

20 (8) 32 (19.5) 34 (21.5) 22 (10) 28 (16) 

23 (11.5) 39 (23) 43 (24.5) 23 (11.5) 31 (18) 

24 (13) 43 (24.5) 52 (28) 34 (21.5) 49 (27) 

30 (17) 

32 (19.5) 

46 (26) 

COMPUTATIONS: Note that nc = 8 and N*

 = 20 since ni = 5 for i = 1,...,4. Compute 

( ) 304.5 210 94.5

2

1 1 * *

1 1

0 = − = + =∑∑ −

−

= =

N N FW r

k

j

n

i

ij

i

. 

STEP 1. Null Hypothesis: H0 :pondi = pondc , for i = 1,K,4

STEP 2. Alternative Hypothesis: HA : pondi ≥ pondc , for i = 1,K,4 (at least one strict inequality) 

STEP 3. Test Statistic: Since min(nc, N*

) = 8 ≤ 20, then the test statistic is FW0 = 94.5. 

STEP 4. a) Critical Value: Since min(nc, N*

) ≤ 20, Table A-8 is used to find w0.05 = 47. 

STEP 4. b) p-value: Since min(nc, N*

) ≤ 20, using Table A-8 gives a p-value of 

 P( ) Wrs < 8 ⋅ 20 − 94.5 = P(Wrs < 65.5) > 0.10 (the exact 

 p-value is 0.2380.) 

STEP 5. a) Conclusion: Since min(nc, N*

) ≤ 20, and FW = < = ⋅ − = ncN − wα * 0 94.5 113 8 20 47 , we 

 fail to reject the null hypothesis of no difference between ponds. 

STEP 5. b) Conclusion: Since p-value = 0.2380 > 0.05 = significance level, we fail to reject the null 

hypothesis of no difference between population ponds. 
EPA QA/G-9S 93 February 2006 

CHAPTER 4 

STEP 4: VERIFY THE ASSUMPTIONS OF THE STATISTICAL METHOD 

Step 4: Verify the Assumptions of the Statistical Test

• Determine approach for verifying assumptions.

 Identify any strong graphical evidence from the preliminary data review.

 Review (or develop) the statistical model for the data.

 Select the methods for verifying assumptions.

• Perform tests of assumptions.

 Adjust for distributional assumption if warranted.

 Perform the calculations required for the tests.

• If necessary, determine corrective actions.

 Determine whether data transformations will correct the problem.

 If data are missing, explore the feasibility of using theoretical justification or of 

collecting new data.

 Consider robust procedures or nonparametric hypothesis tests.

THE DATA QUALITY ASSESSMENT PROCESS

Review DQOs and Sampling Design

Conduct Preliminary Data Review

Select the Statistical Method 

Verify the Assumptions 

Draw Conclusions from the Data 

VERIFY THE ASSUMPTIONS OF THE

 STATISTICAL TEST

Purpose

Examine the underlying assumptions of the statistical

method in light of the environmental data. 

 Activities

Determine Approach for Verifying Assumptions

Perform Tests of Assumptions

Determine Corrective Actions 

Tools

Tests of distributional assumptions

Tests for independence and trends

Tests for dispersion assumptions
EPA QA/G-9S 94 February 2006 

List of Boxes 

Page

Box 4-1: Directions for Studentized Range Test ......................................................................... 99 

Box 4-2: Example of the Studentized Range Test..................................................................... 100 

Box 4-3: Directions for Geary’s Test......................................................................................... 101 

Box 4-4: Example of Geary's Test............................................................................................. 101 

Box 4-5: Directions for the Test for a Correlation Coefficient and an Example....................... 105 

Box 4-6: Upper Triangular Computations for Basic Mann-Kendall Trend Test with a Single 

Measurement at Each Time Point............................................................................... 106 

Box 4-7: Directions for the Mann-Kendall Trend Test for Small Sample Sizes....................... 106 

Box 4-8: An Example of Mann-Kendall Trend Test for Small Sample Sizes........................... 106 

Box 4-9: Directions for the Mann-Kendall Test Using a Normal Approximation.................... 108 

Box 4-10: An Example of Mann-Kendall Trend Test by Normal Approximation.................... 109 

Box 4-11: Data for Multiple Times and Multiple Stations........................................................ 111 

Box 4-12: Testing for Comparability of Stations and an Overall Monotonic Trend................. 112 

Box 4-13: Directions for the Wald-Wolfowitz Runs Test......................................................... 114 

Box 4-14: An Example of the Wald-Wolfowitz Runs Test....................................................... 115 

Box 4-15: Directions for the Extreme Value Test (Dixon's Test) ............................................. 118 

Box 4-16: An Example of the Extreme Value Test (Dixon's Test) ........................................... 118 

Box 4-17: Directions for the Discordance Test ......................................................................... 119 

Box 4-18: An Example of the Discordance Test ....................................................................... 119 

Box 4-19: Directions for Rosner's Test for Outliers.................................................................. 120 

Box 4-20: An Example of Rosner's Test for Outliers................................................................ 121 

Box 4-21: Directions for Walsh's Test for Large Sample Sizes ................................................ 122 

Box 4-22: Directions for Constructing Confidence Intervals and Confidence Limits for the 

Sample Variance and Standard Deviation with an Example .................................... 123 

Box 4-23: Directions for an F-Test to Compare Two Population Variances with an Example 124 

Box 4-24: Directions for Bartlett's Test..................................................................................... 125 

Box 4-25: An Example of Bartlett's Test................................................................................... 125 

Box 4-26: Directions for Levene's Test ..................................................................................... 126

Box 4-27: An Example of Levene's Test ................................................................................... 127 

Box 4-28: Directions for Transforming Data and an Example.................................................. 129 

Box 4-29: Directions for Aitchison’s Method to Adjust Means and Variances........................ 131 

Box 4-30: An Example of Aitchison’s Method ......................................................................... 132 

Box 4-31: Directions for Cohen's Method................................................................................. 133 

Box 4-32: An Example of Cohen's Method............................................................................... 133 

Box 4-33: Example of Double Linear Interpolation.................................................................. 134 

Box 4-34: Directions for Selecting Between Cohen’s Method or Aitchison’s Method ............ 134 

Box 4-35: Example of Determining Between Cohen’s Method and Aitchison’s Method ........ 135 

Box 4-36: Directions for the Rank von Neumann Test ............................................................. 137 

Box 4-37: An Example of the Rank von Neumann Test ........................................................... 138 
EPA QA/G-9S 95 February 2006 

List of Tables 

Page

Table 4-1. Data for Examples in Section 4.2............................................................................... 96 

Table 4-2. Tests for Normality..................................................................................................... 97 

Table 4-3. Recommendations for Selecting a Statistical Test for Outliers................................ 117 

Table 4-4. Guidelines for Analyzing Data with Non-Detects.................................................... 130 

Table 4-5. Guidelines for Recommended Parameters for Different Coefficient of Variations 

and Censoring ........................................................................................................... 136 
EPA QA/G-9S 96 February 2006 

CHAPTER 4 

STEP 4: VERIFY THE ASSUMPTIONS OF THE STATISTICAL METHOD 

4.1 OVERVIEW AND ACTIVITIES 

In this step, the analyst should assess the validity of the statistical method chosen in 

step 3 by examining its underlying assumptions or determine that the data support the underlying 

assumptions necessary for the selected method, or if a different statistical method should be used. 

If it is determined that one or more of the assumptions is not met, then an alternative 

action is required. Typically, this means the selection of a different statistical method. Each 

method of Chapter 3 provides a detailed list of alternative methods. 

Parametric tests also have difficulty dealing with outliers and non-detects. If either is 

found in the data, then a corrective action would be to use the corresponding nonparametric 

method. In general, nonparametric methods handle outliers and non-detects better than 

parametric methods. If a trend in the data is detected or the data are found to be dependent, then 

the methods of Chapter 3 should not be applied. Time series or geostatistical methods may be 

required and a statistician should be consulted. For a more extensive discussion of the overview 

and activities of this step, see Data Quality Assessment: A Reviewer’s Guide (EPA QA/G-9R) 

(U.S.EPA 2004). 

4.2 TESTS FOR DISTRIBUTIONAL ASSUMPTIONS 

Many statistical tests and models are only appropriate for data that follow a particular 

distribution. This section will aid in determining if a distributional assumption of a statistical test 

is satisfied, in particular, the assumption of normality. Two of the most important distributions 

for tests involving environmental data are the normal distribution and the lognormal distribution, 

both of which are discussed in this section. To test if the data follow a distribution other than the 

normal distribution or the lognormal distribution, apply the chi-square test discussed in 

Section 4.2.5 or consult a statistician. 

There are many methods available for verifying the assumption of normality ranging 

from simple to complex. This section discusses methods based on graphs, sample moments 

(kurtosis and skewness), sample ranges, the Shapiro-Wilk test and closely related tests, and 

goodness-of-fit tests. Discussions for the simplest tests contain step-by-step directions and 

examples based on the data in Table 4-1. These tests are summarized in Table 4-2. This section 

ends with a comparison of the tests to help the analyst select a test for normality. 

Table 4-1. Data for Examples in Section 4.2 

These are 10 observations of dust from commercial air filters in ppm 

15.63 11.00 11.75 10.45 13.18 10.37 10.54 11.55 11.01 10.23 

X =11.57 

s = 1.677 

n = 10
EPA QA/G-9S 97 February 2006 

Table 4-2. Tests for Normality 

Test Section Sample 

Size Recommended Use 

Shapiro Wilk W Test 4.2.2 ≤ 5000 Highly recommended. 

Filliben's Statistic 4.2.2 ≤ 100 Highly recommended, especially when used in 

conjunction with a normal probability plot. 

Skewness and Kurtosis Tests 4.2.3 > 50 Useful for large sample sizes. 

Studentized Range Test 4.2.4 ≤ 1000 Highly recommended (with some conditions). 

Geary's Test 4.2.4 > 50 Useful when tables for other tests are not 

available. 

Chi-Square Test 4.2.5 Largea Useful for grouped data and when the 

comparison distribution is known. 

Lilliefors 

Kolmogorov- Smirnoff Test 4.2.5 > 50 Useful when tables for other tests are not available. 

a The necessary sample size depends on the number of groups formed when implementing this test. Each group 

should contain at least 5 observations.

The normal distribution is one of the most common probability distributions in the 

analysis of environmental data. It can often be used to approximate other probability 

distributions and in some instances more complex distributions can be transformed to an 

appropriate normal distribution. Additionally, as the sample size becomes larger, the sample 

mean has an approximate normal distribution hence the common assumption associated with 

parametric tests is that the data follows a normal distribution. 

The graph of a normally distributed random 

variable is bell-shaped (see Figure 4-1) with the 

highest point located at the mean which is equal to the 

median. A normal curve is symmetric about the 

mean, hence the part to the left of the mean is a mirror 

image of the part to the right. In environmental data, 

random errors occurring during the measurement 

process may be normally distributed. 

Environmental data commonly exhibit 

frequency distributions that are non-negative and 

skewed with heavy or long right-tails. Several 

standard parametric probability models have these properties, including the Weibull, gamma, and 

lognormal distributions. The lognormal distribution (Figure 4-1) is a commonly used 

distribution for modeling environmental contaminant data. The advantage to this distribution is 

that a simple (logarithmic) transformation will transform a lognormal distribution into a normal 

distribution. Therefore, the methods for testing for normality described in this section can be 

used to test for lognormality if a logarithmic transformation has been used. It should be noted 

that as the shape parameter of the lognormal approaches zero, the lognormal approaches a 

normal. When the shape parameter is large, the lognormal becomes more positively skewed. 

0 5 10 15 20 25

0

0.1

0.2

0.3

Lognormal Distribution

Normal Distribution

Figure 4-1. Density Plots for the Normal 

and Lognormal Distributions 
EPA QA/G-9S 98 February 2006 

4.2.1 Graphical Methods 

Graphical methods (Section 2.3) present detailed information about data sets that may not 

be apparent from a test statistic. Histograms, stem-and-leaf plots, and normal probability plots 

are some graphical methods that are useful for determining whether or not data follow a normal 

curve. Both the histogram and stem-and-leaf plot of a normal distribution are bell-shaped. The 

normal probability plot of a normal distribution follows a straight line. For non-normally 

distributed data, there will be large deviations in the tails or middle of a normal probability plot. 

Using a plot to decide if the data are normally distributed is a qualitative judgment. For 

extremely non-normal data, it is easy to make this determination; however, in many cases the 

decision is not straightforward. Therefore, formal test procedures are usually necessary to test 

the assumption of normality. 

4.2.2 Normal Probability Plot Tests 

One of the most powerful tests for normality is the Shapiro-Wilk W test. This test is 

similar to computing a correlation between the quantiles of the standard normal distribution and 

the ordered values of a data set. If the normal probability plot is approximately linear (i.e., the 

data follow a normal curve), the test statistic will be relatively high. If the normal probability 

plot is nonlinear, then the test statistic will be relatively low. 

The W test is recommended in several EPA guidance documents and in many statistical 

texts. Tables of critical values for sample sizes up to 50 have been developed for determining 

the significance of the test statistic. However, many software packages can perform the W test 

for data sets with sample sizes as large as 5000. This test is difficult to compute by hand since it 

requires two different sets of tabled values and a large number of summations and 

multiplications. Therefore, directions for implementing this test are not given in this document. 

Several tests related to the Shapiro-Wilk test have been proposed. D’Agostino’s test for 

sample sizes between 50 and 1000 and Royston’s test for sample sizes up to 2000 are two such 

tests that approximate some of the key quantities or parameters of the W test. 

Another related test is the Filliben statistic, also called the probability plot correlation 

coefficient. This test measures the linearity of the points on the normal probability plot. Similar 

to the W test, if the normal probability plot is approximately linear, then the correlation 

coefficient will be relatively high. Although easier to compute that the W test, the Filliben 

statistic is still difficult to compute by hand. Therefore, directions for implementing this test are 

not given in this guidance. 

4.2.3 Coefficient of Skewness/Coefficient of Kurtosis Tests 

The degree of symmetry (or asymmetry) displayed by a data set is measured by the 

coefficient of skewness (g3). The coefficient of kurtosis, g4, measures the degree of flatness of a 

probability distribution near its center. Several test methods have been proposed using these 

coefficients to test for normality. One method tests for normality by adjusting the coefficients of 
EPA QA/G-9S 99 February 2006 

skewness and kurtosis to approximate a standard normal distribution for sample sizes greater 

than 50. 

Two other tests based on these coefficients include a combined test based on a chisquared (χ

2

) distribution and Fisher’s cumulant test. Fisher’s cumulant test computes the exact 

sampling distribution of g3 and g4; therefore, it is more powerful than previous methods which 

assume that the distributions of the two coefficients are normal. Fisher’s cumulant test requires a 

table of critical values, and these tests require a sample size of greater than 50. Tests based on 

skewness and kurtosis are rarely used as they are less powerful than many alternatives. 

4.2.4 Range Tests 

Nearly 100% of the area of a normal curve lies within ±5 standard deviations from the 

mean. Tests for normality have been developed based on this fact. Two such tests are the 

Studentized Range test and Geary’s test. Both of these tests use a ratio of an estimate of the 

sample range to the sample standard deviation. Very large and very small values of the ratio 

imply that the data are not well modeled by a normal distribution. 

4.2.4.1 The Studentized Range Test 

This test compares the sample range to the sample standard deviation. Tables of critical 

values for sample sizes up to 1000 (Table A-3 of Appendix A) are available for determining 

whether the absolute value of this ratio is significantly large. Directions for implementing this 

method are given in Box 4-1 along with an example. The studentized range test does not 

perform well if the data are asymmetric or if the tails of the data are heavier than the normal 

distribution. In addition, this test may be sensitive to extreme values. Many environmental data 

sets are positively skewed (have a long tail of high values) and are similar to a lognormal 

distribution.. If the data appear to be lognormally distributed, then this test should not be used. 

In most cases, the studentized range test performs as well as the Shapiro-Wilk test and is easier 

to apply. 

Box 4-1: Directions for Studentized Range Test 

COMPUTATIONS: Using Section 2.2.3, calculate sample range, w, and sample standard deviation, s. 

STEP 1. Null Hypothesis: H0: The underlying distribution of the data is a normal distribution. 

STEP 2. Null Hypothesis: HA: The underlying distribution of the data is not a normal distribution. 

STEP 3. Test Statistic: R = w/s 

STEP 4. a) Critical Value: Use Table A-3 to find the critical values a and b. 

STEP 5. a) Conclusion: If R falls outside the two critical values, then reject the null hypothesis that 

 the underlying distribution is normal. 

Directions for the studentized range test are contained in Box 4-1, with an example in Box 4-2. 
EPA QA/G-9S 100 February 2006 

Box 4-2: Example of the Studentized Range Test 

Using a significance level of 0.01, determine if the underlying distribution of the data from Table 4-1 can be 

modeled using a normal distribution. 

COMPUTATIONS: w = X(10) – X(1) = 15.63 – 10.23 = 5.40 and s = 1.677. 

STEP 1. Null Hypothesis: H0: The underlying distribution of the data is a normal distribution. 

STEP 2. Null Hypothesis: HA: The underlying distribution of the data is not a normal distribution. 

STEP 3. Test Statistic: R = w/s = 5.4 / 1.677 = 3.22. 

STEP 4. a) Critical Value: Using Table A-3, the critical values a and b are 2.51 and 3.88. 

STEP 5. a) Conclusion: Since R falls between the two critical values, we fail to reject the null 

 hypothesis that the underlying distribution is normal. 

4.2.4.2 Geary’s Test 

Geary’s test compares the sum of the absolute deviations from the mean to the sum of the 

squares. If the ratio is too large or too small, then the underlying distribution of the data should 

not be modeled as a normal distribution. Directions for implementing this method are given in 

Box 4-3 and an example is given in Box 4-4. This test does not perform as well as the ShapiroWilk test or the Studentized Range test. 

 Directions for Geary’s test are given in Box 4-3, with an example in Box 4-4. 

4.2.5 Goodness-of-Fit Tests 

Goodness-of-fit tests are used to test whether data follow a specific distribution, i.e., how 

"good" a specified distribution fits the data. In verifying assumptions of normality, one would 

compare the data to a normal distribution with a specified mean and variance. 

4.2.5.1 Chi-Square Test 

One classic goodness-of-fit test is the chi-square test which involves breaking the data 

into groups and comparing these groups to the expected groups from the known distribution. 

There are no fixed methods for selecting these groups and this test also requires a large sample 

size since at least 5 observations per group is needed to implement this test. In addition, the chisquare test does not have the power of the Shapiro-Wilk test or some of the other tests mentioned 

above. However, it is more flexible since the data can be compared to probability distributions 

other than the normal but the application of goodness-of-fit tests to non-normal data is beyond 

the scope of this guidance. 
EPA QA/G-9S 101 February 2006 

Box 4-3: Directions for Geary’s Test 

COMPUTATIONS: Calculate the sample mean, X , the sample sum of squares, SSS, and the sum of 

absolute deviations, SAD: 

X

n

X

i=

n

= ∑ i

1

1

, SSS X

n

X ,

i=

n

i

i=

n

= − i

⎛

⎝

⎜

⎜

⎞

⎠

⎟ ∑ ∑ ⎟ 1

2

1

2

1 and SAD X X

i=

n

= − ∑ i

1

. 

STEP 1. Null Hypothesis: H0: The underlying distribution of the data is a normal distribution. 

STEP 2. Null Hypothesis: HA: The underlying distribution of the data is not a normal distribution. 

STEP 3. Test Statistic: 

n

.

a . z 0 2123

0 7979

0

− = , where a

SAD

n SSS = ⋅ . 

STEP 4. a) Critical Value: Use Table A-1 to find z1-α/2. 

STEP 4. b) p-value: Use Table A-1 to find ( ) 2 P Z > z0 ⋅ . 

STEP 5. a) Conclusion: If z0 > z1−α / 2 , then reject the null hypothesis that the underlying 

 distribution is normal. 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis that the underlying 

 distribution is normal. 

Box 4-4: Example of Geary's Test 

Using a significance level of 0.05, determine if the underlying distribution of the data from Table 4-1 can be 

modeled using a normal distribution. 

COMPUTATIONS: X = 11.571, 694 SAD = 11. , and 298 SSS = 25. . 

STEP 1. Null Hypothesis: H0: The underlying distribution of the data is a normal distribution. 

STEP 2. Null Hypothesis: HA: The underlying distribution of the data is not a normal distribution. 

STEP 3. Test Statistic: 0.9339

10

0.2123

0 7352 0 7979

0 = − − = . . z , since 0 7352

10 25 298

11694 . .

. a = ⋅ = . 

STEP 4. a) Critical Value: Using Table A-1, z0.975 = 1.96. 

STEP 4. b) p-value: Using Table A-1, 2⋅P(Z > 0.9339) = 2⋅(1 – 0.8238) = 0.3524. 

STEP 5. a) Conclusion: Since z0 = 0.9339 < 1.96 = z1−α , we fail to reject the null hypothesis 

 that the underlying distribution is normal. 

STEP 5. b) Conclusion: Since p-value = 0.3524 > 0.05 = α, we fail to reject the null hypothesis 

 that the underlying distribution is normal. 
EPA QA/G-9S 102 February 2006 

4.2.5.2 Tests Based on the Empirical Distribution Function 

Various methods have been used to measure the discrepancy between the empirical 

distribution function and the theoretical cumulative distribution function (cdf). These measures 

are referred to as empirical distribution function statistics. The best known is the KolmogorovSmirnov (K-S) statistic. The K-S approach is appropriate if the sample size exceeds 50 or if the 

theoretical cumulative density function is a specific distribution with known parameters. A 

modification to the test, called the Lilliefors K-S test, can be used to test that the data (n > 50) 

comes from a normal distribution with mean and variance equal to the sample values. 

Unlike the K-S type statistics, most empirical distribution function statistics are based on 

integrated or average values between the empirical distribution function and hypothesized 

cumulative distribution functions. The two most powerful are the Cramer-von Mises and 

Anderson-Darling statistics. Extensive simulations show that the Anderson-Darling empirical 

distribution function statistic is as good as any test, including the Shapiro-Wilk test, when testing 

for normality. However, the Shapiro-Wilk test is applicable only for the normal distribution, 

while the Anderson-Darling method applies to other distributions. 

4.2.6 Recommendations 

Tests for normality with small samples have very little statistical power. Therefore, for 

small sample sizes it is common for a nonparametric statistical test be selected during Step 3 of 

the DQA in order to avoid incorrectly assuming the data are normally distributed when there is 

simply not enough information. 

This guidance recommends using the Shapiro-Wilk W test, wherever practicable. The 

Shapiro-Wilk W test is one of most powerful tests for normality. This test is difficult to 

implement by hand but can be applied easily using a statistical software package. If the ShapiroWilk W test is not feasible, then this guidance recommends using either Filliben's statistic 

together with inspection of the normal probability plot, or the studentized range test. Filliben's 

statistic performs similarly to the Shapiro-Wilk test. The studentized range is a simple test to 

perform, but, it is not applicable for non-symmetric data with large tails. If the data are not 

highly skewed and the tails are not significantly large compared to a normal distribution, then the 

studentized range provides a simple and powerful test that can be calculated by hand. The 

Lilliefors Kolmogorov-Smirnoff test is statistically powerful but is also more difficult to apply 

and uses specialized tables not readily available. 

4.3 TESTS FOR TRENDS 

4.3.1 Introduction 

This section presents statistical tools for detecting and estimating trends in environmental 

data. The detection and estimation of temporal or spatial trends are important for many 

environmental studies or monitoring programs. In cases where temporal or spatial patterns are 

strong, simple procedures such as time plots or linear regression over time can reveal trends. In 

more complex situations, sophisticated statistical models and procedures may be needed. For 
EPA QA/G-9S 103 February 2006 

example, the detection of trends may be complicated by the overlaying of long- and short-term 

trends, cyclical effects (e.g., seasonal or weekly systematic variations), autocorrelations, or 

impulses or jumps (e.g., due to interventions or procedural changes). 

The graphical representations of Section 2.3.7 are recommended as the first step to 

identify possible trends. A Time Plot and/or Lag Plot is recommended for temporal data as it 

may reveal long-term trends, seasonal behavior, or impulses. A posting or bubble plot is 

recommended for spatial data to reveal spatial trends such as areas of high concentration or 

inaccessible areas. 

For most of the statistical tools presented below, the focus is on monotonic long-term 

trends (i.e., a trend that is exclusively increasing or decreasing), as well as other sources of 

systematic variation, such as seasonality. The investigations of trend in this section are limited 

to one-dimensional domains, e.g., trends in a pollutant concentration over time. The current 

edition of this document does not address spatial trends or trends over space and time, which 

may involve sophisticated geostatistical techniques such as kriging and require the assistance of 

a statistician. Section 4.3.2 discusses estimating and testing for trends using regression 

techniques. Section 4.3.3 discusses nonparametric trend estimation procedures, and Section 

4.3.4 discusses hypothesis tests for detecting trends under several types of situations. 

4.3.2 Regression-Based Methods for Estimating and Testing for Trends 

4.3.2.1 Estimating a Trend Using the Slope of the Regression Line 

The classic procedures for assessing linear trends involve regression. Linear regression is 

a commonly used procedure in which calculations are performed on a data set containing pairs of 

observations (Xi, Yi), so as to obtain the slope and intercept of a line that best fits the data. For 

temporal data, the Xi values represent time and the Yi values represent the observations. An 

estimate of the magnitude of trend can be obtained by performing a regression of the data versus 

time and using the slope of the regression line as the measure of the strength of the trend. 

Regression procedures are easy to apply. All statistical software packages and 

spreadsheet programs will calculate the slope and intercept of the best fitting line, as well as the 

correlation coefficient r (see Section 2.2.4). However, regression entails several limitations and 

assumptions. First of all, simple linear regression (the most commonly used method) is designed 

to detect linear relationships between two variables; other types of regression models are 

generally needed to detect non-linear relationships such as cyclical or non-monotonic trends. 

Regression is very sensitive to outliers and presents difficulties in handling data below the 

detection limit, which are commonly encountered in environmental studies. Hypothesis testing 

for linear regression also relies on two key assumptions: normally distributed errors, and constant 

variance. It may be difficult or burdensome to verify these assumptions in practice, so the 

accuracy of the slope estimate may be suspect. Moreover, the analyst must ensure that time plots 

of the data show no cyclical patterns, outlier tests show no extreme data values, and data 

validation reports indicate that nearly all the measurements were above detection limits. Due to 

these drawbacks, linear regression is not recommended as a general tool for estimating and 
EPA QA/G-9S 104 February 2006 

detecting trends, although it may be useful as an informal and quick screening tool for 

identifying strong linear trends. 

4.3.2.2 Testing for Trends Using Regression Methods 

For simple linear regression, the statistical test of whether the slope is significantly 

different from zero is equivalent to testing if the correlation coefficient is significantly different 

from zero. This test assumes a linear relation between Y and X with independent normally 

distributed errors and constant variance across all the X values. Non-detects and outliers may, 

however, invalidate the test. 

Directions for this test are given in Box 4-5 along with an example. 

4.3.3 General Trend Estimation Methods 

4.3.3.1 Sen's Slope Estimator 

Sen’s Slope Estimate is a nonparametric alternative for estimating a slope. This approach 

involves computing slopes for all the pairs of time points and then using the median of these 

slopes as an estimate of the overall slope. As such, it is insensitive to outliers and can handle a 

moderate number of values below the detection limit and missing values. Assume that there are 

n time points, and let Xi denote the data value for the i

th time point. If there are no missing data, 

there will be n(n-1)/2 possible pairs of time points (i, j) in which i < j. The slope for such a pair 

is bij = (Xj – Xi) / (j - i). Sen's slope estimator is then the median of the n(n-1)/2 pairwise slopes. 

If there is no underlying trend, there would be an approximately equal number of positive and 

negative slopes, and thus the median would be near zero. 

4.3.3.2 Seasonal Kendall Slope Estimator 

If the data exhibit cyclic trends, then Sen's slope estimator can be modified to account for 

the cycles. For example, if data are available for each month for a number of years, 12 separate 

sets of slopes would be determined; similarly, if daily observations exhibit weekly cycles, seven 

sets of slopes would be determined. In these estimates, the above pairwise slope is calculated for 

each time period and the median of all of the slopes is an estimator of the slope for a long-term 

trend. This is known as the seasonal Kendall slope estimator. 

4.3.4 Hypothesis Tests for Detecting Trends 

Most of the trend tests treated in this section involve the Mann-Kendall test or extensions 

of it. The Mann-Kendall test does not assume any particular distributional form and 

accommodates values below the detection limit by assigning them a common value. The test can 

also be modified to deal with multiple observations, multiple sampling locations, and 

seasonality. 
EPA QA/G-9S 105 February 2006 

Box 4-5: Directions for the Test for a Correlation Coefficient and an Example 

COMPUTATIONS: Calculate the correlation coefficient, r (Section 2.2.4). 

STEP 1. Null Hypothesis: H0: The correlation coefficient is zero. 

STEP 2. Alternative Hypothesis: HA: The correlation coefficient is different from zero. 

STEP 3. Test Statistic: t r

r

n

0 2 1

2

=

−

−

. 

STEP 4. a) Critical Value: Use Table A-2 to find n−2,1−α / 2 t . 

STEP 4. b) p-value: Use Table A-2 to find ( ) 2 P 2 0 t t ⋅ n− > . 

STEP 5. a) Conclusion: If 0 > n−2,1−α / 2 t t , then reject the null hypothesis that the 

 correlation coefficient is zero. 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis that the correlation 

 coefficient is zero. 

Example: Using a significance level of 0.10, determine if the following data set (in ppb) has significant 

correlation between it’s two variables: 

Sample Number 1 2 3 4 

Arsenic 8.0 6.0 2.0 1.0 

Lead 8.0 7.0 7.0 6.0 

COMPUTATIONS: In Section 2.2.4, the correlation coefficient r for this data was calculated to be 0.865. 

STEP 1. Null Hypothesis: H0: The correlation coefficient is zero. 

STEP 2. Alternative Hypothesis: HA: The correlation coefficient is different from zero. 

STEP 3. Test Statistic: 2.438

4 2

1 0.865

0 865

2 0 =

−

−

= . t . 

STEP 4. a) Critical Value: Using Table A-2, 303 4. t2, 0.975 = . 

STEP 4. b) p-value: Using Table A-2, 0.10< p-value < 0.20 (the exact 

 p-value = 0.1350) 

STEP 5. a) Conclusion: Since 0 2, 0.975 t < t , we fail to reject the null hypothesis that the 

 correlation coefficient is zero. 

STEP 5. b) Conclusion: Since p-value = 0.1350 > 0.10 (α) we fail to reject the null hypothesis 

 that the correlation coefficient is zero. 
EPA QA/G-9S 106 February 2006 

4.3.4.1 One Observation per Time Period for One Sampling Location 

The Mann-Kendall test involves computing a statistic S, which is the difference between 

the number of pairwise differences that are positive minus the number that are negative. If S is a 

large positive value, then there is evidence of an increasing trend in the data. If S is a large 

negative value, then there is evidence of a decreasing trend in the data. The null hypothesis or 

baseline condition for this test is that there is no temporal trend in the data values. The 

alternative hypothesis is that of either an upward trend or a downward trend. 

Box 4-6: Upper Triangular Computations for Basic Mann-Kendall 

Trend Test with a Single Measurement at Each Time Point 

Original Time 

Measurement 

t1

X1

t2

X2

t3

X3

t4

X4

. . 

. 

. . 

. 

tn-1

Xn-1

tn

Xn

X1 Y12 Y13 Y32

. . 

. Y1,n-1 Y1n #(plusses) #(minuses) 

X2 Y23 Y32

. . 

. Y2,n-1 Y2n #(plusses) #(minuses) 

. 

. 

. . 

. 

. 

. 

. 

. 

. 

. 

. 

. 

. 

. 

Xn-2 Yn-2,n-1 Yn-2,n #(plusses) #(minuses) 

Xn-1 Yn-1,n #(plusses) #(minuses) 

 Total 

#(plusses) 

Total 

#(minuses) 

where Yij = sign (Xj-Xi) = + if Xj > Xi 

= 0 if Xj = Xi 

= - if Xj < Xi 

Box 4-7: Directions for the Mann-Kendall Trend Test for Small Sample Sizes 

COMPUTATIONS: Create the upper triangular table of pairwise differences as described in Box 4-6. 

STEP 1. Null Hypothesis: H0: There is no trend. 

STEP 2. Null Hypothesis: i) HA: There is a downward trend. 

 ii) HA: There is an upward trend. 

STEP 3. Test Statistic: S = Total #(plusses) – Total #(minuses) 

STEP 4. a) Critical Value: Use Table A-12a to find the critical value. 

STEP 4. b) p-value: Use Table A-12b to find the p-value. 

STEP 5. a) Conclusion: If |S| ≥ the critical value, then reject the null hypothesis of no trend. 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis of no trend. 
EPA QA/G-9S 107 February 2006 

The computations for the basic Mann-Kendall trend test are depicted in Box 4-6. Assign 

a value of DL/2 to all non-detects. The test statistic is the difference between the number of 

strictly positive differences and the number of strictly negative differences. Differences of zero 

are not included in the test statistic (and should be avoided, if possible, by recording data to 

sufficient accuracy). The steps for conducting the Mann-Kendall test for small sample sizes 

(n < 10) are contained in Box 4-7 and an example is contained in Box 4-8. For sample sizes 

greater than 10, there is a normal approximation for the Mann-Kendall test. Directions for this 

approximation are contained in Box 4-9 with an example given in Box 4-10. 

Box 4-8: An Example of Mann-Kendall Trend Test for Small Sample Sizes 

Consider 5 measurements ordered by the time of their collection: 5, 6, 11, 8, and 10. This data will be used to test for 

an upward trend at a significance level of 0.05. 

COMPUTATIONS: A triangular table was used to construct the pairwise differences.

STEP 1. Null Hypothesis: H0: There is no trend. 

STEP 2. Null Hypothesis: HA: There is an upward trend. 

STEP 3. Test Statistic: S = Total #(plusses) – Total #(minuses) = 8 - 2 = 6 

STEP 4. a) Critical Value: Using Table A-12a, the critical value is 8. 

STEP 4. b) p-value: Using Table A-12b, the p-value is 0.117. 

STEP 5. a) Conclusion: As S = 6 < 8 (critical value), fail to reject the null hypothesis of no trend 

STEP 5. b) Conclusion: Since p-value = 0.117 > 0.05(significance level), we fail to reject the null 

 hypothesis of no trend.

Time 

Data 

1 

5 

2 

6 

3 

11 

4 

8 

5 

10 

No. of 

+ Signs 

No. of - 

Signs 

5 

6 

11 

8 

+ + 

+ 

+ 

+ 

- 

+ 

+ 

- 

+ 

4 

3 

0 

1

8 

0 

0 

2 

0

2 
EPA QA/G-9S 108 February 2006 

Box 4-9: Directions for the Mann-Kendall Test Using a Normal Approximation 

COMPUTATIONS: If the sample size is 10 or more, a normal approximation to the Mann-Kendall procedure may 

be used. Compute S as described in Box 4-7 above. 

STEP 1. Null Hypothesis: H0: There is no trend. 

STEP 2. Null Hypothesis: i) HA: There is a downward trend. 

 ii) HA: There is an upward trend. 

STEP 3. Test Statistic: ( )

( ) S

S S

z

V

sign

0

− = , where ( ) ( )( ) ( )( )

⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

= − + −∑ − +

=

1 2 5 1 2 5

18

1 V

1

j j j

g

j

S n n n t t t , 

 g is the number of tied groups, and tj is the number of points in the j

th group. 

 Note that sign(S) = 1 if S > 0, 0 if S = 0 and –1 if S < 0. 

STEP 4. a) Critical Value: Use Table A-1 to find z1-α. 

STEP 4. b) p-value: Use Table A-1 to find ( ) P 0 Z > z . 

STEP 5. a) Conclusion: If 0 > 1−α z z , then reject the null hypothesis of no trend. 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis of no trend. 
EPA QA/G-9S 109 February 2006 

 Box 4-10: An Example of Mann-Kendall Trend Test by Normal Approximation 

A test for an upward trend with α = 0.05 will be based on the 11 weekly measurements is shown below. 

COMPUTATIONS: Using Box 4-6, a triangular table was constructed of the pairwise slopes. “0” indicates a tie. 

Data 

10 

10 

10 

5 

10 

20 

18 

17 

15 

24 

1 

10

2 

10

 0 

3 

10

 0 

0 

4 

 5

- 

- 

- 

5 

10

0 

0 

0 

+ 

6 

20

+ 

+ 

+ 

+ 

+ 

7 

18

+ 

+ 

+ 

+ 

+ 

- 

8 

17

+ 

+ 

+ 

+ 

+ 

- 

- 

9 

15

+ 

+ 

+ 

+ 

+ 

- 

- 

- 

10 

24

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

11 

15

+ 

+ 

+ 

+ 

+ 

- 

- 

- 

0 

- 

No. of 

+ Signs

6 

6 

6 

7 

6 

1 

1 

1 

1 

0

35 

No. of 

- Signs

1 

1 

1 

0 

0 

4 

3 

2 

0 

1

13 

S = (sum of + signs) - (sum of - signs) = 35 - 13 = 22. 

STEP 1. Null Hypothesis: H0: There is no trend. 

STEP 2. Null Hypothesis: HA: There is an upward trend. 

STEP 3. Test Statistic: There are several observations tied at 10 and 15. Thus, the formula for tied 

 values will be used. In this formula, g = 2, t1 = 4 for tied values of 10, and 

 t2 = 2 for tied values of 15. Therefore, 

( ) { } 11( )( ) ( )( ) ( )( ) 11 1 2 11 5 [ ] 4 4 1 2 4 5 2 2 1 2 2 5 155 33

18

1 V S = − ⋅ + − − ⋅ + + − ⋅ + = .

and ( )

( ) 1.685

155.33

22 1

V

sign

0 = − = − = S

S S

z . 

STEP 4. a) Critical Value: Using Table A-1, z0.95 = 1.645. 

STEP 4. b) p-value: Using Table A-1, p-value = P(Z > 1.685) = 0.046. 

STEP 5. a) Conclusion: Since test statistic = 1.685 > 1.645 = critical value, we reject the null 

 hypothesis of no trend. 

STEP 5. b) Conclusion: Since p-value = 0.046 < 0.05 = α, we reject the null hypothesis of no trend. 

4.3.4.2 Multiple Observations per Time Period for One Sampling Location 

Often, more than one sample is collected for each time period. There are two ways to 

deal with multiple observations per time period. One method is to compute a summary statistic, 

such as the median, for each time period and to apply one of the Mann-Kendall trend tests of 

Section 4.3.4.1 to the summary statistic. Therefore, instead of using the individual data points in 

the triangular table, the summary statistic would be used. Then the steps given in Box 4-7 or 

Box 4-9 could be applied to the summary statistics. 
EPA QA/G-9S 110 February 2006 

An alternative approach is to consider all the multiple observations within a given time 

period as being essentially taken at the same time within that period. The S statistic is computed 

as before with n being the total of all observations. The variance of the S statistic (calculated in 

step 2 of Box 4-9) is changed to: 

( )( ) ( )( ) ( )( )

⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

= − + −∑ − + −∑ − +

= =

1 2 5 1 2 5 1 2 5

18

1 VAR

1 1

k k k

h

k

j j j

g

j

(S) n n n t t t u u u

( )( ) ( )( )

( )( )

( ) ( )

2 ( ) 1

1 1

9 1 2

1 2 1 2

1 1 1 1

−

− ⋅ −

+ − −

− − ⋅ − −

+

∑ ∑ ∑ ∑ = = = =

n n

t t u u

n n n

t t t u u u

h

k

k k

g

j

j j

h

k

k k k

g

j

j j j

where g represents the number of tied groups, tj represents the number of data points in the j

th

group, h is the number of time periods which contain multiple data, and uk is the sample size in 

the k

th time period. 

The preceding variance formula assumes that the data are not correlated. If correlation 

within single time periods is suspected, it is preferable to use a summary statistic (e.g., the 

median) for each time period and then apply either Box 4-7 or Box 4-9 to the summary statistics. 

4.3.4.3 Multiple Sampling Locations with Multiple Observations 

The preceding methods involve a single sampling location (station). However, 

environmental data often consist of sets of data collected at several sampling locations (see 

Box 4-11). For example, data are often systematically collected at several fixed sites on a lake or 

river, or within a region or basin. The data collection plan (or experimental design) must be 

systematic in the sense that approximately the same sampling times should be used at all 

locations. In this situation, it is desirable to express the results by an overall regional summary 

statement across all sampling locations. However, there must be consistency in behavioral 

characteristics across sites over time in order for a single summary statement to be valid across 

all sampling locations. A useful plot to assess the consistency requirement is a single time plot 

(Section 2.3.7.1) of the measurements from all stations where a different symbol is used to 

represent each station. 

If the stations exhibit approximate trends in the same direction with comparable slopes, 

then a single summary statement across stations is valid and this implies two relevant sets of 

hypotheses should be investigated: 

Comparability of stations. H0: Similar dynamics affect all K stations vs. HA: At least 

two stations exhibit different dynamics. 

Testing for overall monotonic trend. H0*: Contaminant levels do not change over time 

vs. HA*: There is an increasing (or decreasing) trend consistent across all stations. 
EPA QA/G-9S 111 February 2006 

Therefore, the analyst must first test for homogeneity of stations, and then, if 

homogeneity is confirmed, test for an overall monotonic trend. Directions for the test are 

contained in Box 4-11 and ideally, the stations in Box 4-11 should have equal sample sizes. 

However, the numbers of observations at the stations can differ slightly, because of isolated 

missing values, but the overall time periods spanned must be similar. This guidance 

recommends that for less than 3 time periods, an equal number of observations (a balanced 

design) are required. For 4 or more time periods, up to 1 missing value per sampling location 

may be tolerated. 

Box 4-11: Data for Multiple Times and Multiple Stations 

Let i = 1, 2, ..., n represent time, k = 1, 2, ..., K represent sampling locations, and Xik represent the measurement 

at time i for location k. This data can be summarized in matrix form, as shown below. 

Stations 

1 2 . . . K 

Time 

1 

2 

. 

. 

. 

n 

X11

X21

. 

. 

. 

Xn1

X12

X22

. 

. 

. 

Xn2

. . . 

. . . 

. . . 

. . . 

. . . 

. . . 

X1K

X2K

. 

. 

. 

XnK

S1

V(S1) 

Z1

S2

V(S2) 

Z2

 . . 

. 

 . . 

. 

 . . 

. 

SK

V(SK) 

ZK

where Sk = Mann-Kendall statistic for station k (see STEP 3, Box 4-7), 

V(Sk) = variance for S statistic for station k (see STEP 3, Box 4-9), and 

( ) k

k k

k S

S S

z

V

− sign( ) =

a. One Observation per Time Period. When only one measurement is taken for 

each time period for each station, a generalization of the Mann-Kendall statistic can be used to 

test the above hypotheses. This procedure is described in Box 4-12. 

b. Multiple Observations per Time Period. If multiple measurements are taken at 

some times and stations, then the previous approaches are still applicable. However, the 

variance of the statistic Sk must be calculated using the equation for calculating V(S) given in 

Section 4.3.4.2. Note that Sk is computed for each station, so n, tj, g, h, and uk are all stationspecific. 
EPA QA/G-9S 112 February 2006 

Box 4-12: Testing for Comparability of Stations and an Overall Monotonic Trend 

Let i = 1, 2, ..., n represent time, k = 1, 2, ..., K represent sampling locations, and Xik represent the 

measurement at time i for location k. Let α represent the significance level for testing homogeneity and α* 

represent the significance level for testing for an overall trend. 

COMPUTATIONS: Calculate the Mann-Kendall statistic Sk and its variance V(Sk) for each of the K stations using 

the methods of Box 4-9. Now, calculate 

( )

( ) k

k k

k S

S S Z

V

− sign = , for k = 1,...,K and Z

K

Zk

k

K

=

=

∑

1

1

. 

Test of Homogeneity

STEP 1. Null Hypothesis: H0: Similar dynamics affect all K stations. 

STEP 2. Null Hypothesis: HA: At least two stations exhibit different dynamics. 

STEP 3. Test Statistic: χh k

k

K

Z KZ 2 2

1

2 = −

=

∑ . 

STEP 4. a) Critical Value: Use Table A-9 to find 2 χ K −1, 1−α . 

STEP 4. b) p-value: Use Table A-9 to find ( ) 2 2 P χ K −1 > χ h . 

STEP 5. a) Conclusion: If 2

1, 1

2 χ h > χ K − −α , then reject the null hypothesis that similar dynamics affect 

 all K stations. 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis that similar dynamics affect all 

 K stations. 

If H0 is not rejected, then proceed to test of overall trend. Otherwise, individual α*-level Mann-Kendall tests 

should be conducted using the methods presented in Section 4.3.4.1. 

Test of Overall Trend

STEP 1. Null Hypothesis: H0*: Contaminant levels do not change over time. 

STEP 2. Null Hypothesis: HA*: There is an increasing (or decreasing) trend consistently 

 exhibited across all stations. 

STEP 3. Test Statistic: 2 2 χ o = K ⋅ Z . 

STEP 4. a) Critical Value: Use Table A-9 to find 2 χ1, 1−α * . 

STEP 4. b) p-value: Use Table A-9 to find ( ) 2 2 P χ1 > χ o . 

STEP 5. a) Conclusion: If 2

1, 1 * 2 χ h > χ −α , then reject the null hypothesis that contaminant levels do 

 not change over time. 

STEP 5. b) Conclusion: If p-value < α, then reject the null hypothesis .
EPA QA/G-9S 113 February 2006 

4.3.4.4 One Observation for One Station with Multiple Seasons 

Temporal data are often collected over extended periods of time. Within the time 

variable, data may exhibit periodic cycles, which are patterns in the data that repeat over time. 

For example, temperature and humidity may change with the season or month, and may affect 

environmental measurements. (For more information on seasonal cycles, see Section 2.3.7). In 

the following discussion, the term season represents one time point in the periodic cycle, such as 

a month within a year or an hour within a day.

If seasonal cycles are anticipated, then two approaches for testing for trends are the 

seasonal Kendall test and Sen's test for trends. The seasonal Kendall test may be used for large 

sample sizes, and Sen's test for trends may be used for small sample sizes. If different seasons 

manifest similar slopes (rates of change) but possibly different intercepts, then the Mann-Kendall 

technique of Section 4.3.4.3 is applicable, replacing time by year and replacing station by season. 

 

The seasonal Kendall test, which is an extension of the Mann-Kendall test, involves 

calculating the Mann-Kendall test statistic, S, and its variance separately for each season. The 

sum of the S's and the sum of their variances are then used to form an overall test statistic that is 

assumed to be approximately normally distributed for larger size samples. 

For data at a single site, collected at multiple seasons within multiple years, the 

techniques of Section 4.3.4.3 can be applied to test for homogeneity of time trends across 

seasons. The methodology follows Boxes 4-11 and 4-12 exactly except that ‘station’ is replaced 

by ‘season’ and the inferences refer to seasons. 

4.3.5 A Discussion on Tests for Trends 

This section discusses some further considerations for choosing among the many tests for 

trend. Mann-Kendall type nonparametric trend tests and estimates use ordinal time (ranks) rather 

than cardinal time (actual time values) and this restricts the interpretation of measured trends. 

All of the Mann-Kendall Trend Tests presented are based on certain pairwise differences in 

measurements at different time points. The only information about these differences that is used 

in the Mann-Kendall calculations is the sign, and can be regarded as generalizations of the signed 

rank test. However, since information about the magnitudes of the differences is not used, this 

can adversely affect the statistical power when only limited amounts of data are available. 

There are nonparametric methods based on ranks that take such magnitudes into account 

and still retain the benefit of robustness to outliers. These procedures can be thought of as 

replacing the data with their ranks and then conducting parametric analyses. These include the 

Wilcoxon Rank Sum test and its many generalizations. These methods are more resistant to 

outliers than parametric methods. Rank-based methods which make fuller use of the information 

in the data than the Mann-Kendall methods are not as robust with respect to outliers as the signed 

rank test and the Mann-Kendall tests, but they have more statistical power. This kind of tradeoff 
EPA QA/G-9S 114 February 2006 

between power and robustness shows the need for an evaluation process leading to the selection 

of the best statistical procedure for the situation. 

4.3.6 Testing for Trends in Sequences of Data 

There are cases where it is desirable to see if a sequence of data (for example, readings 

from a monitoring station) could be considered random variation or correlated in some way. One 

test to make this determination is the Wald-Wolfowitz test. This test can only be used if the data 

are binary, i.e., there are only two potential values. For example, the data could be either 

‘Yes/No’ or an investigation of persistent violation of a permitted limit by a pollution control 

process where a violation equals 1, with 0 for not in violation. Directions for the WaldWolfowitz test are given in Box 4-13 and an example in Box 4-14. 

Box 4-13: Directions for the Wald-Wolfowitz Runs Test 

Consider a sequence of binary values. Let m and n denote the number of observations for the two values with 

n<m. This test is used to test the null hypothesis that the sequence is random against the alternative hypothesis 

that the values in the sequence are correlated or may come from different populations. 

COMPUTATIONS: List the data in the order collected and compute T, the number of runs in the sequence. For 

example, if the data sequence is AAABAABBBBBBBABB, then T = 6. 

STEP 1. Null Hypothesis: H0: data sequence is random. 

STEP 2. Alternative Hypothesis: HA: data sequence is non-random. 

STEP 3. Test Statistic: If either m or n is less than 10, then T. 

 If both m and n are at least 10, then 

r

T r z σ− μ

0 = , where 

 1 2

+

+ = m n

mn

μ r and ( )

( )( ) 1

2 2

2 + + −

− − = m n m n

mn mn m n σ r

STEP 4. a) Critical Value: If either m or n is less than 10, then use Table A-13 to find wα/2. 

 If both m and n are both at least 10, then use Table A-1 to find z1-α/2. 

STEP 4. a) p-value: If both m and n are both at least 10, then use Table A-1 to find ( ) 0 P Z > z . 

STEP 5. a) Conclusion: If either m or n is less than 10 and if T < wα/2 or T > 2 μr - wα/2, then reject 

 the null hypothesis that the data sequence is random. 

 If both m and n are at least 10 and if |z0| > z1-α/2, then reject the null 

 hypothesis that the data sequence is random. 

STEP 5. b) Conclusion: If p-value < α, the reject the null hypothesis that the data sequence is 

 random. 
EPA QA/G-9S 115 February 2006 

Box 4-14: An Example of the Wald-Wolfowitz Runs Test 

The main discharge station at a chemical manufacturing plant is under a monitoring program. The permit states 

that the discharge should have a pH of 7.0 and should never be less than 5.0. So the plant manager has decided 

to use a pH of 6.0 to an indicate potential problems. In a four-week period the following values were recorded: 

6.5 6.6 6.4 6.2 5.9 5.8 5.9 6.2 6.2 6.3 6.6 6.6 6.7 6.4 

6.2 6.3 6.2 5.8 5.9 5.8 6.1 5.9 6.0 6.2 6.3 6.2 

Since the plant manager has decided that a pH of 6.0 will indicate trouble the data have been replaced with a 

binary indicator. If the value is greater than 6.0, the value will be replaced by a 1; otherwise the value will be 

replaced by a 0. So the binary sequence is: 

1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 0 1 0 0 1 1 1 

As there are 8 values of ‘0’and 19 values of ‘1’, n = 8 and m = 19 and the number of runs is T = 7. Test at a 

significance level of 0.10 whether the data sequence is random. 

STEP 1. Null Hypothesis: H0: data sequence is random. 

STEP 2. Alternative Hypothesis: HA: data sequence is non-random. 

STEP 3. Test Statistic: Since n is less than 10, the test statistic is T = 7. 

STEP 4. a) Critical Value: Since n is less than 10, Table A-13 is used to find w0.05 = 9. 

STEP 5. a) Conclusion: Since n is less than 10 and T =7 < 9 = w0.05, we reject H0. 

4.4 OUTLIERS 

4.4.1 Background 

Potential outliers are measurements that are extremely large or small relative to the rest of 

the data and, therefore, are suspected of misrepresenting the population from which they were 

collected. Potential outliers may result from transcription errors, data-coding errors, or 

measurement system problems. However, outliers may also represent true extreme values of a 

distribution (for instance, hot spots) and indicate more variability in the population than was 

expected. Failure to remove true outliers or the removal of false outliers both lead to a 

distortion of estimates of population parameters and if it is recommended that the QA Project 

Plan or Sampling and Analysis Plan be reviewed for anomalies that could account for the 

potential outlier. 

Statistical outlier tests give the analyst probabilistic evidence that an extreme value does 

not "fit" with the distribution of the remainder of the data and is therefore a statistical outlier. 

These tests should only be used to identify data points that require further investigation. The 

tests alone cannot determine whether a statistical outlier should be discarded or corrected within 

a data set. This decision should be based on judgmental or scientific grounds. 
EPA QA/G-9S 116 February 2006 

There are 5 steps involved in treating extreme values or outliers: 

1. Identify extreme values that may be potential outliers; 

2. Apply statistical test; 

3. Scientifically review statistical outliers and decide on their disposition; 

4. Conduct data analyses with and without statistical outliers; and 

5. Document the entire process. 

Potential outliers may be identified through the graphical representations of Chapter 2 (step 1 

above). Graphs such as the box and whisker plot, ranked data plot, normal probability plot, and 

time plot can all be used to identify observations that are much larger or smaller than the rest of 

the data. If potential outliers are identified, the next step is to apply one of the statistical tests 

described in the following sections. Section 4.4.2 provides recommendations on selecting a 

statistical test for outliers.

If a data point is found to be an outlier, the analyst may either: 1) correct the data point; 

2) discard the data point from analysis; or 3) use the data point in all analyses. This decision 

should be based on scientific reasoning in addition to the results of the statistical test. For 

instance, data points containing transcription errors should be corrected, whereas data points 

collected while an instrument was malfunctioning may be discarded. Discarding an outlier from 

a data set should be done with extreme caution, particularly for environmental data sets, which 

often contain legitimate extreme values. If an outlier is discarded from the data set, all statistical 

analysis of the data should be applied to both the full and truncated data set so that the effect of 

discarding observations may be assessed. If scientific reasoning does not explain the outlier, it 

should not be discarded from the data set. 

If any data points are found to be statistical outliers through the use of a statistical test, 

this information will need to be documented along with the analysis of the data set, regardless of 

whether any data points are discarded. If no data points are discarded, document the 

identification of any statistical outliers by documenting the statistical test performed and the 

possible scientific reasons investigated. If any data points are discarded, document each data 

point, the statistical test performed, the scientific reason for discarding each data point, and the 

effect on the analysis of deleting the data points. This information is critical for effective peer 

review. 

4.4.2 Selection of a Statistical Test for Outliers 

There are several statistical tests for determining whether or not one or more 

observations are statistical outliers. Step by step directions for implementing some of these 

tests are described in Sections 4.4.3 through 4.4.6. Section 4.4.7 describes statistical tests for 

multivariate outliers. 

If the data are approximately normally distributed, this guidance recommends Rosner's 

test when the sample size is greater than 25 and the Extreme Value test when the sample size is 

less than 25. If only one outlier is suspected, then the Discordance test may be substituted for 

either of these tests. If the data are not normally distributed, or if the data cannot be transformed 
EPA QA/G-9S 117 February 2006 

so that the transformed data are normally distributed, then the analyst should either apply a 

nonparametric test (such as Walsh's test) or consult a statistician. A summary of these 

recommendations is contained in Table 4-3. 

Table 4-3. Recommendations for Selecting a Statistical Test for Outliers 

Sample 

Size Test Section Assumes Normality 

Multiple 

Outliers 

n ≤ 25 Extreme Value Test 4.4.3 Yes No/Yes 

n ≤ 50 Discordance Test 4.4.4 Yes No 

n ≥ 25 Rosner's Test 4.4.5 Yes Yes 

n ≥ 50 Walsh's Test 4.4.6 No Yes 

4.4.3 Extreme Value Test (Dixon's Test) 

Dixon's Extreme Value test can be used to test for statistical outliers when the sample 

size is less than or equal to 25. This test considers both extreme values that are much smaller 

than the rest of the data (case 1) and extreme values that are much larger than the rest of the data 

(case 2). This test assumes that the data without the suspected outlier are normally distributed; 

therefore, it is necessary to perform a test for normality on the data without the suspected outlier 

before applying this test. If the data are not normally distributed, then either transform the data, 

apply a different test, or consult a statistician. Directions for the Extreme Value test are 

contained in Box 4-15; an example of this test is contained in Box 4-16. 

This guidance recommends using this test when only one outlier is suspected in the data. 

If more than one outlier is suspected, the Extreme Value test may lead to masking where two or 

more outliers close in value "hide" one another. Therefore, if the analyst decides to use the 

Extreme Value test for multiple outliers, apply the test to the least extreme value first. 

4.4.4 Discordance Test 

The Discordance test can be used to test if one extreme value is an outlier. The 

Discordance test assumes that the data without the suspected outlier are approximately normally 

distributed. Therefore, it is necessary to check for normality before applying this test. 

Directions and an example of the Discordance test are contained in Box 4-17 and Box 4-18. 
EPA QA/G-9S 118 February 2006 

Box 4-15: Directions for the Extreme Value Test (Dixon's Test) 

Let X(1), X(2), . . . , X(n) represent the data ordered from smallest to largest. Check that the data without the 

suspect outlier are normally distributed, using one of the methods of Section 4.2. If normality fails, either 

transform the data or apply a different outlier test. 

STEP 1. Null Hypothesis: H0: There are no outliers in the data. 

STEP 2. Alternative Hypothesis: i) HA: X(1) is an outlier. 

 ii) HA: X(n) is an outlier. 

STEP 3. Test Statistic: i) Compute the test statistic C, where 

3 ≤ n ≤ 7 8 ≤ n ≤ 10 11 ≤ n ≤ 13 14 ≤ n ≤ 25 

( ) ()

( ) () 1

2 1

X X

X X

C

n −

− = ( ) ()

( ) () 1 1

2 1

X X

X X

C

n −

− =

−

( ) ()

( ) () 1 1

3 1

X X

X X

C

n −

− =

−

( ) ()

( ) () 2 1

3 1

X X

X X

C

n −

− =

−

 ii) Compute the test statistic C, where 

3 ≤ n ≤ 7 8 ≤ n ≤ 10 11 ≤ n ≤ 13 14 ≤ n ≤ 25 

() ( )

( ) () 1

1

X X

X X

C

n

n n

−

− = − () ( )

() () 2

1

X X

X X

C

n

n n

−

− = − () ( )

() () 2

2

X X

X X

C

n

n n

−

− = − () ( )

() () 3

2

X X

X X

C

n

n n

−

− = −

STEP 4. a) Critical Value: Use Table A-4 to find dα. 

STEP 5. a) Conclusion: If C > dα, then reject the null hypothesis that there are no outliers in the data. 

Box 4-16: An Example of the Extreme Value Test (Dixon's Test) 

The data in order of magnitude from smallest to largest are (in ppm): 

82.39, 86.62, 91.72, 98.37, 103.46, 104.93, 105.52, 108.21, 113.23, 150.55. 

As the value 150.55 is much larger than the other values, it is suspected that this data point might be an outlier. 

The Studentized Range test (Section 4.2.6) shows that there is no reason to suspect that the data are not 

normally distributed. 

STEP 1. Null Hypothesis: H0: There are no outliers in the data. 

STEP 2. Alternative Hypothesis: HA: X(n) is an outlier. 

STEP 3. Test Statistic: Since n = 10, 0 584

150 55 86 62

150 55 113 23

( ) (2)

( ) ( 1) . . .

. .

X - X

X - X C

n

n n = −

− = = − . 

STEP 4. a) Critical Value: Using Table A-4, d0.05 = 0.477. 

STEP 5. a) Conclusion: Since C = 0.584 > 0.477 = d0.05, we reject the null hypothesis of no outliers in 

 the data. 
EPA QA/G-9S 119 February 2006 

Box 4-17: Directions for the Discordance Test 

Let X(1), X(2), . . . , X(n) represent the data ordered from smallest to largest. Check that the data without the 

suspect outlier are normally distributed, using one of the methods of Section 4.2. If normality fails, either 

transform the data or apply a different outlier test. 

COMPUTATIONS: Compute the sample mean, X , and the sample standard deviation, s. 

STEP 1. Null Hypothesis: H0: There are no outliers in the data. 

STEP 2. Alternative Hypothesis: i) HA: X(1) is an outlier. 

 ii) HA: X(n) is an outlier. 

STEP 3. Test Statistic: i) ( )

s

X X

D − 1 = ii) ( )

s

X X

D n − =

STEP 4. a) Critical Value: Use Table A-5 to find dα. 

STEP 5. a) Conclusion: If D > dα, then reject the null hypothesis that there are no outliers in the data. 

Box 4-18: An Example of the Discordance Test 

The data in order of magnitude from smallest to largest are (in ppm): 

82.39, 86.62, 91.72, 98.37, 103.46, 104.93, 105.52, 108.21, 113.23, 150.55. 

It is suspected that the data point 150.55 might be an outlier. The Studentized Range test (Section 4.2.6) shows 

that there is no reason to suspect that the data are not normally distributed. 

COMPUTATIONS: X = 104.5 ppm and s = 18.922 ppm. 

STEP 1. Null Hypothesis: H0: There are no outliers in the data. 

STEP 2. Alternative Hypothesis: HA: X(n) is an outlier. 

STEP 3. Test Statistic: ( ) 2.43

18.922

150.55 104.5 = − = − = s

X X

D n

STEP 4. a) Critical Value: Using Table A-5, d0.05 = 2.176. 

STEP 5. a) Conclusion: Since D = 2.43 > 2.176 = dα, we reject the null hypothesis of no outliers. 

4.4.5 Rosner's Test 

A parametric test developed by Rosner can be used to detect up to 10 outliers for sample 

sizes of 25 or more. This test assumes that the data without the suspected outliers are normally 

distributed. Therefore, it is necessary to perform a test for normality before applying this test. If 

the data are not normally distributed, then either transform the data, apply a different test, or 

consult a statistician. Directions for Rosner's test are contained in Box 4-19 and an example is 

contained in Box 4-20. 
EPA QA/G-9S 120 February 2006 

Box 4-19: Directions for Rosner's Test for Outliers 

Let X1, X2, . . . , Xn represent the ordered data points. By inspection, identify the maximum number of possible 

outliers, 1 ≤ r0 ≤ 10. Check that the data are normally distributed without the suspected outlier(s) using one of the 

methods of Section 4.2. 

COMPUTATIONS: Using the entire data set, compute the sample mean, (0) X , and the sample standard 

deviation, (0) s . Determine the observation farthest from X( ) 0 and label it y( ) 0 . Delete y( ) 0 from the data and 

compute the sample mean, X( ) 1 , and the sample standard deviation, s( ) 1 . Now determine the observation 

farthest from X( ) 1 and label it y( ) 1 . Delete (1) y from the data and compute the sample mean, (2) X , and the 

sample standard deviation, (2) s . Continue this process until r0 extreme values have been eliminated. After this 

process the analyst should have: 

{ } { } { } (0) (0) (0) (1) (1) (1) ( 1) ( 1) ( 1) 0 0 0 , , , , , , , , , r − r − r − X s y X s y K X s y

STEP 1. Null Hypothesis: H0: There are no outliers in the data. 

STEP 2. Alternative Hypothesis: HA: There are as many as r0 outliers in the data. 

Steps 3 through 5 of this test are iterative. First, test if there are r0 outliers. If not, then test if there are r0 - 1 

outliers. Continue, until it is determined that either there are a certain number of outliers or that there are no 

outliers. 

STEP 3. Test Statistic: r

r r

R r

y X

s =

− − −

−

() ()

( )

1 1

1 , where r starts at r0 and runs through 1. 

STEP 4. a) Critical Value: Use Table A-6 to find λr. 

STEP 5. a) Conclusion: If Rr > λr, then conclude that there are r outliers. Otherwise, return to step 3 

to 

 test for r-1 outliers. 

4.4.6 Walsh's Test 

A nonparametric test was developed by Walsh to detect multiple outliers in a data set. 

This test requires a large sample size: n > 220 for a significance level of α = 0.05, and n > 60 for 

a significance level of α = 0.10. However, since the test is a nonparametric test, it may be used 

when the data is not normally distributed. It should be noted that this test is used infrequently 

with environmental data. Directions for the test for large sample sizes are given in Box 4-21. 

4.4.7 Multivariate Outliers 

Multivariate analysis, such as factor analysis and principal components analysis, involves 

the analysis of several variables simultaneously. Outliers in multivariate analysis are then values 

that are extreme in relationship to either one or more variables. As the number of variables 

increases, identifying potential outliers using graphical representations becomes more difficult. 

In addition, special procedures are required to test for multivariate outliers and details of these 

procedures are beyond the scope of this guidance. 
EPA QA/G-9S 121 February 2006 

Box 4-20: An Example of Rosner's Test for Outliers 

Consider the following 32 ordered data points (in ppm): 2.07, 40.55, 84.15, 88.41, 98.84, 100.54, 115.37, 121.19, 

122.08, 125.84, 129.47, 131.90, 149.06, 163.89, 166.77, 171.91, 178.23, 181.64, 185.47, 187.64, 193.73, 199.74, 

209.43, 213.29, 223.14, 225.12, 232.72, 233.21, 239.97, 251.12, 275.36, 395.67. 

A normal probability plot excluding the suspected outliers shows that there is no reason to suspect that the data is 

not normally distributed. In addition, this graph identified four potential outliers: 2.07, 40.55, 275.36, and 395.67. 

Rosner's test at a significance level of 0.05 will be applied to see if there are r0 = 4 or fewer outliers. 

COMPUTATIONS: The summary statistics and suspected outliers are listed in the following table: 

i X( )i s( )i y( )i

0 169.923 75.133 395.67 

1 162.640 63.872 2.07 

2 167.993 57.460 40.55 

3 172.387 53.099 275.36 

STEP 1. Null Hypothesis: H0: There are no outliers in the data. 

STEP 2. Alternative Hypothesis: HA: There are as many as 4 outliers in the data. 

STEP 3. Test Statistic: 1.939

53.099

275.36 172.387

(3)

(3) (3)

4 = − =

−

= s

y X

R

STEP 4. a) Critical Value: Using Table A-6, λ4 = 2.89. 

STEP 5. a) Conclusion: Since R4 = 1.939 < 2.89 = λ4, there aren’t 4 outliers. Therefore, test if there 

 are 3 outliers by computing 

STEP 3. Test Statistic: 2.218

57.46

40.55 167.993

(2)

(2) (2)

3 = − =

−

= s

y X

R

STEP 4. a) Critical Value: Using Table A-6, λ3 = 2.91. 

STEP 5. a) Conclusion: Since R3 = 2.218 < 2.91 = λ3, there aren’t 3 outliers. Therefore, test if there 

 are 2 outliers by computing 

STEP 3. Test Statistic: 2.514

63.872

2.07 162.64

(1)

(1) (1)

2 = − =

−

= s

y X

R

STEP 4. a) Critical Value: Using Table A-6, λ2 = 2.92. 

STEP 5. a) Conclusion: Since R2 = 2.514 < 2.92 = λ2, there aren’t 2 outliers. Therefore, test if there 

 are 1 outlier by computing 

STEP 3. Test Statistic: 3.005

75.133

395.67 169.923

(0)

(0) (0)

1 = − =

−

= s

y X

R

STEP 4. a) Critical Value: Using Table A-6, λ1 = 2.94. 

STEP 5. a) Conclusion: Since R1 = 3.005 > 2.94 = λ1, we conclude there is one outlier, 395.67. 
EPA QA/G-9S 122 February 2006 

Box 4-21: Directions for Walsh's Test for Large Sample Sizes 

Let X(1), X(2), . . ., X(n) represent the ordered data. If n ≤ 60, do not apply this test. If 60 < n ≤ 220, then α = 0.10. 

 If n > 220, then α = 0.05. 

COMPUTATIONS: Identify the number of possible outliers, r. Compute 

c = ceiling( 2n ), k = r + c , α b2 = 1 , and ( ) ( )

1

1 1

2

2

− −

+ − − = c b

b c b c

a

where ceiling( ) indicates rounding the value to the next largest integer. 

STEP 1. The r smallest points are outliers (with an α level of significance) if 

X(r ) − (1+ a)X(r +1) + aX(k ) < 0

STEP 2. The r largest points are outliers (with an α level of significance) if 

X(n+1−r ) − (1+ a)X(n−r ) + aX(n+1−k ) > 0

STEP 3. If both of the inequalities are true, then both small and large outliers are indicated.

4.5 TESTS FOR DISPERSIONS 

Many statistical tests make assumptions about the dispersion (variance) of data. This 

section considers some of the most commonly used statistical tests for comparing variances. 

Section 4.5.1 constructs a confidence interval for a population variance. Section 4.5.2 provides a 

test for comparing two population variances. Section 4.5.3 (Bartlett’s test) and Section 4.5.4 

(Levene’s test) describe tests that compare two or more population variances. The analyst 

should be aware that many statistical tests only require the approximate equality of variances and 

that many of the tests remain valid unless there is gross inequality in the population variances. 

4.5.1 Confidence Intervals for a Single Variance 

This section discusses confidence intervals for a single variance or standard deviation. 

The method described in Box 4-22 can be used to find a two-sided 100(1-α)% confidence 

interval. The upper end point of a two-sided 100(1-α)% confidence interval is also a 

100(1-α/2)% upper confidence limit, and the lower end point is also a 100(1-α/2)% lower 

confidence limit. Since the standard deviation is the square root of the variance, a confidence 

interval for the variance can be converted to a confidence interval for the standard deviation by 

taking the square roots of the endpoints of the interval. The confidence interval procedure 

assumes the data are a random sample from a normally distributed population and can be highly 

sensitive to outliers or to departures from normality. 
EPA QA/G-9S 123 February 2006 

Box 4-22: Directions for Constructing Confidence Intervals and 

Confidence Limits for the Sample Variance and Standard Deviation with an Example 

Let X1, X2, . . . , Xn represent the n data points. 

COMPUTATIONS: Calculate the sample variance s2

 (Section 2.2.3). 

A 100(1-α)% confidence interval for the population variance, σ2

, is ( ) ( )

⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧ − −

− − −

2

1, / 2

2

2

1, 1 / 2

2 1 , 1

χ n α χ n α

n s n s , where Table A-9 is 

used to find 2 χ n−1,α / 2 and 2 χ n−1,1−α / 2 . A 100(1-α)% confidence interval for the population standard deviation is 

the square root of the interval above. 

Example: Ten samples were analyzed for lead (in ppb): 46.4, 46.1, 45.8, 47, 46.1, 45.9, 45.8, 46.9, 45.2, 46. 

COMPUTATIONS: The sample variance is s2

 = 0.286. 

A 95% confidence interval for the population variance, σ2

, is 

( ) ( ) { } 0.135 , 0.953

2.70

9 0.286 , 19.02

10 1 0.286 9 0.286 , 10 1 0.286

2

9, 0.025

2

9, 0.975

=

⎭

⎬

⎫

⎩

⎨

⎧ ⋅ ⋅ =

⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧ − ⋅ − ⋅

χ χ

4.5.2 The F-Test for the Equality of Two Variances 

An F-test may be used to compare two population variances. The assumptions 

underlying the F-test are that the two samples are independent random samples from two normal 

populations. The F-test for equality of variances is highly sensitive to departures from 

normality. Directions for implementing an F-test with an example are given in Box 4-23. 

4.5.3 Bartlett's Test for the Equality of Two or More Variances 

Bartlett's test is a means of testing whether two or more population variances of normal 

distributions are equal. In the case of only two variances, Bartlett's test is equivalent to the Ftest. 

Often in practice unequal variances and non-normality occur together and Bartlett's test is itself 

sensitive to departures from normality. With long-tailed distributions, the test too often rejects 

equality of the variances. Rejecting the equality of variances does not mean that one or more is 

significantly different from the others. It simply implies the variances are unequal as a group. 

Directions for Bartlett's test are given in Box 4-24 and an example is given in Box 4-25. 

4.5.4 Levene's Test for the Equality of Two or More Variances 

Levene's test provides an alternative to Bartlett's test for comparing population variances. 

 Levene's test is less sensitive to departures from normality than Bartlett's test and has greater 

power than Bartlett's for non-normal data. In addition, Levene's test has power nearly as great as 

Bartlett's test for normally distributed data. Directions and an example of Levene's test are 

contained in Box 4-26 and Box 4-27, respectively. 
EPA QA/G-9S 124 February 2006 

Box 4-23: Directions for an F-Test to Compare Two Population Variances with an Example 

Let X1, X2, . . . , Xm represent the m data points from population 1 and Y1, Y2, . . . , Yn represent the n data 

points from population 2. 

COMPUTATIONS: Calculate the sample variances sX

2 and sY2 (Section 2.2.3). 

STEP 1. Null Hypothesis: H0: 2 2 σ X = σY . 

STEP 2. Alternative Hypothesis: HA: 2 2 σ X ≠ σY . 

STEP 3. Test Statistic: ( ) ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ = = 2

2

2

2

0 max , max ,

X

Y

Y

X

X Y

s

s

s

s F F F . If F0 = FX, then let k = m – 1 and 

 q = n – 1. If F0 = FY , then let k = n – 1 and q = m – 1. 

STEP 4. a) Critical Value: Use Table A-10 to find Fk, q, 1-α/2. 

STEP 4. b) p-value: Use Table A-10 to find 2⋅P(Fk, q > F0). 

STEP 5. a) Conclusion: If F0 > Fk, q, 1-α/2, then reject the null hypothesis of equal population variances. 

STEP 5. a) Conclusion: If p-value < α, then reject the null hypothesis of equal population variances. 

Example: Manganese concentrations (in ppm) were collected from 2 wells. A 0.05-level F-test will be used to 

test if the population variances are equal. 

well X: 50, 73, 244, 202 

well Y: 272, 171, 32, 250, 53 

COMPUTATIONS: The sample variances are sX

2 = 9076 and sY2 = 12125. 

STEP 1. Null Hypothesis: H0: 2 2 σ X = σY . 

STEP 2. Alternative Hypothesis: HA: 2 2 σ X ≠ σY . 

STEP 3. Test Statistic: ( ) 1.336

9076

12125 , 12125

9076

max , max , max 2

2

2

2

0 ⎟ =

⎠

⎞ ⎜

⎝

⎛ =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ = =

X

Y

Y

X

X Y

s

s

s

s F F F . 

 Also, k = 5 – 1 = 4 and q = 4 – 1 = 3. 

STEP 4. a) Critical Value: Using Table A-10, F4, 3, 0.975 = 15.1. 

STEP 4. b) p-value: Using Table A-10, p-value = 2⋅P(F4, 3 > 1.336) which exceeds 0.20. 

 (Using software, the exact p-value is 0.8454) 

STEP 5. a) Conclusion: Since F0 = 1.336 < 15.1 = F4, 3, 0.975, we fail to reject the null 

 hypothesis of equal population variances. 

STEP 5. a) Conclusion: Since p-value = 0.8454 > 0.05 = α, we fail to reject the null hypothesis 

 of equal population variances. 
EPA QA/G-9S 125 February 2006 

Box 4-24: Directions for Bartlett's Test 

Consider k independent random samples with a sample size of ni for the i

th group and let N = n1 +L+ nk . 

COMPUTATIONS: For each of the k groups, calculate the sample variance, 2 si (Section 2.2.3). Also, compute 

the pooled variance: ∑ ( ) = − − = k

i i p ni s N k

s

1

2 2 1 1 . 

STEP 1. Null Hypothesis: H0: 2 2 σ 1 = L = σ k (The variances are equal) 

STEP 2. Alternative Hypothesis: HA: The variances are not equal. 

STEP 3. Test Statistic: B ( ) N k ( ) s ( ) n ( ) si i

k

i

p

2

1

2

0 = − ⋅ ln −∑ − 1 ⋅ ln

=

. 

STEP 4. a) Critical Value: Use Table A-9 to find 2 χ k −1, 1−α . 

STEP 4. b) p-value: Use Table A-9 to find ( ) 0

2 P χ k −1 > B . 

STEP 5. a) Conclusion: If B0 > 2 χ k −1, 1−α , then reject the null hypothesis of equal variances. 

STEP 5. a) Conclusion: If p-value < α, then reject the null hypothesis of equal variances. 

Box 4-25: An Example of Bartlett's Test 

Manganese concentrations were collected from 6 wells over a 4 month period. It is important to determine if the 

variances of the six wells are equal. Bartlett's test at a significance level of 0.05 will be used. 

Sampling Date Well 1 Well 2 Well 3 Well 4 Well 5 Well 6 

January 1 50 272 

February 1 73 171 68 

March 1 244 46 32 34 48 991 

April 1 202 77 53 3940 54 54 

ni (N=17) 4 2 4 2 2 3 

Xi 142.25 61.50 132 1987 51 371 

si

2 9076.25 480.5 12454 7628418 18 288349 

COMPUTATIONS: The pooled variance is [ ] () () 4 1 9076.25 3 1 288349 751836.84

17 6

2 1 − ⋅ + + − ⋅ = − s p = L

STEP 1. Null Hypothesis: H0: 2 2 σ 1 = L = σ k . 

STEP 2. Alternative Hypothesis: HA: The variances are not equal. 

STEP 3. Test Statistic: ( ) 17 6 ln(751836.84) [(4 1) ln(9076.25) ( )( ) 3 1 ln 288349 ] 0 B = − ⋅ − − ⋅ +L+ − ⋅

 15 = 43.

STEP 4. a) Critical Value: Using Table A-9, 07 11. 2 χ 5, 0.95 = . 

STEP 4. b) p-value: Using Table A-9, p-value = ( 43.15 ) 2 P χ 5 > < 0.005. 

 (Using statistical software, the exact p-value is almost 0). 

STEP 5. a) Conclusion: Since B0 = 43.15 > 11.07 = 2 χ k −1, 1−α , we reject the null hypothesis of equal 

 population variances. 

STEP 5. a) Conclusion: Since p-value = almost 0 < 0.05 = α, we reject the null hypothesis of equal 

 population variances. 
EPA QA/G-9S 126 February 2006 

Box 4-26: Directions for Levene's Test 

Consider k independent random samples with a sample size of ni for the i

th group and let N = n1 +L+ nk . 

COMPUTATIONS: For each of the k groups, calculate the group mean, Xi . Then compute the absolute 

residuals, z X ij ij = − Xi , where Xij represents the j

th value of the ith group. For each group, calculate the mean 

absolute residual, ∑ = = ni

j ij

i

i z

n

z

1

1 . Next, calculate the overall mean absolute residual, 

i i

k

i

ij

n

j

k

i

n z

N z N

1

z

i

∑ ∑ ∑ = = =

= =

1 1 1

1 . 

Finally, compute the following sums of squares for the absolute residuals: 

2

1 1

2 SS z N z

k

i

n

j

TOTAL ij

i

=∑∑ − ⋅

= =

, 2

1

2 SS n z N z

k

i

GROUPS i i =∑ − ⋅

=

, and SSERROR = SSTOTAL − SSGROUPS . 

STEP 1. Null Hypothesis: H0: 2 2 σ 1 = L = σ k . 

STEP 2. Alternative Hypothesis: HA: The variances are not equal. 

STEP 3. Test Statistic: ( )

SS / ( ) N k

SS / k F

ERROR

GROUPS

−

− = 1

0 . 

STEP 4. a) Critical Value: Use Table A-10 to find Fk-1, N-k, 1-α. 

STEP 4. b) p-value: Use Table A-10 to find P(Fk-1, N-k > F0). 

STEP 5. a) Conclusion: If F0 > Fk-1, N-k, 1-α, then reject the null hypothesis of equal population 

 variances. 

STEP 5. a) Conclusion: If p-value < α, then reject the null hypothesis of equal population variances. 

4.6 TRANSFORMATIONS 

Most statistical tests and procedures contain assumptions about the data. For example, 

some common assumptions are that the data is normally distributed; variance components of a 

statistical model are additive; two independent data sets have equal variance; and a data set has 

no trends over time or space. If the data do not satisfy such assumptions, then the results of a 

statistical procedure or test may be biased or incorrect. Fortunately, data that do not satisfy 

statistical assumptions may often be converted or transformed mathematically into a form that 

allows standard statistical tests to perform adequately. 

It is not recommended to transform data for estimation purposes. Transforming, 

estimating, and then transforming the estimate back to the original domain will, in general, lead 

to biased estimates. A better approach to estimation here is to use a nonparametric procedure. 
EPA QA/G-9S 127 February 2006 

Box 4-27: An Example of Levene's Test 

Four months of data on arsenic concentration (ppm) were collected from six wells at a Superfund site. This data 

set is shown in the table below. Before analyzing this data, it is important to determine if the variances of the six 

wells are equal. Levene's test at a significance level of 0.05 will be used to make this determination. 

COMPUTATIONS: The table below contains the data values, the absolute residuals (labeled res), the sample 

means and the absolute residual means. 

well 1 well 2 well 3 well 4 well 5 well 6 

Month 

value res value res value res value res value res value res 

1 22.90 6.43 2.00 13.76 2.0 27.6 7.84 3.42 24.90 11.41 0.34 1.95 

2 3.09 13.38 1.25 14.51 109.4 79.8 9.30 1.96 1.30 12.19 4.78 2.49 

3 35.70 19.23 7.80 7.96 4.5 25.1 25.90 14.64 0.75 12.74 2.85 0.56 

4 4.18 12.29 52.00 36.24 2.5 27.1 2.00 9.26 27.00 13.51 1.20 1.09 

X1 = 16.47 X2 = 15.76 X3 = 29.60 X4 = 11.26 X5 = 13.49 X6 = 2.29

z1 = 12.83 z2 = 18.12 z3 = 39.90 z4 = 7.32 z5 = 12.46 z6 = 1.52

The overall absolute residual mean is z = (12.83 + 18.12 + 39.9 + 7.32 + 12.46 + 1.52)/6 = 15.36. 

The sum of squares are: SSTOTAL = 630089 . , SSWELLS = 352290. , and SSERROR = 277799. . 

STEP 1. Null Hypothesis: H0: 2

6

2 σ1 = L = σ . 

STEP 2. Alternative Hypothesis: HA: The variances are not equal. 

STEP 3. Test Statistic: ( )

( )

( )

( ) 4.56

2777.99 24 6

1 3522.9 6 1

0 = −

− = −

− = SS / N k

SS / k F

ERROR

GROUPS . 

STEP 4. a) Critical Value: Using Table A-10, F5, 18, 0.95 = 2.77. 

STEP 4. b) p-value: Using Table A-10, 0.01 < p-value < 0.001. 

 Using software, P(F5, 18 > 4.56) = 0.0073. 

STEP 5. a) Conclusion: Since F0 = 4.56 > 2.77 = F5, 18, 0.95, we reject the null hypothesis of equal 

 population variances. 

STEP 5. a) Conclusion: Since p-value = 0.0073 < 0.05 = α, we reject the null hypothesis of equal 

 population variances. 

4.6.1 Types of Data Transformations 

Any mathematical function that is applied to every point in a data set is called a 

transformation. Some commonly used transformations include: 

Logarithmic (Log X or Ln X): This transformation is best suited for data that is rightskewed which may occur when the measurement follows a lognormal distribution. The 

transformation is also helpful when the variance at each level of the data is proportional to the 

square of the mean of the data points at that level. For example, if the variance of data collected 

around 50 ppm is approximately 250, but the variance of data collected around 100 ppm is 

approximately 1000, then a logarithmic transformation may be useful. This situation is often 
EPA QA/G-9S 128 February 2006 

characterized by having a constant coefficient of variation (standard deviation divided by the 

mean) over all possible data values. 

If some of the original values are zero or negative, it is customary to add a small quantity 

to make the data value non-zero since the logarithm of zero or a negative number does not exist. 

The size of the small quantity depends on the magnitude of the non-zero data. As a working 

point, a value of one tenth the smallest non-zero value could be selected. It does not matter 

whether a natural (ln) or base 10 (log) transformation is used because the two transformations are 

related by the expression ln(X) = 2.303 log(X). Directions for applying a logarithmic 

transformation with an example are given in Box 4-28. 

Square Root ( X ): This transformation may be used when dealing with small whole 

numbers, such as bacteriological counts, or the occurrence of rare events, such as violations of a 

standard over the course of a year. The underlying assumption is that the original data follow a 

Poisson-like distribution in which case the mean and variance of the data are equal. It should be 

noted that the square root transformation overcorrects when very small values and zeros appear 

in the original data. In these cases, X +1 is often used as a transformation. 

Inverse Sine ( Arcsine X): This transformation may be used for binomial proportions 

based on count data to achieve stability in variance. The resulting transformed data are 

expressed in radians (angular degrees). Special tables must be used to transform the proportions 

into degrees. 

Box-Cox Transformations: This transformation is a complex power transformation that 

takes the original data and raises each data observation to the power lambda (λ). A logarithmic 

transformation is a special case of the Box-Cox transformation. The rationale is to find λ such 

that the transformed data have the best possible additive model for the variance structure, the 

errors are normally distributed, and the variance is as constant as possible over all possible 

concentration values. The Maximum Likelihood technique is used to find λ such that the 

residual error from fitting the theorized model is minimized. In practice, the exact value of λ is 

often rounded to a convenient value for ease in interpretation (for example, λ = 0.45 would be 

rounded to 0.5 as it would then have the interpretation of a square root transform). One of the 

drawbacks of the Box-Cox transformation is the difficulty in physically interpreting the 

transformed data. 

4.6.2 Reasons for Transforming Data 

By transforming the data, assumptions that are not satisfied in the original data can be 

satisfied by the transformed data. For instance, a right-skewed distribution can be transformed to 

be approximately Gaussian (normal) by using a logarithmic or square-root transformation. Then 

the normal-theory procedures can be applied to the transformed data. If data are lognormally 

distributed, then apply procedures to logarithms of the data. However, selecting the correct 

transformation may be difficult. If standard transformations do not apply, it is suggested that the 

data user consult a statistician. 
EPA QA/G-9S 129 February 2006 

Box 4-28: Directions for Transforming Data and an Example 

Let X1, X2, . . . , Xn represent the n data points. To apply a transformation, simply apply the transforming 

function to each data point. When a transformation is implemented to make the data satisfy some 

statistical assumption, it will need to be verified that the transformed data satisfy this assumption. 

Example: Transforming Lognormal Data

A logarithmic transformation is particularly useful for pollution data. Pollution data are often right-skewed, 

thus the log-transformed data will tend to be symmetric. Consider the data set shown below with 15 data 

points. A histogram of this data below shows that the data are possibly lognormally distributed. The 

transformed data are shown in column 2. A histogram of the transformed data below shows that the 

transformed data appear to be normally distributed. 

 Observed Transformed Observed Transformed

 X → ln(X) X → ln(X) 

 0.22 → -1.51 0.47 → -0.76 

 3.48 → 1.25 0.67 → -0.40 

 6.67 → 1.90 0.75 → -0.29 

 2.53 → 0.93 0.60 → -0.51 

 1.11 → 0.10 0.99 → -0.01 

 0.33 → -1.11 0.90 → -0.11 

 1.64 → 0.50 0.26 → -1.35 

 1.37 → 0.31 

-1 0 1 2 3 4 5 6 7

0

1

2

3

4

5

6

7

Observed Values

Number of Observations

-4 -3 -2 -1 0 1 2 3 4

0

1

2

3

4

5

6

7

Transformed Values

Number of Observations

While transformations are useful for dealing with data that do not satisfy statistical 

assumptions, they can also be used for various other purposes. For example, transformations are 

useful for consolidating data that may be spread out or that have several extreme values. In 

addition, transformations can be used to derive a linear relationship between two variables on the 

newly transformed data, so that linear regression analysis can be applied. They can also be used 

to efficiently estimate quantities such as the mean and variance of a lognormal distribution. 
EPA QA/G-9S 130 February 2006 

Transformations may also make the analysis of data easier by changing the scale into one that is 

more familiar or easier to work with. 

Once the data have been transformed, all statistical analysis should be performed on the 

transformed data. Rarely should an attempt made to transform the data back to the original form 

because this can lead to biased estimates. For example, estimating quantities such as means, 

variances, confidence limits, and regression coefficients in the transformed scale typically leads 

to biased estimates when transformed back into original scale. However, it may be difficult to 

understand or apply results of statistical analysis expressed in the transformed scale. Therefore, 

if the transformed data do not give noticeable benefits to the analysis, it is better to use the 

original data. There is no point in working with transformed data unless it adds value to the 

analysis. 

4.7 VALUES BELOW DETECTION LIMITS 

Data generated from chemical analysis may fall below the detection limit (DL) of the 

analytical procedure. These measurement data are generally described as non-detects rather 

than as zero or not present and the appropriate limit of detection is usually reported. In cases 

where measurement data are described as non-detects, the concentration of the chemical is 

unknown although it lies somewhere between zero and the detection limit. Data that includes 

both detected and non-detected results are called censored data in the statistical literature. 

There are a variety of ways to evaluate data that includes values below the detection 

limit. However, there are no general procedures that are applicable in all cases. Some general 

guidelines are presented in Table 4-4. Although these guidelines are usually adequate, they 

should be implemented cautiously. 

Table 4-4. Guidelines for Analyzing Data with Non-Detects 

Approximate Percentage 

of Non-Detects Section Statistical Analysis Method

< 15% 4.7.1 Replace non-detects with 0, DL/2, DL., 

Cohen’s Method. 

15% - 50% 4.7.2 Trimmed mean, Cohen's Method, 

Winsorized mean and standard deviation. 

> 50% - 90% 4.7.3 Tests for proportions (Section 3.2.1.5) 

All of the suggested procedures for analyzing data with non-detects depend upon the 

amount of data below the detection limit. For relatively small amounts below detection limit 

values, replacing the non-detects with a small number and proceeding with the usual analysis 

may be satisfactory depending on the purpose of the analysis. For moderate amounts of data 

below the detection limit, a more detailed adjustment is appropriate. In situations where 

relatively large amounts of data fall below the detection limit, one may need only to consider 

whether or not the chemical was detected above some level. Table 4-4 provides percentages to 
EPA QA/G-9S 131 February 2006 

assist the user in evaluating their particular situation. However, it should be recognized that 

these percentages are not hard and fast rules. 

In addition, sample size influences which procedures should be used to evaluate the data. 

 For example, the case where 1 sample out of 4 is not detected should be treated differently from 

the case where 25 samples out of 100 are non-detects. Therefore, this guidance suggests that the 

data analyst consult a statistician for the most appropriate way to evaluate data containing values 

below the detection level. 

4.7.1 Approximately less than 15% Non-detects - Substitution Methods 

If a small proportion of the observations are non-detects, then these may be replaced with 

a small number, usually DL/2, and the usual analysis performed. Alternative substitution values 

are 0 (see Aitchison’s Method below) or the detection limit. It should be noted that Cohen’s 

Method (section 4.7.2.1) will also work with small amounts of non-detects. 

4.7.1.1 Aitchison’s Method 

Later adjustments to the mean and variance assume that the data values really were 

present but could not be recorded since they were below the detection limit. However, there are 

cases where the data values are below the detection limit because they are actually zero, i.e., the 

contaminant or chemical of concern being entirely absent. Such data sets typically contain a 

mixture of zero values and present, but nondetected values. Aitchison’s Method is simply 

adjustment formulas for the mean and variance if 0 values are substituted for non-detects. 

Directions for Aitchison’s method are contained in Box 4-29 with an example in Box 4-30. 

Box 4-29: Directions for Aitchison’s Method to Adjust Means and Variances 

Let X1, X2, . ., Xm, Xm+1,. . . , Xn represent the data points where the first m values are above the detection limit 

and the remaining n-m data points are below the detection limit. 

COMPUTATIONS: Using the data above the detection level, compute the sample mean and sample variance: 

∑ = = m

i

Xd Xi 1 and 

⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧

⎟

⎠

⎞ ⎜

⎝

⎛ − − = ∑ = ∑ =

2

1 1

2 2 1

1

1 m

i i

m

i d i X

m

X

m

s . 

Compute the adjusted sample mean and sample variance, Xd n

m X = and ( )

( )

2 2 2

1 1

1

d Xd n n

m n m

s

n

m

s −

− + −− = . 
EPA QA/G-9S 132 February 2006 

Box 4-30: An Example of Aitchison’s Method 

The following data consist of 10 Methylene Chloride samples: 1.9,1.3, <1, 2.0, 1.9, <1, <1, <1, 1.6, 1.7. There are 

6 values above the detection limit and 4 below, so m = 6 and n-m = 4. Aitchison’s method will be used to estimate 

the mean and sample variance of this data. 

COMPUTATIONS: Compute the mean and variance for the 6 values above the detection limit 

Xd = 1.733 and 0.0667 2 sd = . 

The adjusted sample mean and sample variance are: 1.733 1.04

10

6 X = ⋅ = and 

( ) 1.7333 0.8382

10 10 1

6 4 0.0667

10 1

2 6 1 2 ⋅ = ⋅ −

⋅ ⋅ + −− s = . 

4.7.2 Between Approximately 15% - 50% Non-detects 

4.7.2.1 Cohen's Method 

Cohen's method provides adjusted estimates of the sample mean and standard deviation 

that accounts for data below the detection level. The adjusted estimates are based on the 

statistical technique of maximum likelihood estimation of the mean and variance so that the nondetects are accounted for. Care has to be taken when using the adjusted mean and variance in 

statistical tests. If the percentage of data below the detection limit is relatively small (e.g., less 

than 20%), the significance level and power of the statistical test are approximately correct. As 

the proportion of data below detection increases, power declines and the true significance level 

increases dramatically. This is mainly attributable to the lack of independence between the 

adjusted mean and adjusted variance. If more than 50% of the observations are not detected, 

Cohen's method should not be used. In addition, this method requires that the data without the 

non-detects be normally distributed and that the detection limit is always the same. Directions 

for Cohen's method are contained in Box 4-31 with an example in Box 4-32. 

4.7.2.2 Selecting Between Aitchison’s Method and Cohen’s Method 

Cohen’s underlying model is that the population contains a normal distribution, but we 

cannot see the values below the censoring point. Aitchison’s underlying model is that the 

population consists of a proportion following a normal distribution together with a proportion of 

values at zero. The difference in concepts becomes relevant depending on the types of 

inferences made. For example, in estimating upper quantiles, the analyst may use only the 

normal portion for the statistics, adjusting the quantile to account for the estimated proportion at 

zero. If a confidence interval for the mean was required a simple substitution of zero for all data 

below detection would suffice. To determine if a data set is better adjusted by Cohen’s method 

or Aitchison’s method, a simple graphical procedure using a Normal Probability Plot 

(Section 2.3.5) can be used. Directions for this procedure are given in Box 4-34 with an example 

in Box 4-35. 
EPA QA/G-9S 133 February 2006 

Box 4-31: Directions for Cohen's Method 

Let X1, X2, . ., Xm, Xm+1,. . . , Xn represent the data points where the first m values are above the detection limit 

(DL) and the remaining n-m data points are below the detection limit. 

COMPUTATIONS: Using the data above the detection level, compute the sample mean and sample variance: 

∑=

=

m

i

d Xi

m

X

1

1

 and 

⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧ ⎟

⎠

⎞ ⎜

⎝

⎛ − − = ∑ ∑ = =

2

1 1

2 2 1

1

1 m

i

i

m

i

d i X

m

X

m

s . 

Compute h n m

n = − and ( )2 X DL

s

d

2

d

−

γ = . Use h, γ, and Table A-11 to determine $λ . If the exact values of h and 

γ do not appear in the table, use double linear interpolation (Box 4-33) to estimate $λ . 

Estimate the corrected sample mean, X , and sample variance, s2

: 

X X (X DL) = d − λ d − ˆ

( )

2 2 2

s = sd + λ

ˆ X d − DL . 

Box 4-32: An Example of Cohen's Method 

Sulfate concentrations (mg/L) were measured for 24 data points with 3 values falling below the detection limit of 

1450 mg/L. The 24 values are: 

1850, 1760, <1450, 1710, 1575, 1475, 1780, 1790, 1780, <1450, 1790, 1800, 

<1450, 1800, 1840, 1820, 1860, 1780, 1760, 1800, 1900, 1770, 1790, 1780. 

Cohen's Method will be used to adjust the sample mean and sample variance for use in a t-test to determine if the 

mean is greater than 1600 mg/L. 

COMPUTATIONS: The sample mean and sample variance of the m = 21 values above the detection level are 

Xd = 1771.9 and 8593.69 2 sd = . 

The values of h and γ are: 0.125

24

24 21 = − h = and ( ) 0.083

1771.9 1450

8593.69

2 = − γ = . Table A-11 does not contain 

the exact entries for h and γ, double linear interpolation was used to estimate 0.149839 ˆ

λ = (see Box 4-33). 

The adjusted sample mean and sample variance are: 

X = X d − λ

ˆ( ) X d − DL = 1771.9 − 0.149839 ⋅(1771.9 −1450) = 1723.67 = X

s sd ( ) X d DL . . ( ) . s

2 2 2 2 2 8593 69 0149839 1771.9 1450 24119 95 ˆ = + λ − = + ⋅ − = =
EPA QA/G-9S 134 February 2006 

Box 4-33: Example of Double Linear Interpolation 

The details of the double linear interpolation are provided to assist in the use of Table A-11. The desired 

value for $

λ corresponds to γ = 0.083 and h = 0.125 from Box 4-32. The values from Table A-11 used for 

interpolation are: 

h 

γ

c1 = 0.10 c2 = 0.15 

r1 = 0.05 x11 = 0.11431 x12 = 0.17925 

r2 = 0.10 x21 = 0.11804 x22 = 0.18479 

We first interpolate between columns: 

( ) ( ) 0.17925 0.11431 0.14678

0.15 0.10

0.125 0.10

12 11 0.11431

2 1

1

1 11 ⋅ − = −

− ⋅ − = + −

− = + x x

c c

c c

y x

( ) ( ) 0.18479 0.11804 0.151415

0.15 0.10

0.125 0.10

22 21 0.11804

2 1

1

2 21 ⋅ − = −

− ⋅ − = + −

− = + x x

c c

c c

y x

Now we interpolate between the rows: 

( ) ( ) 0.151415 0.14678 0.149839

0.10 0.05

0.083 0.05 0.14678 ˆ 2 1

2 1

1

1 ⋅ − = −

− ⋅ − = + −

− = + y y

r r

r r λ y

Box 4-34: Directions for Selecting Between Cohen’s Method or Aitchison’s Method 

Let X1, X2, . .,Xm, . . . , Xn represent the data points with the first m values are above the detection limit (DL) and 

the remaining n-m data points are below the DL. 

STEP 1: Use Box 2-17 to construct a Normal Probability Plot of all the data but only plot the values belonging 

 to those above the detection level. This is called the Censored Plot. 

STEP 2: Use Box 2-17 to construct a Normal Probability Plot of only those values above the detection level. 

This called the Detects only Plot. 

STEP 3: If the Censored Plot is more linear than the Detects Only Plot, use Cohen’s Method to estimate the 

sample mean and variance. If the Detects Only Plot is more linear than the Censored Plot, then use 

Aitchison’s Method to estimate the sample mean and variance. 

4.7.3 Greater than Approximately 50% Non-detects - Test of Proportions 

If more than 50% of the data are below the detection limit but at least 10% of the 

observations are quantified, then the best option is a test of proportions. Thus, if the parameter 

of interest is a mean, consider switching the parameter of interest to some percentile greater than 

the percent of data below the detection limit. For example, if 67% of the data are below the DL, 

consider switching the parameter of interest to the 75th percentile. Then the method described in 

3.2.1.5 can be applied to test a hypothesis concerning the 75th percentile. It is important to note 

that the tests of proportions may not be applicable for composite samples. In this case, the data 

analyst should consult a statistician before proceeding with analysis. 
EPA QA/G-9S 135 February 2006 

Box 4-35: Example of Determining Between Cohen’s Method and Aitchison’s Method 

Readings of Chlorobenzene were obtained from a monitoring well: 

<1, <1, <1, 1.2, 1.25, 1.3, 1.45, 1.35, 1.55, 1.6, 1.85, 2.1 

Step 1: Using the directions in Box 2-17 the following is the Censored Plot: 

STEP 2: Using the directions in Box 2-17 the following is the Detects only Plot: 

STEP 3: Since the Censored Plots is more linear than the Detects Only Plot, Cohen’s Method should be used 

to estimate the sample mean and variance.

4.7.4 Greater than Approximately 90% Non-detects 

If very few quantified values are found, a method based on the Poisson distribution may 

be used as an alternative approach. However, with a large proportion of non-detects in the data, 

the data analyst should consult with a statistician before proceeding with analysis. 

4.7.5 Recommendations 

If the number of sample observations is small (n < 20), Cohen’s and other maximum 

likelihood methods can produce biased results since it is difficult to assure that the underlying 

distribution is appropriate and the solutions to the likelihood equation are statistically consistent 

only if the number of samples is large. Additionally, most methods will yield estimated 

parameters with large estimation variance, which reduces the power to detect import differences 

from standards or between populations. While these methods can be applied to small data sets, 

the user should be cautioned that they will only be effective in detecting large departures from 

the null hypothesis. 

If the degree of censoring is relatively low, reasonably good estimates of means, 

variances and upper percentiles can be obtained. However, if the rate of censoring is very high 

(greater than 50%) then little can be done statistically except to focus on some upper quantile of 
EPA QA/G-9S 136 February 2006 

the contaminant distribution, or on some proportion of measurements above a certain critical 

level that is at or above the censoring limit. 

When the numerical standard is at or below one of the censoring levels and a one-sample 

test is used, the most useful statistical method is to test whether the proportion of a population 

that is above (below) the standard is too large, or to test whether and upper quantile of the 

population distribution is above the numerical standard. Table 4-5 gives some recommendation 

on which statistical parameter to use when censoring is present in data sets for different sizes of 

the coefficient of variation. 

Table 4-5. Guidelines for Recommended Parameters for Different 

Coefficient of Variations and Censoring 

Approximate Proportion of Data Below the Detection Limit

Approximate Coefficient 

of Variation (CV) Low 

(<50%) 

High 

(>50%) 

Large: CV > 1.5 Mean or 

Upper Percentile Upper Percentile 

Medium: 0.5 < CV < 1.5 Mean or 

Upper Percentile Upper Percentile 

Small: CV < 0.5 Mean or 

Median Median 

When comparing two data sets with different censoring levels (i.e., different detection 

limits), it is recommended that all data be censored at the highest censoring value present and a 

nonparametric test such as the Wilcoxon Rank Sum Test (Section 3.3.2.1.1) used to compare the 

two data sets. There is a corresponding loss of statistical power but to a certain extent this can be 

minimized through the use of large sample sizes. 

4.8 INDEPENDENCE 

When data are truly independent, the correlation between data points is by definition zero 

and the selected statistical tests attains the desired decision error rates (given the appropriate 

assumptions have been satisfied). When correlation exists, the effectiveness of statistical tests is 

diminished. Environmental data are particularly susceptible to correlation problems due to the 

fact that such environmental data are collected under a spatial pattern or sequentially over time. 

If observations are positively correlated over time or space, then the effective sample size 

for a test tends to be smaller than the actual sample size—i.e., each additional observation does 

not provide as much ‘new’ information because its value is partially determined by the value of 

adjacent observations. This smaller effective sample size means that the degrees of freedom for 

the test statistic is smaller, or equivalently, the test is not as powerful as originally thought. In 

addition to affecting the false acceptance error rate, applying the usual tests to correlated data 

tends to result in a test whose actual significance level is larger than the nominal error rate. 
EPA QA/G-9S 137 February 2006 

When observations are correlated, the estimate of the variance in the test statistic formula 

is often understated. For example, consider the mean of a series of n temporally-ordered 

observations. If these observations are independent, then the variance of the mean is σ

2

/n, where 

σ

2

 is the variance of an individual observation. However, if the observations are not independent 

and the correlation between successive observations is ρ, then the variance of the mean is 

var( ) X = (1+ q) 2 σ where ∑( )

−

=

= −

1

1

2 n

k

k

n k

n

q ρ , 

which will tend to be larger than σ

2

/n if the correlations are positive. If one conducts a t-test at a 

certain significance level using the usual formula for the estimated variance, then the actual 

significance level can be as much as double what was expected even for low values of ρ. 

One of the most effective ways to determine statistical independence is through use of the 

Rank von Neumann Test. Directions for this test are given in Box 4-36 with an example in 

Box 4-37. Compared to other tests of statistical independence, the Rank von Neumann test has 

been shown to be more powerful over a wide variety of cases. It is also a reasonable test when 

the data follow a normal distribution. 

Box 4-36: Directions for the Rank von Neumann Test 

Let X1, X2, . . . , Xn represent the data values collected in sequence over equally spaced periods of time. 

COMPUTATIONS: Order the data measurements from smallest to largest and assign rank ri to measurement Xi. 

If measurements are tied, then assign the average rank. 

STEP 1. Null Hypothesis: H0: The data are independent. 

STEP 2. Alternative Hypothesis: HA: The data are not independent. 

STEP 3. Test Statistic: ( )∑( ) =

− − − =

n

i

i i r r

n n

v

2

2

1 2 0 1

12

STEP 4. a) Critical Value: Use Table A-16 to find vn, α. 

STEP 5. a) Conclusion: If v0 < vn, α, then reject the null hypothesis that the data are independent. 

NOTE: If the Rank von Neumann ratio test indicates significant evidence of dependence in the data, then a 

statistician should be consulted before further analysis is performed. If ranks are tied, the power of the statistical 

test is diminished 
EPA QA/G-9S 138 February 2006 

Box 4-37: An Example of the Rank von Neumann Test 

The following are hourly readings from a discharge monitor: hourly readings from a discharge monitor. 

COMPUTATIONS: 

Time 12:00 13:00 14:00 15:00 16:00 17:00 18:00 19:00 20:00 21:00 22:00 23:00 24:00 

Reading 6.5 6.6 6.7 6.4 6.3 6.4 6.2 6.2 6.3 6.6 6.8 6.9 7.0 

Rank 7 8.5 10 5.5 3.5 5.5 1.5 1.5 3.5 8.5 11 12 13 

STEP 1. Null Hypothesis: H0: The data are independent. 

STEP 2. Alternative Hypothesis: HA: The data are not independent. 

STEP 3. Test Statistic: ( ){( )( ) ( ) 8.5 7 10 8.5 13 12 } 0.473

13 13 1

12 2 2 2

2 0 − + − + + − = ⋅ − v = L

STEP 4. a) Critical Value: Using Table A-16, v13, 0.05 = 1.14. 

STEP 5. a) Conclusion: Since v0 = 0.473 < 1.14 = v13, 0.05, we reject the null hypothesis that the data 

 are independent. 
EPA QA/G-9S 139 February 2006 

CHAPTER 5 

STEP 5: DRAW CONCLUSIONS FROM THE DATA 

Step 5: Draw Conclusions from the Data

• Perform the calculations for the statistical method.

 Perform the calculations and document them clearly.

 If anomalies or outliers are present in the data set, perform the calculations with 

and without the questionable data.

• Evaluate the results and draw conclusions.

 If the null hypothesis is rejected, then draw the conclusions and document the 

analysis.

 If the null hypothesis is not rejected, verify whether the tolerable limits on false 

acceptance decision errors have been satisfied. If so, draw conclusions and 

document the analysis; if not, determine corrective actions, if any.

 Interpret the results of the test or confidence interval.

• Evaluate the performance of the sampling design if the design is to be used again.

 Evaluate the statistical power of the design over the full range of parameter values; 

consult a statistician as necessary.

THE DATA QUALITY ASSESSMENT PROCESS

Review DQOs and Sampling Design

Conduct Preliminary Data Review 

Select the Statistical Method 

Verify the Assumptions 

Draw Conclusions from the Data 

DRAW CONCLUSIONS FROM THE DATA

Purpose

Perform the statistical procedure and interpret the

results in the context of the data user's objectives.

Activities

Perform the Statistical Procedure

Draw Study Conclusions. 

Evaluate Performance of the Sampling Design 

Tools

Issues in hypothesis testing related to 

communicating the results of the DQA 
EPA QA/G-9S 140 February 2006 

List of Boxes 

Page

Box 5-1: Checking Adequacy of Sample Size for a One-Sample t-Test for Simple 

Random Sampling....................................................................................................... 142 

Box 5-2: Example of Power Calculations for the One-Sample Test of a Single Proportion..... 144 
EPA QA/G-9S 141 February 2006 

CHAPTER 5 

STEP 5: DRAW CONCLUSIONS FROM THE DATA 

5.1 OVERVIEW AND ACTIVITIES 

In this final step of the DQA, the analyst performs the statistical hypothesis test or 

computes the confidence interval and draws conclusions that address the data user's objectives. 

5.2 PERFORM THE STATISTICAL METHOD 

The goal of this activity is to conduct the statistical hypothesis test or compute the 

confidence interval procedure chosen in Chapter 3. The calculations for the procedure should be 

clearly documented and easily verifiable. In addition, documentation of the results should be 

understandable so they can be communicated effectively to those who may hold a stake in the 

resulting decision. If computer software is used to perform the calculations, then the procedures 

should be adequately documented. 

5.3 DRAW STUDY CONCLUSIONS 

5.3.1 Hypothesis Tests 

The goal of this activity is to translate the results of the statistical hypothesis test so that 

the data user may draw a conclusion from the data; the results being either: 

(a) reject the null hypothesis, in which case there is significant evidence in favor of 

the alternative hypothesis. The decision can be made with sufficient confidence 

and without further analysis. This is because the statistical tests described in this 

document inherently control the false rejection error rate within the data user's 

tolerable limits when the underlying assumptions are valid. 

(b) fail to reject the null hypothesis, in which case there is not significant evidence for 

the alternative hypothesis. The analyst is concerned about a possible false 

acceptance error. The most thorough procedure for verifying whether the false 

acceptance error limits have been satisfied is to compute the estimated power of 

the statistical test. 

Alternatively the sample size required to satisfy the data user's objectives can be 

calculated retrospectively using an estimate of the variance or an upper confidence limit on 

variance obtained from the actual data. If this theoretical sample size is less than or equal to the 

number of samples actually taken, then the test is probably sufficiently powerful. The equations 

required to perform these calculations have been provided in the instructions for many of the 

hypothesis test procedures in Chapter 3. An example of this method is contained in Box 5-1, but 

it is emphasized that this only gives an estimate of power, not an absolute determination. 
EPA QA/G-9S 142 February 2006 

Box 5-1: Checking Adequacy of Sample Size for a One-Sample t-Test for Simple Random Sampling 

In Box 3-3, the one-sample t-test was used to test the hypothesis H0: μ ≤ 95 ppm vs. HA: μ > 95 ppm. DQOs 

specified that the test should limit the false rejection error rate to 5% and the false acceptance error rate to 20% if 

the true mean was 105 ppm. 

A random sample of size n = 9 had sample mean X = 99.38 ppm and sample standard deviation s = 10.41 

ppm. The null hypothesis was not rejected. Assuming that the true value of the standard deviation was equal to 

the sample estimate of 10.41 ppm, it was found that a sample size of 9 would be required. This validated the 

sample size of 9 which had actually been used. 

In such a case it makes sense to build in some conservatism, for example, by using an upper 90% confidence 

limit for σ in the sample size calculation of Box 3-3. Using Box 4-22, it is found that an upper 90% confidence 

limit for the true standard deviation is 

15 76

3.49

8 10 41 1 . . U

n

s = = − . 

Using this value for s in the sample size calculation of Box 3-3 leads to the sample size estimate of 17. Hence, a 

sample size of at least 17 should be used to be 90% sure of achieving the DQOs. 

Since it is generally desirable to avoid the need for additional sampling, it is advisable to conservatively estimate 

sample size in the first place. In cases where DQOs depend on a variance estimate, this conservatism is 

achieved by intentionally overestimating the variance. 

5.3.2 Confidence Intervals or Limits 

A confidence interval is simply an interval estimate for the population parameter of 

interest. The interval’s width is dependent upon the variance of the point estimate, the sample 

size, and the confidence level. More specifically, the width is large if the variance is large, the 

sample size is small, or the confidence level is large. 

The interpretation of a confidence interval makes use of probability in an intuitive sense. 

When a confidence interval has been constructed using the data, there is still a chance that the 

interval does not include the true value of the parameter estimated. For example, consider this 

confidence interval statement: “the 95% confidence interval for the unknown population mean is 

43.5 to 48.9”. It is interpreted as, “I can be 95% certain that the interval 43.5 to 48.9 captures the 

unknown mean.” Notice how there is a 5% chance that the interval does not capture the mean. 

The confidence level is the ‘confidence’ we have that the population parameter lies 

within the interval. This concept is analogous to the false rejection error rate. The width of the 

interval is related to statistical power, or the false acceptance error rate. Rather than specifying a 

desired false acceptance error rate, the desired interval width can be specified. 

A confidence interval can be used to make to decisions and in some situations a test of 

hypothesis is set up as a confidence interval. Confidence intervals are analogous to two-sided 

hypothesis tests. If the threshold value lies outside of the interval, then there is evidence that the 

population parameter differs from the threshold value. In a similar manner, confidence limits 

can also be related to one-sided hypothesis tests. If the threshold value lies above (below) an 
EPA QA/G-9S 143 February 2006 

upper (lower) confidence bound, then there is evidence that the population parameter is less 

(greater) than the threshold. 

5.3.3 Tolerance Intervals or Limits 

A tolerance interval is an interval estimate for a certain proportion of the population. The 

interval’s width is dependent upon the variance of the population, the sample size, the desired 

proportion of the population, and the confidence level. More specifically, the width is large if 

the variance is large, the sample size is small, the proportion is large, or the confidence level is 

large. 

When a tolerance interval has been constructed using the data, there is still a chance that 

the interval does not include the desired proportion of the population. For example, consider this 

tolerance interval statement: “the 99% tolerance interval for 90% of the population is 7.5 to 9.9”. 

 It is interpreted as, “I can be 99% certain that the interval 7.5 to 9.9 captures 90% of the 

population.” Notice how there is a 1% chance that the interval does not capture at least the 

desired proportion. 

The confidence level is the ‘confidence’ we have that the desired proportion of the 

population lies within the interval. This concept is analogous to the false rejection error rate. 

The width of the interval is related to statistical power, or the false acceptance error rate. Rather 

than specifying a desired false acceptance error rate, the desired interval width can be specified. 

A tolerance interval can be used to make to decisions, and in some situations a test of 

hypothesis can be set up as a tolerance interval. Tolerance intervals are analogous to two-sided 

hypothesis tests. If the threshold value lies outside of the interval, then there is evidence that the 

desired proportion of the population differs from the threshold value. In a similar manner, 

tolerance limits can also be related to one-sided hypothesis tests. If the threshold value lies 

above (below) an upper (lower) tolerance limit, then there is evidence that the desired proportion 

of the population is less (greater) than the threshold. 

5.4 EVALUATE PERFORMANCE OF THE SAMPLING DESIGN 

If the sampling design is to be used again, either in a later phase of the current study or in 

a similar study, the analyst will be interested in evaluating the overall performance of the design. 

 To evaluate the sampling design, the analyst performs a statistical power analysis that describes 

the estimated power of the statistical test over the range of possible parameter values. The 

estimated power is computed for all parameter values under the alternative hypothesis to create a 

power curve. A power analysis helps the analyst evaluate the adequacy of the sampling design 

when the true parameter value lies in the vicinity of the action level. In this manner, the analyst 

may determine how well a statistical test performed and compare this performance with that of 

other tests. Box 5-2 illustrates power calculations for a test of a single proportion. 
EPA QA/G-9S 144 February 2006 

Box 5-2: Example of Power Calculations for the One-Sample Test of a Single Proportion 

This box illustrates power calculations for the test of H0: P ≥ 0.20 vs. HA: P < 0.20, with a false rejection error 

rate of 5% when P = 0.20 presented in Box 3-13. The power of the test will be calculated assuming P1 = 0.15 

and before data is available. Since nP0 and n(1-P0) both exceed 5, the normal approximation may be used. 

STEP 1: Determine the general conditions for rejection of the null hypothesis. In this case, the null hypothesis 

is rejected if the sample proportion is sufficiently smaller than P0. Using Box 3-11, H0 is rejected if 

( ) z

P P n

P n p

< − −α −

+ −

1

0 0

0

1 /

0.5

 or n

P Q P z n p 0 0 0 1 0.5 + < − −α , 

where p is the sample proportion and – z1−α is the standard normal critical value. 

STEP 2: Determine the specific conditions for rejection of the null hypothesis if P1 is the true value of the 

proportion P. Using the equations above, rejection occurs if 

( )

( )

( ) 0 55

0.15 0.85 / 85

0 20 015 1645 0.2 0.8 / 85

1 /

1 /

1 /

0.5

1 1

0 1 1 0 0

1 1

1 . . . .

P P n

P P z P P n

P P n

P n p = − ⋅

− − ⋅ = −

− − − < −

+ − −α

STEP 3: Find the probability of rejection if P1 is the true proportion. The quantity on the left-hand side of the 

above inequality is a standard normal variable. Hence the power at P1 = 0.15 is the probability that a 

standard normal variable is less than -0.55. Using Table A-1, this probability is approximately 0.3, 

which is fairly small. 

5.5 INTERPRET AND COMMUNICATE THE RESULTS 

At this point, the analyst has performed the applicable statistical procedure and has drawn 

conclusions. In many cases, the conclusions are straightforward and convincing so they lead to 

an unambiguous path forward for the project. In other cases, however, it is advantageous to 

consider these conclusions in a broader context in order to determine a course of action, see Data 

Quality Assessment: A Reviewer’s Guide (EPA QA/G-9R) (U.S. EPA 2004). 
EPA QA/G-9S 145 February 2006 

APPENDIX A: 

STATISTICAL TABLES
EPA QA/G-9S 146 February 2006 

List of Tables 

Page

TABLE A-1. STANDARD NORMAL DISTRIBUTION ........................................................ 147 

TABLE A-2. CRITICAL VALUES OF STUDENT'S-t DISTRIBUTION .............................. 149 

TABLE A-3. CRITICAL VALUES FOR THE STUDENTIZED RANGE TEST................... 150 

TABLE A-4. CRITICAL VALUES FOR THE EXTREME VALUE TEST 

(DIXON'S TEST) ................................................................................................ 151 

TABLE A-5. CRITICAL VALUES FOR DISCORDANCE TEST ......................................... 152 

TABLE A-6. APPROXIMATE CRITICAL VALUES λr FOR ROSNER'S TEST.................. 153 

TABLE A-7. QUANTILES OF THE WILCOXON SIGNED RANKS TEST......................... 155 

TABLE A-8. CRITICAL VALUES FOR THE WILCOXON RANK-SUM TEST................. 156 

TABLE A-9. PERCENTILES OF THE CHI-SQUARE DISTRIBUTION .............................. 159 

TABLE A-10. PERCENTILES OF THE F-DISTRIBUTION.................................................. 160 

TABLE A-11. VALUES OF THE PARAMETER $

λ FOR COHEN'S ESTIMATES 

ADJUSTING FOR NONDETECTED VALUES.............................................. 164 

TABLE A-12a: CRITICAL VALUES FOR THE MANN-KENDALL TEST FOR TREND.. 165 

TABLE A-12b: PROBABILITIES FOR THE SMALL-SAMPLE MANN-KENDALL 

TEST FOR TREND......................................................................................... 165 

TABLE A-13. QUANTILES FOR THE WALD-WOLFOWITZ TEST FOR RUNS.............. 166 

TABLE A-14. CRITICAL VALUES FOR THE SLIPPAGE TEST ........................................ 170 

TABLE A-15. DUNNETT’S TEST (ONE TAILED) ............................................................... 173 

TABLE A-16. APPROXIMATE α-LEVEL CRITICAL POINTS FOR RANK VON 

NEUMANN RATIO TEST ............................................................................... 175 

TABLE A-17. VALUES FOR COMPUTING A ONE-SIDED CONFIDENCE LIMIT ON 

A LOGNORMAL MEAN ................................................................................. 176 

TABLE A-18. CRITICAL VALUES FOR THE SIGN TEST.................................................. 178 

TABLE A-19. CRITICAL VALUES FOR THE QUANTILE TEST....................................... 179 
EPA QA/G-9S 147 February 2006 

TABLE A-1. STANDARD NORMAL DISTRIBUTION 

Table values are P(Z ≤ zp) = p. 

zp .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 

-0.0 .5000 .4960 .4920 .4880 .4840 .4801 .4761 .4721 .4681 .4641 

-0.1 .4602 .4562 .4522 .4483 .4443 .4404 .4364 .4325 .4286 .4247 

-0.2 .4207 .4168 .4129 .4090 .4052 .4013 .3974 .3936 .3897 .3859 

-0.3 .3821 .3783 .3745 .3707 .3669 .3632 .3594 .3557 .3520 .3483 

-0.4 .3446 .3409 .3372 .3336 .3300 .3264 .3228 .3192 .3156 .3121 

-0.5 .3085 .3050 .3015 .2981 .2946 .2912 .2877 .2843 .2810 .2776 

-0.6 .2743 .2709 .2676 .2643 .2611 .2578 .2546 .2514 .2483 .2451 

-0.7 .2420 .2389 .2358 .2327 .2296 .2266 .2236 .2206 .2177 .2148 

-0.8 .2119 .2090 .2061 .2033 .2005 .1977 .1949 .1922 .1894 .1867 

-0.9 .1841 .1814 .1788 .1762 .1736 .1711 .1685 .1660 .1635 .1611 

-1.0 .1587 .1562 .1539 .1515 .1492 .1469 .1446 .1423 .1401 .1379 

-1.1 .1357 .1335 .1314 .1292 .1271 .1251 .1230 .1210 .1190 .1170 

-1.2 .1151 .1131 .1112 .1093 .1075 .1056 .1038 .1020 .1003 .0985 

-1.3 .0968 .0951 .0934 .0918 .0901 .0885 .0869 .0853 .0838 .0823 

-1.4 .0808 .0793 .0778 .0764 .0749 .0735 .0721 .0708 .0694 .0681 

-1.5 .0668 .0655 .0643 .0630 .0618 .0606 .0594 .0582 .0571 .0559 

-1.6 .0548 .0537 .0526 .0516 .0505 .0495 .0485 .0475 .0465 .0455 

-1.7 .0446 .0436 .0427 .0418 .0409 .0401 .0392 .0384 .0375 .0367 

-1.8 .0359 .0351 .0344 .0336 .0329 .0322 .0314 .0307 .0301 .0294 

-1.9 .0287 .0281 .0274 .0268 .0262 .0256 .0250 .0244 .0239 .0233 

-2.0 .0228 .0222 .0217 .0212 .0207 .0202 .0197 .0192 .0188 .0183 

-2.1 .0179 .0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .0143 

-2.2 .0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110 

-2.3 .0107 .0104 .0102 .0099 .0096 .0094 .0091 .0089 .0087 .0084 

-2.4 .0082 .0080 .0078 .0075 .0073 .0071 .0069 .0068 .0066 .0064 

-2.5 .0062 .0060 .0059 .0057 .0055 .0054 .0052 .0051 .0049 .0048 

-2.6 .0047 .0045 .0044 .0043 .0041 .0040 .0039 .0038 .0037 .0036 

-2.7 .0035 .0034 .0033 .0032 .0031 .0030 .0029 .0028 .0027 .0026 

-2.8 .0026 .0025 .0024 .0023 .0023 .0022 .0021 .0021 .0020 .0019 

-2.9 .0019 .0018 .0018 .0017 .0016 .0016 .0015 .0015 .0014 .0014 

-3.0 .0013 .0013 .0013 .0012 .0012 .0011 .0011 .0011 .0010 .0010 

-3.1 .0010 .0009 .0009 .0009 .0008 .0008 .0008 .0008 .0007 .0007 

-3.2 .0007 .0007 .0006 .0006 .0006 .0006 .0006 .0005 .0005 .0005 

-3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0004 .0004 .0004 .0003 

-3.4 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0002 
EPA QA/G-9S 148 February 2006 

TABLE A-1. STANDARD NORMAL DISTRIBUTION (CONT.) 

Table values are P(Z ≤ zp) = p. 

zp .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 

0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 

0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 

0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 

0.3 .6179 .6247 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 

0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 

0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 

0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 

0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 

0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 

0.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389 

1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621 

1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830 

1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8897 .9015 

1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177 

1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319 

1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441 

1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545 

1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633 

1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706 

1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767 

2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817 

2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857 

2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9889 .9890 

2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9137 .9916 

2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936 

2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952 

2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964 

2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974 

2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981 

2.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986 .9986 

3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990 .9990 

3.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993 .9993 

3.2 .9993 .9993 .9994 .9994 .9994 .9994 .9994 .9995 .9995 .9995 

3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996 .9997 

3.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9998 
EPA QA/G-9S 149 February 2006 

t1-α

TABLE A-2. CRITICAL VALUES OF STUDENT'S-t DISTRIBUTION 

1 - α

Degrees of 

 Freedom 0.70 0.75 0.80 0.85 0.90 0.95 0.975 0.99 0.995 

1 

2 

3 

4 

5 

0.727 

0.617 

0.584 

0.569 

0.559 

1.000 

0.816 

0.765 

0.741 

0.727 

1.376 

1.061 

0.978 

0.941 

0.920 

1.963 

1.386 

1.250 

1.190 

1.156 

3.078 

1.886 

1.638 

1.533 

1.476 

6.314 

2.920 

2.353 

2.132 

2.015 

12.706 

4.303 

3.182 

2.776 

2.571 

31.821 

6.965 

4.541 

3.747 

3.365 

63.657 

9.925 

5.841 

4.604 

4.032 

6 

7 

8 

9 

10 

0.553 

0.549 

0.546 

0.543 

0.542 

0.718 

0.711 

0.706 

0.703 

0.700 

0.906 

0.896 

0.889 

0.883 

0.879 

1.134 

1.119 

1.108 

1.100 

1.093 

1.440 

1.415 

1.397 

1.383 

1.372 

1.943 

1.895 

1.860 

1.833 

1.812 

2.447 

2.365 

2.306 

2.262 

2.228 

3.143 

2.998 

2.896 

2.821 

2.764 

3.707 

3.499 

3.355 

3.250 

3.169 

11 

12 

13 

14 

15 

0.540 

0.539 

0.538 

0.537 

0.536 

0.697 

0.695 

0.694 

0.692 

0.691 

0.876 

0.873 

0.870 

0.868 

0.866 

1.088 

1.083 

1.079 

1.076 

1.074 

1.363 

1.356 

1.350 

1.345 

1.34 

1.796 

1.782 

1.771 

1.761 

1.753 

2.201 

2.179 

2.160 

2.145 

2.131 

2.718 

2.681 

2.650 

2.624 

2.602 

3.106 

3.055 

3.012 

2.977 

2.947 

16 

17 

18 

19 

20 

0.535 

0.534 

0.534 

0.533 

0.533 

0.690 

0.689 

0.688 

0.688 

0.687 

0.865 

0.863 

0.862 

0.861 

0.860 

1.071 

1.069 

1.067 

1.066 

1.064 

1.337 

1.333 

1.330 

1.328 

1.325 

1.746 

1.740 

1.734 

1.729 

1.725 

2.120 

2.110 

2.101 

2.093 

2.086 

2.583 

2.567 

2.552 

2.539 

2.528 

2.921 

2.898 

2.878 

2.861 

2.845 

21 

22 

23 

24 

25 

0.532 

0.532 

0.532 

0.531 

0.531 

0.686 

0.686 

0.685 

0.685 

0.684 

0.859 

0.858 

0.858 

0.857 

0.856 

1.063 

1.061 

1.060 

1.059 

1.058 

1.323 

1.321 

1.319 

1.318 

1.316 

1.721 

1.717 

1.714 

1.711 

1.708 

2.080 

2.074 

2.069 

2.064 

2.060 

2.518 

2.508 

2.500 

2.492 

2.485 

2.831 

2.819 

2.807 

2.797 

2.787 

26 

27 

28 

29 

30 

0.531 

0.531 

0.530 

0.530 

0.530 

0.684 

0.684 

0.683 

0.683 

0.683 

0.856 

0.855 

0.855 

0.854 

0.854 

1.058 

1.057 

1.056 

1.055 

1.055 

1.315 

1.314 

1.313 

1.311 

1.310 

1.706 

1.703 

1.701 

1.699 

1.697 

2.056 

2.052 

2.048 

2.045 

2.042 

2.479 

2.473 

2.467 

2.462 

2.457 

2.779 

2.771 

2.763 

2.756 

2.750 

40 

60 

120 

∞

0.529 

0.527 

0.526 

0.524 

0.681 

0.679 

0.677 

0.674 

0.851 

0.848 

0.845 

0.842 

1.050 

1.046 

1.041 

1.036 

1.303 

1.296 

1.289 

1.282 

1.684 

1.671 

1.658 

1.645 

2.021 

2.000 

1.980 

1.960 

2.423 

2.390 

2.358 

2.326 

2.704 

2.660 

2.617 

2.576 

Note: The last row of the table (∞ degrees of freedom) gives the critical values for a standard normal distribution (Z), 

e.g., t∞, 0.95 = z 0.95 = 1.645. 
EPA QA/G-9S 150 February 2006 

TABLE A-3. CRITICAL VALUES FOR THE STUDENTIZED RANGE TEST 

 Level of Significance α

 0.10 0.05 0.01

n a b a b a b 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 

13 

14 

15 

16 

17 

18 

19 

20 

25 

30 

35 

40 

45 

50 

55 

60 

65 

70 

75 

80 

85 

90 

95 

100 

150 

200 

500 

1000 

1.78 

2.04 

2.22 

2.37 

2.49 

2.59 

2.68 

2.76 

2.84 

2.90 

2.96 

3.02 

3.07 

3.12 

3.17 

3.21 

3.25 

3.29 

3.45 

3.59 

3.70 

3.79 

3.88 

3.95 

4.02 

4.08 

4.14 

4.19 

4.24 

4.28 

4.33 

4.36 

4.40 

4.44 

4.72 

4.90 

5.49 

5.92 

2.00 

2.41 

2.71 

2.95 

3.14 

3.31 

3.45 

3.57 

3.68 

3.78 

3.87 

3.95 

4.02 

4.09 

4.15 

4.21 

4.27 

4.32 

4.53 

4.70 

4.84 

4.96 

5.06 

5.14 

5.22 

5.29 

5.35 

5.41 

5.46 

5.51 

5.56 

5.60 

5.64 

5.68 

5.96 

6.15 

6.72 

7.11 

1.76 

1.98 

2.15 

2.28 

2.40 

2.50 

2.59 

2.67 

2.74 

2.80 

2.86 

2.92 

2.97 

3.01 

3.06 

3.10 

3.14 

3.18 

3.34 

3.47 

3.58 

3.67 

3.75 

3.83 

3.90 

3.96 

4.01 

4.06 

4.11 

4.16 

4.20 

4.24 

4.27 

4.31 

4.59 

4.78 

5.47 

5.79 

2.00 

2.43 

2.75 

3.01 

3.22 

3.40 

3.55 

3.69 

3.80 

3.91 

4.00 

4.09 

4.17 

4.24 

4.31 

4.37 

4.43 

4.49 

4.71 

4.89 

5.04 

5.16 

5.26 

5.35 

5.43 

5.51 

5.57 

5.63 

5.68 

5.73 

5.78 

5.82 

5.86 

5.90 

6.18 

6.39 

6.94 

7.33 

1.74 

1.87 

2.02 

2.15 

2.26 

2.35 

2.44 

2.51 

2.58 

2.64 

2.70 

2.75 

2.80 

2.84 

2.88 

2.92 

2.96 

2.99 

3.15 

3.27 

3.38 

3.47 

3.55 

3.62 

3.69 

3.75 

3.80 

3.85 

3.90 

3.94 

3.99 

4.02 

4.06 

4.10 

4.38 

4.59 

5.13 

5.57 

2.00 

2.45 

2.80 

3.10 

3.34 

3.54 

3.72 

3.88 

4.01 

4.13 

4.24 

4.34 

4.44 

4.52 

4.60 

4.67 

4.74 

4.80 

5.06 

5.26 

5.42 

5.56 

5.67 

5.77 

5.86 

5.94 

6.01 

6.07 

6.13 

6.18 

6.23 

6.27 

6.32 

6.36 

6.64 

6.84 

7.42 

7.80 
EPA QA/G-9S 151 February 2006 

TABLE A-4. CRITICAL VALUES FOR THE EXTREME VALUE TEST 

(DIXON'S TEST)

Level of Significance α

n 0.10 0.05 0.01

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 

13 

14 

15 

16 

17 

18 

19 

20 

21 

22 

23 

24 

25 

 0.886 

 0.679 

 0.557 

 0.482 

 0.434 

 0.479 

 0.441 

 0.409 

 0.517 

 0.490 

 0.467 

 0.492 

 0.472 

 0.454 

 0.438 

 0.424 

 0.412 

 0.401 

 0.391 

 0.382 

 0.374 

 0.367 

 0.360 

 0.941 

 0.765 

 0.642 

 0.560 

 0.507 

 0.554 

 0.512 

 0.477 

 0.576 

 0.546 

 0.521 

 0.546 

 0.525 

 0.507 

 0.490 

 0.475 

 0.462 

 0.450 

 0.440 

 0.430 

 0.421 

 0.413 

 0.406 

 0.988 

 0.889 

 0.780 

 0.698 

 0.637 

 0.683 

 0.635 

 0.597 

 0.679 

 0.642 

 0.615 

 0.641 

 0.616 

 0.595 

 0.577 

 0.561 

 0.547 

 0.535 

 0.524 

 0.514 

 0.505 

 0.497 

 0.489 
EPA QA/G-9S 152 February 2006 

TABLE A-5. CRITICAL VALUES FOR DISCORDANCE TEST 

Level of Significance α Level of Significance α

N 0.01 0.05 n 0.01 0.05

 3 

 4 

 5 

 6 

 7 

 8 

 9 

10 

11 

12 

13 

14 

15 

16 

17 

18 

19 

20 

21 

22 

23 

24 

25 

26 

27 

28 

29 

30 

1.155 

1.492 

1.749 

1.944 

2.097 

2.221 

2.323 

2.410 

2.485 

2.550 

2.607 

2.659 

2.705 

2.747 

2.785 

2.821 

2.854 

2.884 

2.912 

2.939 

2.963 

2.987 

3.009 

3.029 

3.049 

3.068 

3.085 

3.103 

1.153 

1.463 

1.672 

1.822 

1.938 

2.032 

2.110 

2.176 

2.234 

2.285 

2.331 

2.371 

2.409 

2.443 

2.475 

2.504 

2.532 

2.557 

2.580 

2.603 

2.624 

2.644 

2.663 

2.681 

2.698 

2.714 

2.730 

2.745 

31 

32 

33 

34 

35 

36 

37 

38 

39 

40 

41 

42 

43 

44 

45 

46 

47 

48 

49 

50 

3.119 

3.135 

3.150 

3.164 

3.178 

3.191 

3.204 

3.216 

3.228 

3.240 

3.251 

3.261 

3.271 

3.282 

3.292 

3.302 

3.310 

3.319 

3.329 

3.336 

2.759 

2.773 

2.786 

2.799 

2.811 

2.823 

2.835 

2.846 

2.857 

2.866 

2.877 

2.887 

2.896 

2.905 

2.914 

2.923 

2.931 

2.940 

2.948 

2.956 
EPA QA/G-9S 153 February 2006 

TABLE A-6. APPROXIMATE CRITICAL VALUES λr FOR ROSNER'S TEST 

 α α α

n r 0.05 0.01 n r 0.05 0.01 n r 0.05 0.01 

 25 1 

 2 

 3 

 4 

 5 

 10 

 2.82 

 2.80 

 2.78 

 2.76 

 2.73 

 2.59 

 3.14 

 3.11 

 3.09 

 3.06 

 3.03 

 2.85 

 32 1 

2 

3 

4 

5 

10 

2.94 

2.92 

2.91 

2.89 

2.88 

2.78 

3.27 

3.25 

3.24 

3.22 

3.20 

3.09 

 39 1 

2 

3 

4 

5 

10 

 3.03 

3.01 

3.00 

2.99 

2.98 

2.91 

 3.37 

3.36 

3.34 

3.33 

3.32 

3.24 

 26 1 

 2 

 3 

 4 

 5 

 10 

 2.84 

 2.82 

 2.80 

 2.78 

 2.76 

 2.62 

 3.16 

 3.14 

 3.11 

 3.09 

 3.06 

 2.89 

 33 1 

2 

3 

4 

5 

10 

2.95 

2.94 

2.92 

2.91 

2.89 

2.80 

3.29 

3.27 

3.25 

3.24 

3.22 

3.11 

 40 1 

2 

3 

4 

5 

10 

 3.04 

3.03 

3.01 

3.00 

2.99 

2.92 

 3.38 

3.37 

3.36 

3.34 

3.33 

3.25 

 27 1 

2 

3 

4 

5 

10 

 2.86 

 2.84 

 2.82 

 2.80 

 2.78 

 2.65 

 3.18 

3.16 

3.14 

3.11 

3.09 

2.93 

 34 1 

2 

3 

4 

5 

10 

2.97 

2.95 

2.94 

2.92 

2.91 

2.82 

3.30 

3.29 

3.27 

3.25 

3.24 

3.14 

 41 1 

2 

3 

4 

5 

10 

 3.05 

3.04 

3.03 

3.01 

3.00 

2.94 

 3.39 

3.38 

3.37 

3.36 

3.34 

3.27 

 28 1 

2 

3 

4 

5 

10 

 2.88 

2.86 

2.84 

2.82 

2.80 

2.68 

 3.20 

3.18 

3.16 

3.14 

3.11 

2.97 

 35 1 

2 

3 

4 

5 

10 

2.98 

2.97 

2.95 

2.94 

2.92 

2.84 

3.32 

3.30 

3.29 

3.27 

3.25 

3.16 

 42 1 

2 

3 

4 

5 

10 

 3.06 

3.05 

3.04 

3.03 

3.01 

2.95 

3.40 

3.39 

3.38 

3.37 

3.36 

3.29 

 29 1 

2 

3 

4 

5 

10 

 2.89 

2.88 

2.86 

2.84 

2.82 

2.71 

 3.22 

3.20 

3.18 

3.16 

3.14 

3.00 

 36 1 

2 

3 

4 

5 

10 

2.99 

2.98 

2.97 

2.95 

2.94 

2.86 

3.33 

3.32 

3.30 

3.29 

3.27 

3.18 

 43 1 

2 

3 

4 

5 

10 

 3.07 

3.06 

3.05 

3.04 

3.03 

2.97 

 3.41 

3.40 

3.39 

3.38 

3.37 

3.30 

 30 1 

2 

3 

4 

5 

10 

 2.91 

2.89 

2.88 

2.86 

2.84 

2.73 

 3.24 

3.22 

3.20 

3.18 

3.16 

3.03 

 37 1 

2 

3 

4 

5 

10 

3.00 

2.99 

2.98 

2.97 

2.95 

2.88 

3.34 

3.33 

3.32 

3.30 

3.29 

3.20 

 44 1 

 2 

3 

4 

5 

 10 

 3.08 

3.07 

3.06 

3.05 

3.04 

2.98 

 3.43 

3.41 

3.40 

3.39 

3.38 

3.32 

 31 1 

2 

3 

4 

5 

10 

 2.92 

2.91 

2.89 

2.88 

2.86 

2.76 

 3.25 

3.24 

3.22 

3.20 

3.18 

3.06 

 38 1 

2 

3 

4 

5 

10 

 3.01 

3.00 

2.99 

2.98 

2.97 

2.91 

 3.36 

3.34 

3.33 

3.32 

3.30 

3.22 

 45 1 

2 

3 

4 

5 

10 

 3.09 

3.08 

3.07 

3.06 

3.05 

2.99 

3.44 

3.43 

3.41 

3.40 

3.39 

 3.33 
EPA QA/G-9S 154 February 2006 

TABLE A-6. APPROXIMATE CRITICAL VALUES λr FOR ROSNER'S TEST (CONT.) 

 α α α

n r 0.05 0.01 n r 0.05 0.01 n r 0.05 0.01 

 46 1 

2 

3 

4 

5 

10 

 3.09 

3.09 

3.08 

3.07 

3.06 

3.00 

 3.45 

3.44 

3.43 

3.41 

3.40 

3.34 

 70 1 

2 

3 

4 

5 

10 

3.26 

3.25 

3.25 

3.24 

3.24 

3.21 

3.62 

3.62 

3.61 

3.60 

3.60 

3.57 

 250 1 

5 

10 

3.67 

3.67 

3.66 

4.04 

4.04 

4.03 

 47 1 

2 

3 

4 

5 

10 

 3.10 

3.09 

3.09 

3.08 

3.07 

3.01 

 3.46 

3.45 

3.44 

3.43 

3.41 

3.36 

 80 1 

2 

3 

4 

5 

10 

3.31 

3.30 

3.30 

3.29 

3.29 

3.26 

3.67 

3.67 

3.66 

3.66 

3.65 

3.63 

 300 1 

5 

 10 

3.72 

3.72 

3.71 

4.09 

4.09 

4.09 

 48 1 

2 

3 

4 

5 

10 

 3.11 

3.10 

3.09 

3.09 

3.08 

3.03 

 3.46 

3.46 

3.45 

3.44 

3.43 

3.37 

 90 1 

2 

3 

4 

5 

10 

3.35 

3.34 

3.34 

3.34 

3.33 

3.31 

3.72 

3.71 

3.71 

3.70 

3.70 

3.68 

 350 1 

5 

10 

3.77 

3.76 

3.76 

4.14 

4.13 

4.13 

 49 1 

 2 

 3 

 4 

 5 

 10 

 3.12 

3.11 

3.10 

3.09 

3.09 

3.04 

 3.47 

3.46 

3.46 

3.45 

3.44 

3.38 

 100 1 

2 

3 

4 

5 

10 

3.38 

3.38 

3.38 

3.37 

3.37 

3.35 

3.75 

3.75 

3.75 

3.74 

3.74 

3.72 

 400 1 

5 

10 

3.80 

3.80 

3.80 

4.17 

4.17 

4.16 

 50 1 

2 

3 

4 

5 

10 

3.13 

3.12 

3.11 

3.10 

3.09 

3.05 

3.48 

3.47 

3.46 

3.46 

3.45 

3.39 

 150 1 

2 

3 

4 

5 

10 

3.52 

3.51 

3.51 

3.51 

3.51 

3.50 

3.89 

3.89 

3.89 

3.88 

3.88 

3.87 

 450 1 

5 

10 

3.84 

3.83 

3.83 

4.20 

4.20 

4.20 

 60 1 

2 

3 

4 

5 

10 

3.20 

3.19 

3.19 

3.18 

3.17 

3.14 

3.56 

3.55 

3.55 

3.54 

3.53 

3.49 

 200 1 

2 

3 

4 

5 

10 

3.61 

3.60 

3.60 

3.60 

3.60 

3.59 

3.98 

3.98 

3.97 

3.97 

3.97 

3.96 

500 1 

5 

10 

3.86 

3.86 

3.86 

4.23 

4.23 

4.22 
EPA QA/G-9S 155 February 2006 

TABLE A-7. QUANTILES OF THE WILCOXON SIGNED RANKS TEST 

Values in the table are such that P(T+ ≤ wα) is approximately equal to, but less than α. For 

example, if n = 12, then P(T+ ≤ 17) = 0.0461, which is slightly less than 0.05. Note the exact 

probability was computed using the statistical software package R. 

N w0.005 w0.01 w0.025 w0.05 w0.075 w0.10 w0.15 w0.20

4 - - - - 0 0 1 2 

5 - - - 0 1 2 2 3 

6 - - 0 2 2 3 4 5 

7 - 0 2 3 4 5 7 8 

8 0 1 3 5 7 8 9 11 

9 1 3 5 8 9 10 12 14 

10 3 5 8 10 12 14 16 18 

11 5 7 10 13 16 17 20 22 

12 7 9 13 17 19 21 24 27 

13 9 12 17 21 24 26 29 32 

14 12 15 21 25 28 31 35 38 

15 15 19 25 30 33 36 40 44 

16 19 23 29 35 39 42 47 50 

17 23 27 34 41 45 48 53 57 

18 27 32 40 47 51 55 60 65 

19 32 37 46 53 58 62 68 73 

20 37 43 52 60 65 69 76 81 
EPA QA/G-9S 156 February 2006 

TABLE A-8. CRITICAL VALUES FOR THE WILCOXON RANK-SUM TEST 

Table values are the largest x values such that P( Wrs ≤ x ) ≤ α. Therefore, significance levels, α, 

are approximate. If there are ties, then the test is approximate. 

max(m, n) min(m, n) α

2 3 4 5 6 7 8 9 10 

0.010 - - - - - - - - - 

0.025 - - - - - - 0 0 0 

0.050 - - - 0 0 0 1 1 1 

2 

0.100 - 0 0 1 1 1 2 2 3 

0.010 - - - 0 0 0 1 1 

0.025 - - 0 1 1 2 2 3 

0.050 0 0 1 2 2 3 4 4 3 

0.100 1 1 2 3 4 5 5 6 

0.010 - 0 1 1 2 3 3 

0.025 0 0 2 3 4 4 5 

0.050 1 2 3 4 5 6 7 4 

0.100 3 4 5 6 7 9 10 

0.010 1 2 3 4 5 6 

0.025 2 3 5 6 7 8 

0.050 4 5 6 8 9 11 5 

0.100 5 7 8 10 12 13 

0.010 3 4 6 7 8 

0.025 5 6 8 10 11 

0.050 7 8 10 12 14 6 

0.100 9 11 13 15 17 

0.010 6 7 9 11 

0.025 8 10 12 14 

0.050 11 13 15 17 7 

0.100 13 16 18 21 

0.010 9 11 13 

0.025 13 15 17 

0.050 15 18 20 8 

0.100 19 22 25 

0.010 14 16 

0.025 17 20 

0.050 21 24 9 

0.100 25 28 

0.010 13 

0.025 23 

0.050 27 10 

0.100 32 
EPA QA/G-9S 157 February 2006 

TABLE A-8. CRITICAL VALUES FOR THE WILCOXON RANK-SUM TEST (CONT.) 

Table values are the largest x values such that P( Wrs ≤ x ) ≤ α. Therefore, significance levels, α, 

are approximate. If there are ties, then the test is approximate. 

max(m, n) min(m, n) α

11 12 13 14 15 16 17 18 19 20 

0.010 - - 0 0 0 0 0 0 1 1 

0.025 0 1 1 1 1 1 2 2 2 2 

0.050 1 2 2 3 3 3 3 4 4 4 

2 

0.100 3 4 4 4 5 5 6 6 7 7 

0.010 1 2 2 2 3 3 4 4 4 5 

0.025 3 4 4 5 5 6 6 7 7 8 

0.050 5 5 6 7 7 8 9 9 10 11 3 

0.100 7 8 9 10 10 11 12 13 14 15 

0.010 4 5 5 6 7 7 8 9 9 10 

0.025 6 7 8 9 10 11 11 12 13 14 

0.050 8 9 10 11 12 14 15 16 17 18 4 

0.100 11 12 13 15 16 17 18 20 21 22 

0.010 7 8 9 10 11 12 13 14 15 16 

0.025 9 11 12 13 14 15 16 18 19 20 

0.050 12 13 15 16 18 19 20 22 23 25 5 

0.100 15 17 18 20 22 23 25 27 28 30 

0.010 9 11 12 13 15 16 18 19 20 22 

0.025 13 14 16 17 19 21 22 24 25 27 

0.050 16 17 19 21 23 25 26 28 30 32 6 

0.100 19 21 23 25 27 29 31 34 36 38 

0.010 12 14 16 17 19 21 23 24 26 28 

0.025 16 18 20 22 24 26 28 30 32 33 

0.050 19 21 24 26 28 30 33 35 37 39 7 

0.100 23 26 28 31 33 36 38 41 43 46 

0.010 15 17 20 22 24 26 28 30 32 34 

0.025 19 22 24 26 28 31 34 36 38 41 

0.050 23 26 28 31 33 36 39 41 44 47 8 

0.100 27 30 33 36 39 42 45 48 51 54 

0.010 18 21 23 26 28 31 33 36 38 40 

0.025 23 26 28 31 24 37 39 42 45 48 

0.050 27 30 33 36 39 42 45 48 51 54 9 

0.100 31 35 38 41 45 48 52 55 58 62 

0.010 22 24 27 30 33 36 38 41 44 47 

0.025 26 29 33 36 39 42 45 48 52 55 

0.050 31 34 37 41 44 48 51 55 58 62 10 

0.100 36 39 43 47 51 54 58 62 66 70 
EPA QA/G-9S 158 February 2006 

TABLE A-8. CRITICAL VALUES FOR THE WILCOXON RANK-SUM TEST (CONT.) 

Table values are the largest x values such that P( Wrs ≤ x ) ≤ α. Therefore, significance levels, α, 

are approximate. If there are ties, then the test is approximate. 

max(m, n) min(m, n) α

11 12 13 14 15 16 17 18 19 20 

0.010 25 28 31 34 37 41 44 47 50 53 

0.025 30 33 37 40 44 47 51 55 58 62 

0.050 34 38 42 46 50 54 57 61 65 69 

11 

0.100 40 44 48 52 57 61 65 69 73 78 

0.010 31 35 38 52 46 48 53 56 60 

0.025 37 41 45 49 53 57 61 65 69 

0.050 42 47 51 55 60 64 68 72 77 12 

0.100 49 53 58 63 68 72 77 81 86 

0.010 39 43 47 51 55 59 63 67 

0.025 45 50 54 59 63 67 72 76 

0.050 51 56 61 65 70 75 80 84 13 

0.100 58 63 68 74 79 84 89 94 

0.010 47 51 56 60 65 69 73 

0.025 55 59 64 69 74 78 83 

0.050 61 66 71 77 82 87 92 14 

0.100 69 74 80 85 91 97 102 

0.010 56 61 66 70 75 80 

0.025 64 70 75 80 85 90 

0.050 72 77 83 88 94 100 15 

0.100 80 86 92 98 104 110 

0.010 66 71 76 82 87 

0.025 75 81 86 92 98 

0.050 83 89 95 101 107 16 

0.100 93 99 106 112 119 

0.010 77 82 88 93 

0.025 87 93 99 105 

0.050 96 102 109 115 17 

0.100 106 113 120 127 

0.010 88 94 100 

0.025 99 106 112 

0.050 109 116 123 18 

0.100 120 128 135 

0.010 101 107 

0.025 113 119 

0.050 123 130 19 

0.100 135 143 

0.010 114 

0.025 127 

0.050 138 20 

0.100 151 
EPA QA/G-9S 159 February 2006 

2 χ1−α

TABLE A-9. PERCENTILES OF THE CHI-SQUARE DISTRIBUTION

1 - α

df 0.005 0.010 0.025 0.050 0.100 0.900 0.950 0.975 0.990 0.995 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 

13 

14 

15 

16 

17 

18 

19 

20 

21 

22 

23 

24 

25 

26 

27 

28 

29 

30 

40 

50 

60 

70 

80 

90 

100 

0.04

393 

0.0100 

0.072 

0.207 

0.412 

0.676 

0.989 

1.34 

1.73 

2.16 

2.60 

3.07 

3.57 

4.07 

4.60 

5.14 

5.70 

6.26 

6.84 

7.43 

8.03 

8.64 

9.26 

9.89 

10.52 

11.16 

11.81 

12.46 

13.12 

13.79 

20.71 

27.99 

35.53 

43.28 

51.17 

59.20 

67.33 

0.03

157 

0.0201 

0.115 

0.297 

0.554 

0.872 

1.24 

1.65 

2.09 

2.56 

3.05 

3.57 

4.11 

4.66 

5.23 

5.81 

6.41 

7.01 

7.63 

8.26 

8.90 

9.54 

10.20 

10.86 

11.52 

12.20 

12.88 

13.56 

14.26 

14.95 

22.16 

29.71 

37.48 

45.44 

53.54 

61.75 

70.06 

0.03

982 

0.0506 

0.216 

0.484 

0.831 

1.24 

1.69 

2.18 

2.70 

3.25 

3.82 

4.40 

5.01 

5.63 

6.26 

6.91 

7.56 

8.23 

8.91 

9.59 

10.28 

10.98 

11.69 

12.40 

13.12 

13.84 

14.57 

15.31 

16.05 

16.79 

24.43 

32.36 

40.48 

48.76 

57.15 

65.65 

74.22 

0.02

393 

0.103 

0.352 

0.711 

1.145 

1.64 

2.17 

2.73 

3.33 

3.94 

3.57 

5.23 

5.89 

6.57 

7.26 

7.96 

8.67 

9.39 

10.12 

10.85 

11.59 

12.34 

13.09 

13.85 

14.61 

15.38 

16.15 

16.93 

17.71 

18.49 

26.51 

34.76 

43.19 

51.74 

60.39 

69.13 

77.93 

0.0158 

0.211 

0.584 

1.064 

1.61 

2.20 

2.83 

3.49 

4.17 

4.87 

5.58 

6.30 

7.04 

7.79 

8.55 

9.31 

10.09 

10.86 

11.65 

12.44 

13.24 

14.04 

14.85 

15.66 

16.47 

17.29 

18.11 

18.94 

19.77 

20.60 

29.05 

37.69 

46.46 

53.33 

64.28 

73.29 

82.36 

2.71 

4.61 

6.25 

7.78 

9.24 

10.64 

12.02 

13.36 

14.68 

15.99 

17.28 

18.55 

19.81 

21.06 

22.31 

23.54 

24.77 

25.99 

27.20 

28.41 

29.62 

30.81 

32.01 

33.20 

34.38 

35.56 

36.74 

37.92 

39.09 

40.26 

51.81 

63.17 

74.40 

85.53 

96.58 

107.6 

118.5 

3.84 

5.99 

7.81 

9.49 

11.07 

12.59 

14.07 

15.51 

16.92 

18.31 

19.68 

21.03 

22.36 

23.68 

25.00 

26.30 

27.59 

28.87 

30.14 

31.41 

32.67 

33.92 

35.17 

36.42 

37.65 

38.89 

40.11 

41.34 

42.56 

43.77 

55.76 

67.50 

79.08 

90.53 

101.9 

113.1 

124.3 

5.02 

7.38 

9.35 

11.14 

12.83 

14.45 

16.01 

17.53 

19.02 

20.48 

21.92 

23.34 

24.74 

26.12 

27.49 

28.85 

30.19 

31.53 

32.85 

34.17 

35.48 

36.78 

38.08 

39.36 

40.65 

41.92 

43.19 

44.46 

45.72 

46.98 

59.34 

71.42 

83.30 

95.02 

106.6 

118.1 

129.6 

6.63 

9.21 

11.34 

13.28 

15.09 

16.81 

18.48 

20.09 

21.67 

23.21 

24.73 

26.22 

27.69 

29.14 

30.58 

32.00 

33.41 

34.81 

36.19 

37.57 

38.93 

40.29 

41.64 

42.98 

44.31 

45.64 

46.96 

48.28 

49.59 

50.89 

63.69 

76.15 

88.38 

100.4 

112.3 

124.1 

135.8 

7.88 

10.60 

12.84 

14.86 

16.75 

18.55 

20.28 

21.96 

23.59 

25.19 

26.76 

28.30 

29.82 

31.32 

32.80 

34.27 

35.72 

37.16 

38.58 

40.00 

41.40 

42.80 

44.18 

45.56 

46.93 

48.29 

49.64 

50.99 

52.34 

53.67 

66.77 

79.49 

91.95 

104.2 

116.3 

128.3 

140.2 
EPA QA/G-9S 160 

F1-α

TABLE A-10. PERCENTILES OF THE 

F-DISTRIBUTION 

Degrees Degrees of Freedom for Numerator 

Freedom for 

Denominator 1 2 3 4 5 6 7 8 9 10 12 15 20 24 30 6

1 .90 

.95 

.975 

.99 

39.9 

161 648 

4052 

49.5 

200 800 

5000 

53.6 

216 864 

5403 

55.8 

225 900 

5625 

57.2 

230 922 

5764 

58.2 

234 937 

5859 

58.9 

237 948 

5928 

59.4 

239 957 

5981 

59.9 

241 963 

6022 

60.2 

242 969 

6056 

60.7 

244 977 

6106 

61.2 

246 985 

6157 

61.7 

248 993 

6209 

62.0 

249 997 

6235 

62.3 

250 

1001 6261 

62

2

1063

2 .90 

.95 

.975 

.99 

8.53 18.5 38.5 98.5 

9.00 19.0 39.0 99.0 

9.16 19.2 39.2 99.2 

9.24 19.2 39.2 99.2 

9.29 19.3 39.3 99.3 

9.33 19.3 39.3 99.3 

9.35 19.4 39.4 99.4 

9.37 19.4 39.4 99.4 

9.38 19.4 39.4 99.4 

9.39 19.4 39.4 99.4 

9.41 19.4 39.4 99.4 

9.42 19.4 39.4 99.4 

9.44 19.4 39.4 99.4 

9.45 19.5 39.5 99.5 

9.46 19.5 39.5 99.5 

9193999

3 .90 

.95 

.975 

.99 

5.54 10.1 17.4 34.1 

5.46 9.55 16.0 30.8 

5.39 9.28 15.4 29.5 

5.34 9.12 15.1 28.7 

5.31 9.01 14.9 28.2 

5.28 8.94 14.7 27.9 

5.27 8.89 14.6 27.7 

5.25 8.85 14.5 27.5 

5.24 8.81 14.5 27.3 

5.23 8.79 14.4 27.2 

5.22 8.74 14.3 27.1 

5.20 8.70 14.3 26.9 

5.18 8.66 14.2 26.7 

5.18 8.64 14.1 26.6 

5.17 8.62 14.1 26.5 

58142

4 .90 

.95 

.975 

.99 

.999 

4.54 7.71 12.2 21.2 74.1 

4.32 6.94 10.6 18.0 61.2 

4.19 6.59 9.98 16.7 56.2 

4.11 6.39 9.60 16.0 53.4 

4.05 6.26 9.36 15.5 51.7 

4.01 6.16 9.20 15.2 50.5 

3.98 6.09 9.07 15.0 49.7 

3.95 6.04 8.98 14.8 49.0 

3.94 6.00 8.90 14.7 48.5 

3.92 5.96 8.84 14.5 48.1 

3.90 5.91 8.75 14.4 47.4 

3.87 5.86 8.66 14.2 46.8 

3.84 5.80 8.56 14.0 46.1 

3.83 5.77 8.51 13.9 45.8 

3.82 5.75 8.46 13.8 45.4 

3581344

5 .90 

.95 

.975 

.99 

.999 

4.06 6.61 10.0 16.3 47.2 

3.78 5.79 8.43 13.3 37.1 

3.62 5.41 7.76 12.1 33.2 

3.52 5.19 7.39 11.4 31.1 

3.45 5.05 7.15 11.0 29.8 

3.40 4.95 6.98 10.7 28.8 

3.37 4.88 6.85 10.5 28.2 

3.34 4.82 6.76 10.3 27.6 

3.32 4.77 6.68 10.2 27.2 

3.39 4.74 6.62 10.1 26.9 

3.27 4.68 6.52 9.89 26.4 

3.24 4.62 6.43 9.72 25.9 

3.21 4.56 6.33 9.55 25.4 

3.19 4.53 6.28 9.47 25.1 

3.17 4.50 6.23 9.38 24.9 

346924

6 .90 

.95 

.975 

.99 

.999 

3.78 5.99 8.81 22.8 35.5 

3.46 5.14 7.26 10.9 27.0 

3.29 4.76 6.60 9.78 23.7 

3.18 4.53 6.23 9.15 21.9 

3.11 4.39 5.99 8.75 20.8 

3.05 4.28 5.82 8.47 20.0 

3.01 4.21 5.70 8.26 19.5 

2.98 4.15 5.60 8.10 19.0 

2.96 4.10 5.52 7.98 18.7 

2.94 4.06 5.46 7.87 18.4 

2.90 4.00 5.37 7.72 18.0 

2.87 3.94 5.27 7.56 17.6 

2.84 3.87 5.17 7.40 17.1 

2.82 3.84 5.12 7.31 16.9 

2.80 3.81 5.07 7.23 16.7 

23471
EPA QA/G-9S 161 

F1-α

TABLE A-10. PERCENTILES OF THE 

F-DISTRIBUTION (CONT.) 

Degrees Degrees of Freedom for Numerator 

Freedom for 

Denominator 1 2 3 4 5 6 7 8 9 10 12 15 20 24 30 6

7 .90 

.95 

.975 

.99 

.999 

3.59 5.59 8.07 12.2 29.2 

3.26 4.74 6.54 9.55 21.7 

3.07 4.35 5.89 8.45 18.8 

2.96 4.12 5.52 7.85 17.2 

2.88 3.97 5.29 7.46 16.2 

2.83 3.87 5.12 7.19 15.5 

2.78 3.79 4.99 6.99 15.0 

2.75 3.73 4.90 6.84 14.6 

2.72 3.68 4.82 6.72 14.5 

2.70 3.64 4.76 6.62 14.1 

2.67 3.57 4.67 6.47 13.7 

2.63 3.51 4.57 6.31 13.3 

2.59 3.44 4.47 6.16 12.9 

2.58 3.41 4.42 6.07 12.7 

2.56 3.38 4.36 5.99 12.5 

234512

8 .90 

.95 

.975 

.99 

.999 

3.46 5.32 7.57 11.3 25.4 

3.11 4.46 6.06 8.65 18.5 

2.92 4.07 5.42 7.59 15.8 

2.81 3.84 5.05 7.01 14.4 

2.73 3.69 4.82 6.63 13.5 

2.67 3.58 4.65 6.37 12.9 

2.62 3.50 4.53 6.18 12.4 

2.59 3.44 4.43 6.03 12.0 

2.56 3.39 4.36 5.91 11.8 

2.54 3.35 4.30 5.81 11.5 

2.50 3.28 4.20 5.67 11.2 

2.46 3.22 4.10 5.52 10.8 

2.42 3.15 4.00 5.36 10.5 

2.40 3.12 3.95 5.28 10.3 

2.38 3.08 3.89 5.20 10.1 

23359

9 .90 

.95 

.975 

.99 

.999 

3.36 5.12 7.21 10.6 22.9 

3.01 4.26 5.71 8.02 16.4 

2.81 3.86 5.08 6.99 13.9 

2.69 3.63 4.72 6.42 12.6 

2.61 3.48 4.48 6.06 11.7 

2.55 3.37 4.32 5.80 11.1 

2.51 3.29 4.20 5.61 10.7 

2.47 3.23 4.10 5.47 10.4 

2.44 3.18 4.03 5.35 10.1 

2.42 3.14 3.96 5.26 9.89 

2.38 3.07 3.87 5.11 9.57 

2.34 3.01 3.77 4.96 9.24 

2.30 2.94 3.67 4.81 8.90 

2.28 2.90 3.61 4.73 8.72 

2.25 2.86 3.56 4.65 8.55 

22348

10 .90 

.95 

.975 

.99 

.999 

3.29 4.96 6.94 10.0 21.0 

2.92 4.10 5.46 7.56 14.9 

2.73 3.71 4.83 6.55 12.6 

2.61 3.48 4.47 5.99 11.3 

2.52 3.33 4.24 5.64 10.5 

2.46 3.22 4.07 5.39 9.93 

2.41 3.14 3.95 5.20 9.52 

2.38 3.07 3.85 5.06 9.20 

2.35 3.02 3.78 4.94 8.96 

2.32 2.98 3.72 4.85 8.75 

2.28 2.91 3.62 4.71 8.45 

2.24 2.84 3.52 4.56 8.13 

2.20 2.77 3.42 4.41 7.80 

2.18 2.74 3.37 4.33 7.64 

2.16 2.70 3.31 4.25 7.47 

22347

12 .90 

.95 

.975 

.99 

.999 

3.18 4.75 6.55 9.33 18.6 

2.81 3.89 5.10 6.93 13.0 

2.61 3.49 4.47 5.95 10.8 

2.48 3.26 4.12 5.41 9.63 

2.39 3.11 3.89 5.06 8.89 

2.33 3.00 3.73 4.82 8.38 

2.28 2.91 3.61 4.64 8.00 

2.24 2.85 3.51 4.50 7.71 

2.21 2.80 3.44 4.39 7.48 

2.19 2.75 3.37 4.30 7.29 

2.15 2.69 3.28 4.16 7.00 

2.10 2.62 3.18 4.01 6.71 

2.06 2.54 3.07 3.86 6.40 

2.04 2.51 3.02 3.78 6.25 

2.01 2.47 2.96 3.70 6.09 

12235
EPA QA/G-9S 162 

F1-α

TABLE A-10. PERCENTILES OF THE 

F-DISTRIBUTION (CONT.) 

Degrees Degrees of Freedom for Numerator 

Freedom for 

Denominator 1 2 3 4 5 6 7 8 9 10 12 15 20 24 30 6

15 .90 

.95 

.975 

.99 

.999 

3.07 4.54 6.20 8.68 16.6 

2.70 3.68 4.77 6.36 11.3 

2.49 3.29 4.15 5.42 9.34 

2.36 3.06 3.80 4.89 8.25 

2.27 2.90 3.58 4.56 7.57 

2.21 2.79 3.41 4.32 7.09 

2.16 2.71 3.29 4.14 6.74 

2.12 2.64 3.20 4.00 6.47 

2.09 2.59 3.12 3.89 6.26 

2.06 2.54 3.06 3.80 6.08 

2.02 2.48 2.96 3.67 5.81 

1.97 2.40 2.86 3.52 5.54 

1.92 2.33 2.76 3.37 5.25 

1.90 2.29 2.70 3.29 5.10 

1.87 2.25 2.64 3.21 4.95 

12234

20 .90 

.95 

.975 

.99 

.999 

2.97 4.35 5.87 8.10 14.8 

2.59 3.49 4.46 5.85 9.95 

2.38 3.10 3.86 4.94 8.10 

2.25 2.87 3.51 4.43 7.10 

2.16 2.71 3.29 4.10 6.46 

2.09 2.60 3.13 3.87 6.02 

2.04 2.51 3.01 3.70 5.69 

2.00 2.45 2.91 3.56 5.44 

1.96 2.39 2.84 3.46 5.24 

1.94 2.35 2.77 3.37 5.08 

1.89 2.28 2.68 3.23 4.82 

1.84 2.20 2.57 3.09 4.56 

1.79 2.12 2.46 2.94 4.29 

1.77 2.08 2.41 2.86 4.15 

1.74 2.04 2.35 2.78 4.00 

11223

24 .90 

.95 

.975 

.99 

.999 

2.93 4.26 5.72 7.82 14.0 

2.54 3.40 4.32 6.66 9.34 

2.33 3.01 3.72 4.72 7.55 

2.19 2.78 3.38 4.22 6.59 

2.10 2.62 3.15 3.90 5.98 

2.04 2.51 2.99 3.67 5.55 

1.98 2.42 2.87 3.50 5.23 

1.94 2.36 2.78 3.36 4.99 

1.91 2.30 2.70 3.26 4.80 

1.88 2.25 2.64 3.17 4.64 

1.83 2.18 2.54 3.03 4.39 

1.78 2.11 2.44 2.89 4.14 

1.73 2.03 2.33 2.74 3.87 

1.70 1.98 2.27 2.66 3.74 

1.67 1.94 2.21 2.58 3.59 

11223

30 .90 

.95 

.975 

.99 

.999 

2.88 4.17 5.57 7.56 13.3 

2.49 3.32 4.18 5.39 8.77 

2.28 2.92 3.59 4.51 7.05 

2.14 2.69 3.25 4.02 6.12 

2.05 2.53 3.03 3.70 5.53 

1.98 2.42 2.87 3.47 5.12 

1.93 2.33 2.75 3.30 4.82 

1.88 2.27 2.65 3.17 4.58 

1.85 2.21 2.57 3.07 4.39 

1.82 2.16 2.51 2.98 4.24 

1.77 2.09 2.41 2.84 4.00 

1.72 2.01 2.31 2.70 3.75 

1.62 1.93 2.20 2.55 3.49 

1.64 1.89 2.14 2.47 3.36 

1.61 1.84 2.07 2.39 3.22 

11122

60 .90 

.95 

.975 

.99 

.999 

2.79 4.00 5.29 7.08 12.0 

2.39 3.15 3.93 4.98 7.77 

2.18 2.76 3.34 4.13 6.17 

2.04 2.53 3.01 3.65 5.31 

1.95 2.37 2.79 3.34 4.76 

1.87 2.25 2.63 3.12 4.37 

1.82 2.17 2.51 2.95 4.09 

1.77 2.10 2.41 2.82 3.86 

1.74 2.04 2.33 2.72 3.69 

1.71 1.99 2.27 2.63 3.54 

1.66 1.92 2.17 2.50 3.32 

1.60 1.84 2.06 2.35 3.08 

1.54 1.75 1.94 2.20 2.83 

1.51 1.70 1.88 2.12 2.69 

1.48 1.65 1.82 2.03 2.55 

11112
EPA QA/G-9S 163 

F1-α

TABLE A-10. PERCENTILES OF THE 

F-DISTRIBUTION (CONT.) 

Degrees Degrees of Freedom for Numerator 

Freedom for 

Denominator 1 2 3 4 5 6 7 8 9 10 12 15 20 24 30 6

120 .90 

.95 

.975 

.99 

.999 

2.75 3.92 5.15 6.85 11.4 

2.35 3.07 3.80 4.79 7.32 

2.13 2.68 3.23 3.95 5.78 

1.99 2.45 2.89 3.48 4.95 

1.90 2.29 2.67 3.17 4.42 

1.82 2.18 2.52 2.96 4.04 

1.77 2.09 2.39 2.79 3.77 

1.72 2.02 2.30 2.66 3.55 

1.68 1.96 2.22 2.56 3.38 

1.65 1.91 2.16 2.47 3.24 

1.60 1.83 2.05 2.34 3.02 

1.55 1.75 1.95 2.19 2.78 

1.48 1.66 1.82 2.03 2.53 

1.45 1.61 1.76 1.95 2.40 

1.41 1.55 1.69 1.86 2.26 

11111

∞ .90 

.95 

.975 

.99 

.999 

2.71 3.84 5.02 6.63 10.8 

2.30 3.00 3.69 4.61 6.91 

2.08 2.60 3.12 3.78 5.42 

1.94 2.37 2.79 3.32 4.62 

1.85 2.21 2.57 3.02 4.10 

1.77 2.10 

22.41 

2.80 3.74 

1.72 2.01 2.29 2.64 3.47 

1.67 1.94 2.19 2.51 3.27 

1.63 1.88 2.11 2.41 3.10 

1.60 1.83 2.05 2.32 2.96 

1.55 1.75 1.94 2.18 2.74 

1.49 1.67 1.83 2.04 2.51 

1.42 1.57 1.71 1.88 2.27 

1.38 1.52 1.64 1.79 2.13 

1.34 1.46 1.57 1.70 1.99 

11111
EPA QA/G-9S 164 February 2006 

TABLE A-11. VALUES OF THE PARAMETER $λ FOR COHEN'S ESTIMATES 

ADJUSTING FOR NONDETECTED VALUES 

h

γ .01 .02 .03 .04 .05 .06 .07 .08 .09 .10 .15 .20 

.00 

.05 

.10 

.15 

.20 

.25 

.30 

.35 

.40 

.45 

.50 

.55 

.60 

.65 

.70 

.75 

.80 

.85 

.90 

.95 

1.00 

.010100 

.010551 

.010950 

.011310 

.011642 

.011952 

.012243 

.012520 

.012784 

.013036 

.013279 

.013513 

.013739 

.013958 

.014171 

.014378 

.014579 

.014773 

.014967 

.015154 

.015338 

.020400 

.021294 

.022082 

.022798 

.023459 

.024076 

.024658 

.025211 

.025738 

.026243 

.026728 

.027196 

.027849 

.028087 

.028513 

.029927 

.029330 

.029723 

.030107 

.030483 

.030850 

.030902 

.032225 

.033398 

.034466 

.035453 

.036377 

.037249 

.038077 

.038866 

.039624 

.040352 

.041054 

.041733 

.042391 

.043030 

.043652 

.044258 

.044848 

.045425 

.045989 

.046540 

.041583 

.043350 

.044902 

.046318 

.047829 

.048858 

.050018 

.051120 

.052173 

.053182 

.054153 

.055089 

.055995 

.056874 

.057726 

.058556 

.059364 

.060153 

.060923 

.061676 

.062413 

.052507 

.054670 

.056596 

.058356 

.059990 

.061522 

.062969 

.064345 

.065660 

.066921 

.068135 

.069306 

.070439 

.071538 

.072505 

.073643 

.074655 

.075642 

.075606 

.077549 

.078471 

.063625 

.066159 

.068483 

.070586 

.072539 

.074372 

.076106 

.077736 

.079332 

.080845 

.082301 

.083708 

.085068 

.086388 

.087670 

.088917 

.090133 

.091319 

.092477 

.093611 

.094720 

.074953 

.077909 

.080563 

.083009 

.085280 

.087413 

.089433 

.091355 

.093193 

.094958 

.096657 

.098298 

.099887 

.10143 

.10292 

.10438 

.10580 

.10719 

.10854 

.10987 

.11116 

.08649 

.08983 

.09285 

.09563 

.09822 

.10065 

.10295 

.10515 

.10725 

.10926 

.11121 

.11208 

.11490 

.11666 

.11837 

.12004 

.12167 

.12225 

.12480 

.12632 

.12780 

.09824 

.10197 

.10534 

.10845 

.11135 

.11408 

.11667 

.11914 

.12150 

.12377 

.12595 

.12806 

.13011 

.13209 

.13402 

.13590 

.13775 

.13952 

.14126 

.14297 

.14465 

.11020 

.11431 

.11804 

.12148 

.12469 

.12772 

.13059 

.13333 

.13595 

.13847 

.14090 

.14325 

.14552 

.14773 

.14987 

.15196 

.15400 

.15599 

.15793 

.15983 

.16170 

.17342 

.17925 

.18479 

.18985 

.19460 

.19910 

.20338 

.20747 

.21129 

.21517 

.21882 

.22225 

.22578 

.22910 

.23234 

.23550 

.23858 

.24158 

.24452 

.24740 

.25022 

.24268 

.25033 

.25741 

.26405 

.27031 

.27626 

.28193 

.28737 

.29250 

.29765 

.30253 

.30725 

.31184 

.31630 

.32065 

.32489 

.32903 

.33307 

.33703 

.34091 

.34471

h

γ .25 .30 .35 .40 .45 .50 .55 .60 .65 .70 .80 .90 

.00 

.05 

.10 

.15 

.20 

.25 

.30 

.35 

.40 

.45 

.50 

.55 

.60 

.65 

.70 

.75 

.80 

.85 

.90 

.95 

1.00 

.31862 

.32793 

.33662 

.34480 

.35255 

.35993 

.36700 

.37379 

.38033 

.38665 

.39276 

.39679 

.40447 

.41008 

.41555 

.42090 

.42612 

.43122 

.43622 

.44112 

.44592 

.4021 

.4130 

.4233 

.4330 

.4422 

.4510 

.4595 

.4676 

.4735 

.4831 

.4904 

.4976 

.5045 

.5114 

.5180 

.5245 

.5308 

.5370 

.5430 

.5490 

.5548 

.4941 

.5066 

.5184 

.5296 

.5403 

.5506 

.5604 

.5699 

.5791 

.5880 

.5967 

.6061 

.6133 

.6213 

.6291 

.6367 

.6441 

.6515 

.6586 

.6656 

.6724 

.5961 

.6101 

.6234 

.6361 

.6483 

.6600 

.6713 

.6821 

.6927 

.7029 

.7129 

.7225 

.7320 

.7412 

.7502 

.7590 

.7676 

.7781 

.7844 

.7925 

.8005 

.7096 

.7252 

.7400 

.7542 

.7673 

.7810 

.7937 

.8060 

.8179 

.8295 

.8408 

.8517 

.8625 

.8729 

.8832 

.8932 

.9031 

.9127 

.9222 

.9314 

.9406 

.8388 

.8540 

.8703 

.8860 

.9012 

.9158 

.9300 

.9437 

.9570 

.9700 

.9826 

.9950 

1.007 

1.019 

1.030 

1.042 

1.053 

1.064 

1.074 

1.085 

1.095 

.9808 

.9994 

1.017 

1.035 

1.051 

1.067 

1.083 

1.098 

1.113 

1.127 

1.141 

1.155 

1.169 

1.182 

1.195 

1.207 

1.220 

1.232 

1.244 

1.255 

1.287 

1.145 

1.166 

1.185 

1.204 

1.222 

1.240 

1.257 

1.274 

1.290 

1.306 

1.321 

1.337 

1.351 

1.368 

1.380 

1.394 

1.408 

1.422 

1.435 

1.448 

1.461 

1.336 

1.358 

1.379 

1.400 

1.419 

1.439 

1.457 

1.475 

1.494 

1.511 

1.528 

1.545 

1.561 

1.577 

1.593 

1.608 

1.624 

1.639 

1.653 

1.668 

1.882 

1.561 

1.585 

1.608 

1.630 

1.651 

1.672 

1.693 

1.713 

1.732 

1.751 

1.770 

1.788 

1.806 

1.824 

1.841 

1.851 

1.875 

1.892 

1.908 

1.924 

1.940 

2.176 

2.203 

2.229 

2.255 

2.280 

2.305 

2.329 

2.353 

2.376 

2.399 

2.421 

2.443 

2.465 

2.486 

2.507 

2.528 

2.548 

2.568 

2.588 

2.607 

2.626 

3.283 

3.314 

3.345 

3.376 

3.405 

3.435 

3.464 

3.492 

3.520 

3.547 

3.575 

3.601 

3.628 

3.654 

3.679 

3.705 

3.730 

3.754 

3.779 

3.803 

3.827
EPA QA/G-9S 165 February 2006 

TABLE A-12a: CRITICAL VALUES FOR THE 

MANN-KENDALL TEST FOR TREND

Sample size n Significance Level 

α 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 

0.20 4 6 7 7 8 10 11 13 14 16 17 19 20 22 25 27 28 

0.10 6 8 9 11 12 14 17 19 20 24 25 29 30 34 35 39 42 

0.05 6 8 11 13 16 18 21 23 26 28 31 35 38 42 45 49 52 

0.01 - 10 13 17 20 24 27 31 34 40 43 47 52 58 63 67 72 

TABLE A-12b: PROBABILITIES FOR THE SMALL-SAMPLE 

MANN-KENDALL TEST FOR TREND

n n S 4 5 8 9 S 6 7 10 

0 0.625 0.592 0.548 0.540 1 0.500 0.500 0.500 

2 0.375 0.408 0.452 0.460 3 0.360 0.386 0.431 

4 0.167 0.242 0.360 0.381 5 0.235 0.281 0.364 

6 0.042 0.117 0.274 0.306 7 0.136 0.191 0.300 

8 0.042 0.199 0.238 9 0.068 0.119 0.242 

10 0.0083 0.138 0.179 11 0.028 0.068 0.190 

12 0.089 0.130 13 0.0083 0.035 0.146 

14 0.054 0.090 15 0.0014 0.015 0.108 

16 0.031 0.060 17 0.0054 0.078 

18 0.016 0.038 19 0.0014 0.054 

20 0.0071 0.022 21 0.00020 0.036 

22 0.0028 0.012 23 0.023 

24 0.00087 0.0063 25 0.014 

26 0.00019 0.0029 27 0.0083 

28 0.000025 0.0012 29 0.0046 

30 0.00043 31 0.0023 

32 0.00012 33 0.0011 

34 0.000025 35 0.00047 

36 0.0000028 37 0.00018 

 39 0.000058 

41 0.000015 

43 0.0000028 

45 0.00000028 
EPA QA/G-9S 166 

TABLE A-13. QUANTILES FOR THE WALD-WOLFOWITZ TEST FOR RUNS

n m 

W0.01

W0.05

W0.10 n m 

W0.01

W0.05

W0.10 n m 

W0.01

W0.05

W0.10 n m 

4 - - 3 5 3 4 4 6 3 4 5 5 - - 3 6 3 4 4 7 4 5 6 6 - 3 4 7 3 4 5 8 4 5 6 7 - 3 4 8 3 4 5 9 4 5 6 18 - 3 4 9 3 4 5 10 4 6 7 19 - 3 4 10 4 4 5 11 5 6 7 1

10 - 4 5 11 4 5 6 12 5 6 7 111 3 4 5 12 4 5 6 13 5 7 8 112 3 4 5 13 4 5 6 14 5 7 8 113 3 4 5 14 5 6 6 15 5 7 8 114 3 5 6 15 5 6 7 16 6 7 8 115 3 5 6 16 5 6 7 17 6 7 8 116 4 5 6 17 5 6 7 18 6 7 8 117 4 5 6 18 5 6 7 19 6 7 8 

7 

2

18 4 5 6 19 5 6 7 

6 

20 6 7 8 

4 

19 4 5 6 

5 

20 5 6 7 
EPA QA/G-9S 167 

TABLE A-13. QUANTILES FOR THE WALD-WOLFOWITZ TEST FOR RUNS (CON

n m 

W0.01

W0.05

W0.10 n m 

W0.01

W0.05

W0.10 n m 

W0.01

W0.05

W0.10 n m 

20 4 5 6 

8 8 5 6 6 9 9 5 6 7 10 10 7 8 9 11 1

9 5 6 7 10 6 7 8 11 7 8 9 1

10 5 7 7 11 6 7 8 12 7 8 9 111 6 7 8 12 6 7 8 13 7 9 10 112 6 7 8 13 7 8 9 14 8 9 10 113 6 7 8 14 7 8 9 15 8 9 10 114 6 8 8 15 7 9 9 16 8 9 11 115 6 8 9 16 7 9 10 17 8 10 11 116 7 8 9 17 8 9 10 18 9 10 11 117 7 8 9 18 8 9 10 19 9 11 12 218 7 9 10 19 8 10 11 20 9 11 12 19 7 9 10 20 8 10 11 20 7 9 10 
EPA QA/G-9S 168 

TABLE A-13. QUANTILES FOR THE WALD-WOLFOWITZ TEST FOR RUNS (CON

n m 

W0.01

W0.05

W0.10 n m 

W0.01

W0.05

W0.10 n m 

W0.01

W0.05

W0.10 n m 

12 12 7 9 10 13 13 9 10 11 14 14 9 11 12 15 1

13 8 10 10 14 9 11 12 15 9 11 12 114 8 10 10 15 10 11 12 16 10 11 13 115 8 10 11 16 10 11 12 17 10 12 13 116 9 11 12 17 10 11 12 18 10 12 13 117 9 11 12 18 10 11 13 19 11 13 14 218 9 11 12 19 10 12 13 20 11 13 14 19 10 12 13 20 11 12 13 20 10 12 13 

16 16 10 12 13 17 17 11 13 14 18 18 13 14 16 19 1

17 11 13 14 18 11 13 15 19 13 14 16 2
EPA QA/G-9S 169 

TABLE A-13. QUANTILES FOR THE WALD-WOLFOWITZ TEST FOR RUNS (CON

n m 

W0.01

W0.05

W0.10 n m 

W0.01

W0.05

W0.10 n m 

W0.01

W0.05

W0.10 n m 

 18 11 13 14 19 12 14 16 20 14 15 16 

19 12 14 15 20 12 14 16 20 13 14 16 20 2

When n or 

m is greater than 20 the 

Wp quantile is given by: 

( )

( ) ( ) W mn m n z mn mn m n mn mn p p = +

+

+

− −

+ +−

1 2 2 2

1 2 where zp is the appropriate quantile from the standard normal (see Table A-1). 
EPA QA/G-9S 170 February 2006 

TABLE A-14. CRITICAL VALUES FOR THE SLIPPAGE TEST 

LEVEL OF SIGNIFICANCE (α) APPROXIMATELY 0.01

n = size of population 1 (site) 

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 

5 5 6 6 7 8 8 9 9 10 11 11 12 12 13 14 14 

6 5 5 6 6 7 8 8 9 9 10 10 11 11 12 12 13 

7 5 5 6 6 7 7 8 8 9 9 10 10 10 11 11 12 

8 4 5 5 6 6 7 7 8 8 8 9 9 10 10 11 11 

9 4 5 5 5 6 6 7 7 8 8 8 9 9 10 10 10 

10 4 4 5 5 6 6 6 7 7 7 8 8 9 9 9 10 

11 4 4 5 5 5 6 6 6 7 7 7 8 8 9 9 9 

12 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 9 

13 4 4 4 5 5 5 6 6 6 7 7 7 7 8 8 8 

14 4 4 4 4 5 5 5 6 6 6 7 7 7 7 8 8 

15 3 4 4 4 5 5 5 5 6 6 6 7 7 7 7 8 

16 3 4 4 4 4 5 5 5 6 6 6 6 7 7 7 7 

17 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 

18 3 3 4 4 4 5 5 5 5 5 6 6 6 6 7 7 

19 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 7 

m = size of population 2 (background) 

20 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 7 

n = size of population 1 (site) 

25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 

25 7 8 8 9 10 11 12 13 13 14 15 16 17 18 19 19 

30 6 7 7 8 9 10 10 11 12 13 13 14 15 14 16 17 

35 5 6 7 7 8 9 9 10 11 11 12 12 13 14 14 15 

40 5 6 6 7 7 8 9 9 10 10 11 11 12 12 13 13 

45 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 

50 5 5 5 6 6 7 7 8 8 9 9 10 10 10 11 11 

55 4 5 5 6 6 6 7 7 8 8 9 9 9 10 10 11 

60 4 5 5 5 6 6 7 7 7 8 8 8 9 9 10 10 

65 4 4 5 5 5 6 6 7 7 7 8 8 8 9 9 9 

70 4 4 5 5 5 6 6 6 7 7 7 8 8 8 9 9 

75 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 9 

80 4 4 4 5 5 5 6 6 6 6 7 7 7 8 8 8 

85 4 4 4 4 5 5 5 6 6 6 6 7 7 7 8 8 

90 3 4 4 4 5 5 5 5 6 6 6 6 7 7 7 8 

95 3 4 4 4 4 4 5 5 6 6 6 6 7 7 7 7 

m = size of population 2 (background) 

100 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 
EPA QA/G-9S 171 February 2006 

TABLE A-14. CRITICAL VALUES FOR THE SLIPPAGE TEST (CONT.) 

LEVEL OF SIGNIFICANCE (α) APPROXIMATELY 0.05

n = size of population 1 (site) 

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 

5 4 4 5 5 6 6 7 7 8 8 9 9 9 10 10 11 

6 4 4 4 5 5 6 6 6 7 7 8 8 8 9 9 10 

7 3 4 4 5 5 5 6 6 6 7 7 7 8 8 8 9 

8 3 4 4 4 5 5 5 6 6 6 6 7 7 7 8 8 

9 3 3 4 4 4 5 5 5 5 6 6 6 7 7 7 7 

10 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 

11 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 

12 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 

13 3 3 3 3 4 4 4 4 4 5 5 5 5 6 6 6 

14 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 

15 3 3 3 3 3 4 4 4 4 4 4 5 5 5 5 5 

16 2 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5 

17 2 3 3 3 3 3 4 4 4 4 4 4 5 5 5 5 

18 2 3 3 3 3 3 3 4 4 4 4 4 4 5 5 5 

19 2 3 3 3 3 3 3 3 4 4 4 4 4 4 5 5 

m = size of population 2 (background) 

20 2 2 3 3 3 3 3 3 4 4 4 4 4 4 4 5 

n = size of population 1 (site) 

25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 

25 5 5 6 6 7 8 8 9 9 10 10 11 12 12 13 13 

30 4 5 5 6 6 7 7 8 8 9 9 9 10 10 11 11 

35 4 4 5 5 6 6 6 7 7 8 8 8 9 9 10 10 

40 4 4 4 5 5 5 6 6 7 7 7 8 8 8 9 9 

45 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 

50 3 3 4 4 4 5 5 5 6 6 5 7 7 7 7 8 

55 3 3 4 4 4 4 5 5 5 6 6 6 6 7 7 7 

60 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 

65 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 

70 3 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 

75 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 6 

80 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 

85 3 3 3 3 3 3 3 4 4 4 4 5 5 5 5 5 

90 2 3 3 3 3 3 4 4 4 4 4 4 5 5 5 5 

95 2 3 3 3 3 3 3 4 4 4 4 4 4 5 5 5 

m = size of population 2 (background) 

100 2 3 3 3 3 3 3 4 4 4 4 4 4 4 5 5 
EPA QA/G-9S 172 February 2006 

TABLE A-14. CRITICAL VALUES FOR THE SLIPPAGE TEST (CONT.) 

LEVEL OF SIGNIFICANCE (α) APPROXIMATELY 0.10

n = size of population 1 (site) 

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 

5 3 4 4 4 5 5 6 6 6 7 7 7 8 8 9 9 

6 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 

7 3 3 3 4 4 4 5 5 5 5 6 6 6 7 7 7 

8 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 

9 3 3 3 3 4 4 4 4 4 5 5 5 5 6 6 6 

10 2 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 

11 2 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5 

12 2 2 3 3 3 3 3 4 4 4 4 4 5 5 5 5 

13 2 2 3 3 3 3 3 3 4 4 4 4 4 4 5 5 

14 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 5 

15 2 2 2 3 3 3 3 3 3 3 4 4 4 4 4 4 

16 2 2 2 3 3 3 3 3 3 3 4 4 4 4 4 4 

17 2 2 2 2 3 3 3 3 3 3 3 4 4 4 4 4 

18 2 2 2 2 3 3 3 3 3 3 3 3 4 4 4 4 

19 2 2 2 2 2 3 3 3 3 3 3 3 3 4 4 4 

m = size of population 2 (background) 

20 2 2 2 2 2 3 3 3 3 3 3 3 3 3 4 4 

n = size of population 1 (site) 

25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 

25 4 4 5 5 6 6 6 7 7 8 8 9 9 9 10 10 

30 3 4 4 4 5 5 6 6 6 7 7 7 8 8 9 9 

35 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 

40 3 3 3 4 4 4 5 5 5 5 6 6 6 7 7 7 

45 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 

50 3 3 3 3 4 4 4 4 4 5 5 5 5 6 6 6 

55 2 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 

60 2 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5 

65 2 2 3 3 3 3 3 4 4 4 4 4 4 5 5 5 

70 2 2 3 3 3 3 3 3 4 4 4 4 4 4 5 5 

75 2 2 2 3 3 3 3 2 3 4 4 4 4 4 4 5 

80 2 2 2 3 3 3 3 3 3 3 4 4 4 4 4 4 

85 2 2 2 3 3 3 3 3 3 3 4 4 4 4 4 4 

90 2 2 2 2 3 3 3 3 3 3 3 4 4 4 4 4 

95 2 2 2 2 3 3 3 3 3 3 3 3 4 4 4 4 

m = size of population 2 (background) 

100 2 2 2 2 2 3 3 3 3 3 3 3 3 4 4 4 
EPA QA/G-9S 173 

TABLE A-15. DUNNETT’S TEST (ONE TAILED) 

Degrees Total Number of Investigated Groups (

k – 1) of 

Freedom 

α 2 3 4 5 6 7 8 9 10 12 

2 .05 .10 3.80 2.54 4.34 2.92 4.71 3.20 5.08 3.40 5.24 3.57 5.43 3.71 5.60 3.83 5.75 3.94 5.88 4.03 6.11 4.19 3 .05 .10 2.94 2.13 3.28 2.41 3.52 2.61 3.70 2.76 3.85 2.87 3.97 2.97 4.08 3.06 4.17 3.13 4.25 3.20 4.39 3.31 4 .05 .10 2.61 1.96 2.88 2.20 3.08 2.37 3.22 2.50 3.34 2.60 3.44 2.68 3.52 2.75 3.59 2.82 3.66 2.87 3.77 2.97 5 .05 .10 2.44 1.87 2.68 2.09 2.85 2.24 2.98 2.36 3.08 2.45 3.16 2.53 3.24 2.59 3.30 2.65 3.36 2.70 3.45 2.78 6 .05 .10 2.34 1.82 2.56 2.02 2.71 2.17 2.83 2.27 2.92 2.36 3.00 2.43 3.06 2.49 3.12 2.54 3.17 2.59 3.26 2.67 7 .05 .10 2.27 1.78 2.48 1.98 2.82 2.11 2.73 2.22 2.81 2.30 2.89 2.37 2.95 2.42 3.00 2.47 3.05 2.52 3.13 2.59 8 .05 .10 2.22 1.75 2.42 1.94 2.55 2.08 2.66 2.17 2.74 2.25 2.81 2.32 2.87 2.38 2.92 2.42 2.96 2.47 3.04 2.54 9 .05 .10 2.18 1.73 2.37 1.92 2.50 2.05 2.60 2.14 2.68 2.22 2.75 2.28 2.81 2.34 2.86 2.39 2.90 2.43 2.97 2.50 

10 .05 .10 2.15 1.71 2.34 1.90 2.47 2.02 2.56 2.12 2.64 2.19 2.70 2.26 2.76 2.31 2.81 2.35 2.85 2.40 2.92 2.46 12 .05 .10 2.11 1.69 2.29 1.87 2.41 1.99 2.50 2.08 2.58 2.16 2.64 2.22 2.69 2.27 2.74 2.31 2.78 2.35 2.84 2.42 16 .05 .10 2.06 1.66 2.23 1.83 2.34 1.95 2.43 2.04 2.50 2.11 2.56 2.17 2.61 2.22 2.65 2.26 2.69 2.30 2.75 2.36 
EPA QA/G-9S 174 

TABLE A-15. DUNNETT’S TEST (ONE TAILED) (CONT.) 

Degrees Total Number of Investigated Groups (

k – 1) of 

Freedom 

α 2 3 4 5 6 7 8 9 10 12 

20 .05 .10 2.03 1.64 2.19 1.81 2.30 1.93 2.39 2.01 2.46 2.08 2.51 2.14 2.56 2.19 2.60 2.23 2.64 2.26 2.70 2.33 24 .05 .10 2.01 1.63 2.17 2.80 2.28 1.91 2.36 2.00 2.43 2.06 2.48 2.12 2.53 2.17 2.57 2.21 2.60 2.24 2.66 2.30 30 .05 .10 1.99 1.62 2.15 1.79 2.25 1.90 2.34 1.98 2.40 2.05 2.45 2.10 2.50 2.15 2.54 2.19 2.57 2.22 2.63 2.28 40 .05 .10 1.97 1.61 2.13 1.77 2.23 1.88 2.31 1.96 2.37 2.03 2.42 2.08 2.47 2.13 2.51 2.17 2.54 2.20 2.60 2.26 50 .05 .10 1.96 1.61 2.11 1.77 2.22 1.88 2.29 1.96 2.32 2.02 2.41 2.07 2.45 2.12 2.49 2.16 2.52 2.19 2.58 2.25 60 .05 .10 1.95 1.60 2.10 1.76 2.21 1.87 2.28 1.95 2.34 2.01 2.40 2.06 2.44 2.11 2.48 2.15 2.51 2.18 2.57 2.24 70 .05 .10 1.95 1.60 2.10 1.76 2.21 1.87 2.28 1.95 2.34 2.01 2.40 2.06 2.44 2.11 2.48 2.15 2.51 2.18 2.56 2.24 80 .05 .10 1.94 1.60 2.10 1.76 2.20 1.87 2.28 1.95 2.34 2.01 2.39 2.06 2.43 2.10 2.47 2.15 2.50 2.18 2.55 2.23 

90 .05 .10 1.94 1.60 2.09 1.76 2.20 1.86 2.27 1.94 2.33 2.00 2.39 2.06 2.43 2.10 2.47 2.14 2.50 2.17 2.55 2.23 

100 .05 .10 1.93 1.59 2.08 1.75 2.18 1.85 2.27 1.93 2.33 1.99 2.38 2.05 2.42 2.09 2.46 2.14 2.49 2.17 2.54 2.22 

120 .05 .10 1.93 1.59 2.08 1.75 2.18 1.85 2.26 1.93 2.32 1.99 2.37 2.05 2.41 2.09 2.45 2.13 2.48 2.16 2.53 2.22 

∞ .05 

.10 

1.92 1.58 

2.06 1.73 

2.16 1.84 

2.23 1.92 

2.29 1.98 

2.34 2.03 

2.38 2.07 

2.42 2.11 

2.45 2.14 

2.50 2.20 
EPA QA/G-9S 175 February 2006 

TABLE A-16. APPROXIMATE α-LEVEL CRITICAL POINTS FOR 

RANK VON NEUMANN RATIO TEST 

α

n 0.05 0.10 

4 

5 

6 

7 

8 

9 

10 

0.70 

0.80 

0.86 

0.93 

0.98 

1.04 

0.60 

0.97 

1.11 

1.14 

1.18 

1.23 

11 

12 

13 

14 

15 

16 

17 

18 

19 

20 

1.08 

1.11 

1.14 

1.17 

1.19 

1.21 

1.24 

1.26 

1.27 

1.29 

1.26 

1.29 

1.32 

1.34 

1.36 

1.38 

1.40 

1.41 

1.43 

1.44 

21 

22 

23 

24 

25 

26 

27 

28 

29 

30 

1.31 

1.32 

1.33 

1.35 

1.36 

1.37 

1.38 

1.39 

1.40 

1.41 

1.45 

1.46 

1.48 

1.49 

1.50 

1.51 

1.51 

1.52 

1.53 

1.54 

32 

34 

36 

38 

40 

42 

44 

46 

48 

50 

1.43 

1.45 

1.46 

1.48 

1.49 

1.50 

1.51 

1.52 

1.53 

1.54 

1.55 

1.57 

1.58 

1.59 

1.60 

1.61 

1.62 

1.63 

1.63 

1.64 

55 

60 

65 

70 

75 

80 

85 

90 

95 

100 

1.56 

1.58 

1.60 

1.61 

1.62 

1.64 

1.65 

1.66 

1.66 

1.67 

1.66 

1.67 

1.68 

1.70 

1.71 

1.71 

1.72 

1.73 

1.74 

1.74 
EPA QA/G-9S 176 February 2006 

TABLE A-17. VALUES OF H1-α = H0.90 FOR COMPUTING A ONE-SIDED UPPER 90% 

CONFIDENCE LIMIT ON A LOGNORMAL MEAN

n sy

3 5 7 10 12 15 21 31 51 101 

0.10 1.686 1.438 1.381 1.349 1.338 1.328 1.317 1.308 1.301 1.295 

0.20 1.885 1.522 1.442 1.396 1.380 1.365 1.348 1.335 1.324 1.314 

0.30 2.156 1.627 1.517 1.453 1.432 1.411 1.388 1.370 1.354 1.339 

0.40 2.521 1.755 1.607 1.523 1.494 1.467 1.437 1.412 1.390 1.371 

0.50 2.990 1.907 1.712 1.604 1.567 1.532 1.494 1.462 1.434 1.409 

0.60 3.542 2.084 1.834 1.696 1.650 1.606 1.558 1.519 1.485 1.454 

0.70 4.136 2.284 1.970 1.800 1.743 1.690 1.631 1.583 1.541 1.504 

0.80 4.742 2.503 2.119 1.914 1.845 1.781 1.710 1.654 1.604 1.560 

0.90 5.349 2.736 2.280 2.036 1.955 1.880 1.797 1.731 1.672 1.621 

1.00 5.955 2.980 2.450 2.167 2.073 1.985 1.889 1.812 1.745 1.686 

1.25 7.466 3.617 2.904 2.518 2.391 2.271 2.141 2.036 1.946 1.866 

1.50 8.973 4.276 3.383 2.896 2.733 2.581 2.415 2.282 2.166 2.066 

1.75 10.48 4.944 3.877 3.289 3.092 2.907 2.705 2.543 2.402 2.279 

2.00 11.98 5.619 4.380 3.693 3.461 3.244 3.005 2.814 2.648 2.503 

2.50 14.99 6.979 5.401 4.518 4.220 3.938 3.629 3.380 3.163 2.974 

3.00 18.00 8.346 6.434 5.359 4.994 4.650 4.270 3.964 3.697 3.463 

3.50 21.00 9.717 7.473 6.208 5.778 5.370 4.921 4.559 4.242 3.965 

4.00 24.00 11.09 8.516 7.062 6.566 6.097 5.580 5.161 4.796 4.474 

4.50 27.01 12.47 9.562 7.919 7.360 6.829 6.243 5.769 5.354 4.989 

5.00 30.01 13.84 10.61 8.779 8.155 7.563 6.909 6.379 5.916 5.508 

6.00 36.02 16.60 12.71 10.50 9.751 9.037 8.248 7.607 7.048 6.555 

7.00 42.02 19.35 14.81 12.23 11.35 10.52 9.592 8.842 8.186 7.607 

8.00 48.03 22.11 16.91 13.96 12.96 12.00 10.94 10.08 9.329 8.665 

9.00 54.03 24.87 19.02 15.70 14.56 13.48 12.29 11.32 10.48 9.725 

10.00 60.04 27.63 21.12 17.43 16.17 14.97 13.64 12.56 11.62 10.79 

ALUES FOR COMPUTING A ONE-SIDED CONFIDENCE LIMIT ON A LOGNORMAL MEAN 

TABLE A-17. VALUES OF Hα = H0.10 FOR COMPUTING A ONE-SIDED LOWER 10% 

CONFIDENCE LIMIT ON A LOGNORMAL MEAN

n sy

3 5 7 10 12 15 21 31 51 101 

0.10 -1.431 -1.320 -1.296 -1.285 -1.281 -1.279 -1.277 -1.277 -1.278 -1.279 

0.20 -1.350 -1.281 -1.268 -1.266 -1.266 -1.266 -1.268 -1.272 -1.275 -1.280 

0.30 -1.289 -1.252 -1.250 -1.254 -1.257 -1.260 -1.266 -1.272 -1.280 -1.287 

0.40 -1.245 -1.233 -1.239 -1.249 -1.254 -1.261 -1.270 -1.279 -1.289 -1.301 

0.50 -1.213 -1.221 -1.234 -1.250 -1.257 -1.266 -1.279 -1.291 -1.304 -1.319 

0.60 -1.190 -1.215 -1.235 -1.256 -1.266 -1.277 -1.292 -1.307 -1.324 -1.342 

0.70 -1.176 -1.215 -1.241 -1.266 -1.278 -1.292 -1.310 -1.329 -1.349 -1.370 

0.80 -1.168 -1.219 -1.251 -1.280 -1.294 -1.311 -1.332 -1.354 -1.377 -1.403 

0.90 -1.165 -1.227 -1.264 -1.298 -1.314 -1.333 -1.358 -1.383 -1.409 -1.439 

1.00 -1.166 -1.239 -1.281 -1.320 -1.337 -1.358 -1.387 -1.414 -1.445 -1.478 

1.25 -1.184 -1.280 -1.334 -1.384 -1.407 -1.434 -1.470 -1.507 -1.547 -1.589 

1.50 -1.217 -1.334 -1.400 -1.462 -1.491 -1.523 -1.568 -1.613 -1.663 -1.716 

1.75 -1.260 -1.398 -1.477 -1.551 -1.585 -1.624 -1.677 -1.732 -1.790 -1.855 

2.00 -1.310 -1.470 -1.562 -1.647 -1.688 -1.733 -1.795 -1.859 -1.928 -2.003 

2.50 -1.426 -1.634 -1.751 -1.862 -1.913 -1.971 -2.051 -2.133 -2.223 -2.321 

3.00 -1.560 -1.817 -1.960 -2.095 -2.157 -2.229 -2.326 -2.427 -2.536 -2.657 

3.50 -1.710 -2.014 -2.183 -2.341 -2.415 -2.499 -2.615 -2.733 -2.864 -3.007 

4.00 -1.871 -2.221 -2.415 -2.596 -2.681 -2.778 -2.913 -3.050 -3.200 -3.366 

4.50 -2.041 -2.435 -2.653 -2.858 -2.955 -3.064 -3.217 -3.372 -3.542 -3.731 

5.00 -2.217 -2.654 -2.897 -3.126 -3.233 -3.356 -3.525 -3.698 -3.889 -4.100 

6.00 -2.581 -3.104 -3.396 -3.671 -3.800 -3.949 -4.153 -4.363 -4.594 -4.849 

7.00 -2.955 -3.564 -3.904 -4.226 -4.377 -4.549 -4.790 -5.037 -5.307 -5.607 

8.00 -3.336 -4.030 -4.418 -4.787 -4.960 -5.159 -5.433 -5.715 -6.026 -6.370 

9.00 -3.721 -4.500 -4.937 -5.352 -5.547 -5.771 -6.080 -6.399 -6.748 -7.136 

10.00 -4.109 -4.973 -5.459 -5.920 -6.137 -6.386 -6.730 -7.085 -7.474 -7.906 
EPA QA/G-9S 177 February 2006 

TABLE A-17. VALUES OF H1-α = H0.95 FOR COMPUTING A ONE-SIDED UPPER 95% 

CONFIDENCE LIMIT ON A LOGNORMAL MEAN

n sy

3 5 7 10 12 15 21 31 51 101 

0.10 2.750 2.035 1.886 1.802 1.775 1.749 1.722 1.701 1.684 1.670 

0.20 3.295 2.198 1.992 1.881 1.843 1.809 1.771 1.742 1.718 1.697 

0.30 4.109 2.402 2.125 1.977 1.927 1.882 1.833 1.793 1.761 1.733 

0.40 5.220 2.651 2.282 2.089 2.026 1.968 1.905 1.856 1.813 1.777 

0.50 6.495 2.947 2.465 2.220 2.141 2.068 1.989 1.928 1.876 1.830 

0.60 7.807 3.287 2.673 2.368 2.271 2.181 2.085 2.010 1.946 1.891 

0.70 9.120 3.662 2.904 2.532 2.414 2.306 2.191 2.102 2.025 1.960 

0.80 10.43 4.062 3.155 2.710 2.570 2.443 2.307 2.202 2.112 2.035 

0.90 11.74 4.478 3.420 2.902 2.738 2.589 2.432 2.310 2.206 2.117 

1.00 13.05 4.905 3.698 3.103 2.915 2.744 2.564 2.423 2.306 2.205 

1.25 16.33 6.001 4.426 3.639 3.389 3.163 2.923 2.737 2.580 2.447 

1.50 19.60 7.120 5.184 4.207 3.896 3.612 3.311 3.077 2.881 2.713 

1.75 22.87 8.250 5.960 4.795 4.422 4.081 3.719 3.437 3.200 2.997 

2.00 26.14 9.387 6.747 5.396 4.962 4.564 4.141 3.812 3.533 3.295 

2.50 32.69 11.67 8.339 6.621 6.067 5.557 5.013 4.588 4.228 3.920 

3.00 39.23 13.97 9.945 7.864 7.191 6.570 5.907 5.388 4.947 4.569 

3.50 45.77 16.27 11.56 9.118 8.326 7.596 6.815 6.201 5.681 5.233 

4.00 52.31 18.58 13.18 10.38 9.469 8.630 7.731 7.024 6.424 5.908 

4.50 58.85 20.88 14.80 11.64 10.62 9.669 8.652 7.854 7.174 6.590 

5.00 65.39 23.19 16.43 12.91 11.77 10.71 9.579 8.688 7.929 7.277 

6.00 78.47 27.81 19.68 15.45 14.08 12.81 11.44 10.36 9.449 8.661 

7.00 91.55 32.43 22.94 18.00 16.39 14.90 13.31 12.05 10.98 10.05 

8.00 104.6 37.06 26.20 20.55 18.71 17.01 15.18 13.74 12.51 11.45 

9.00 117.7 41.68 29.64 23.10 21.03 19.11 17.05 15.43 14.05 12.85 

10.00 130.8 46.31 32.73 25.66 23.35 21.22 18.93 17.13 15.59 14.26 

TABLE A-17. VALUES OF Hα = H0.05 FOR COMPUTING A ONE-SIDED LOWER 5% 

CONFIDENCE LIMIT ON A LOGNORMAL MEAN

n sy

3 5 7 10 12 15 21 31 51 101 

0.10 -2.130 -1.806 -1.731 -1.690 -1.677 -1.666 -1.655 -1.648 -1.644 -1.642 

0.20 -1.949 -1.729 -1.678 -1.653 -1.646 -1.640 -1.636 -1.636 -1.637 -1.641 

0.30 -1.816 -1.669 -1.639 -1.627 -1.625 -1.625 -1.627 -1.632 -1.638 -1.648 

0.40 -1.717 -1.625 -1.611 -1.611 -1.613 -1.617 -1.625 -1.635 -1.647 -1.662 

0.50 -1.644 -1.594 -1.594 -1.603 -1.609 -1.618 -1.631 -1.646 -1.663 -1.683 

0.60 -1.589 -1.573 -1.584 -1.602 -1.612 -1.625 -1.643 -1.662 -1.685 -1.711 

0.70 -1.549 -1.560 -1.582 -1.608 -1.622 -1.638 -1.661 -1.686 -1.713 -1.744 

0.80 -1.521 -1.555 -1.586 -1.620 -1.636 -1.656 -1.685 -1.714 -1.747 -1.783 

0.90 -1.502 -1.556 -1.595 -1.637 -1.656 -1.680 -1.713 -1.747 -1.785 -1.826 

1.00 -1.490 -1.562 -1.610 -1.658 -1.681 -1.707 -1.745 -1.784 -1.827 -1.874 

1.25 -1.486 -1.596 -1.662 -1.727 -1.758 -1.793 -1.842 -1.893 -1.949 -2.012 

1.50 -1.508 -1.650 -1.733 -1.814 -1.853 -1.896 -1.958 -2.020 -2.091 -2.169 

1.75 -1.547 -1.719 -1.819 -1.916 -1.962 -2.015 -2.088 -2.164 -2.247 -2.341 

2.00 -1.598 -1.799 -1.917 -2.029 -2.083 -2.144 -2.230 -2.318 -2.416 -2.526 

2.50 -1.727 -1.986 -2.138 -2.283 -2.351 -2.430 -2.540 -2.654 -2.780 -2.921 

3.00 -1.880 -2.199 -2.384 -2.560 -2.644 -2.740 -2.874 -3.014 -3.169 -3.342 

3.50 -2.051 -2.429 -2.647 -2.855 -2.953 -3.067 -3.226 -3.391 -3.574 -3.780 

4.00 -2.237 -2.672 -2.922 -3.161 -3.275 -3.406 -3.589 -3.779 -3.990 -4.228 

4.50 -2.434 -2.924 -3.206 -3.476 -3.605 -3.753 -3.960 -4.176 -4.416 -4.685 

5.00 -2.638 -3.183 -3.497 -3.798 -3.941 -4.107 -4.338 -4.579 -4.847 -5.148 

6.00 -3.062 -3.715 -4.092 -4.455 -4.627 -4.827 -5.106 -5.397 -5.721 -6.086 

7.00 -3.499 -4.260 -4.699 -5.123 -5.325 -5.559 -5.886 -6.227 -6.608 -7.036 

8.00 -3.945 -4.812 -5.315 -5.800 -6.031 -6.300 -6.674 -7.066 -7.502 -7.992 

9.00 -4.397 -5.371 -5.936 -6.482 -6.742 -7.045 -7.468 -7.909 -8.401 -8.953 

10.00 -4.852 -5.933 -6.560 -7.168 -7.458 -7.794 -8.264 -8.755 -9.302 -9.918 
EPA QA/G-9S 178 February 2006 

TABLE A-18. CRITICAL VALUES FOR THE SIGN TEST 

The table values are such that for order statistics x[lower] and x[upper], [] [] P(x lower < true median < x upper ) ≥ 1−α . Note 

that the significance levels are for two-sided tests. For one-sided test, divide the significance level in half. 

α

n 0.20 0.10 0.05 0.02 0.01 

lower upper lower upper lower upper lower upper lower upper 

5 1 5 1 5 - - - - - - 

6 1 6 1 6 1 6 - - - - 

7 2 6 1 7 1 7 1 7 - - 

8 2 7 2 7 1 8 1 8 1 8 

9 3 7 2 8 2 8 1 9 1 9 

10 3 8 2 9 2 9 1 10 1 10 

11 3 9 3 9 2 10 2 10 1 11 

12 4 9 3 10 3 10 2 11 2 11 

13 4 10 4 10 3 11 2 12 2 12 

14 5 10 4 11 3 12 3 12 2 13 

15 5 11 4 12 4 12 3 13 3 13 

16 5 12 5 12 4 13 3 14 3 14 

17 6 12 5 13 5 13 4 14 3 15 

18 6 13 6 13 5 14 4 15 4 15 

19 7 13 6 14 5 15 5 15 4 16 

20 7 14 6 15 6 15 5 16 4 17 

21 8 14 7 15 6 16 5 17 5 17 

22 8 15 7 16 6 17 6 17 5 18 

23 8 16 8 16 7 17 6 18 5 19 

24 9 16 8 17 7 18 6 19 6 19 

25 9 17 8 18 8 18 7 19 6 20 

26 10 17 9 18 8 19 7 20 7 20 

27 10 18 9 19 8 20 8 20 7 21 

28 11 18 10 19 9 20 8 21 7 22 

29 11 19 10 20 9 21 8 22 8 22 

30 11 20 11 20 10 21 9 22 8 23 

31 12 20 11 21 10 22 9 23 8 24 

32 12 21 11 22 10 23 9 24 9 24 

33 13 21 12 22 11 23 10 24 9 25 

34 13 22 12 23 11 24 10 25 10 25 

35 14 22 13 23 12 24 11 25 10 26 

36 14 23 13 24 12 25 11 26 10 27 

37 15 23 14 24 13 25 11 27 11 27 

38 15 24 14 25 13 26 12 27 11 28 

39 16 24 14 26 13 27 12 28 12 28 

40 16 25 15 26 14 27 13 28 12 29 
EPA QA/G-9S 179 

TABLE A-19. CRITICAL VALUES FOR THE QUANTILE TEST 

m is the number of background samples, n is the number of site samples, and c is the number values larger than the th b1 quantile. 

If s is greater than or equal to the table value, qα, then reject the null hypothesis of no difference at the given significance level. 

N 

4 5 6 7 8 9 10 11 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

c 4 2 1 4 2 1 5 3 1 5 3 1 6 3 2 6 3 2 7 4 2 7 4 

0.10 7 0.05 4 4 5 5 6 6 7 

4 

0.01 

c 4 2 1 5 3 1 5 3 1 6 3 2 6 3 2 7 4 2 7 4 2 8 4 

0.10 3 0.05 5 6 7 7 8 

5 

0.01 4 5 6 

c 5 3 1 5 3 1 6 3 2 6 3 2 7 4 2 7 4 2 8 4 2 8 4 

0.10 4 3 3 5 6 4 6 4 7 7 0.05 4 3 5 

6 

0.01 5 6 6 7 7 8 8 

c 5 3 1 6 3 2 6 3 2 7 4 2 7 4 2 8 4 2 8 4 2 9 5 

0.10 3 4 4 4 5 0.05 4 3 3 5 6 4 6 7 7 8 

7 

0.01 5 6 7 7 8 8 9 

c 6 3 2 6 3 2 7 4 2 7 4 2 8 4 2 8 4 2 9 5 2 9 5 

0.10 2 4 5 6 4 7 0.05 4 3 3 4 4 4 7 5 8 5 

8 

0.01 5 6 6 7 8 8 9 

c 6 3 2 7 4 2 7 4 2 8 4 2 8 4 2 9 5 2 9 5 2 10 5 

0.10 2 3 7 0.05 4 3 5 5 4 6 4 6 4 7 5 5 8 5 

9 

0.01 4 6 7 7 8 8 9 

c 7 4 2 7 4 2 8 4 2 8 4 2 9 5 2 9 5 2 10 5 2 10 5 

0.10 2 3 2 5 6 4 7 0.05 4 3 6 4 7 5 8 5 8 5 

10 

0.01 4 5 4 6 4 7 7 5 8 9 9 

c 7 4 2 8 4 2 8 4 2 9 5 2 9 5 2 10 5 2 10 5 2 11 6 

0.10 2 3 2 3 6 4 4 7 5 0.05 4 3 5 5 6 4 7 5 8 

11 

0.01 4 4 6 4 7 5 7 5 8 5 8 9 6 

c 8 4 2 8 4 2 9 5 2 9 5 2 10 5 2 10 5 2 11 6 3 11 6 

0.10 3 2 5 2 6 4 4 7 3 5 0.05 4 3 2 4 6 4 7 8 5 8 

m

12 

0.01 4 5 4 6 5 7 5 7 5 8 5 9 6 9 6 
EPA QA/G-9S 180 

TABLE A-19. CRITICAL VALUES FOR THE QUANTILE TEST (CONT.) 

m is the number of background samples, n is the number of site samples, and c is the number values larger than the th b1 quantile. 

If s is greater than or equal to the table value, qα, then reject the null hypothesis of no difference at the given significance level. 

n 

4 5 6 7 8 9 10 11 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

c 8 4 2 9 5 2 9 5 2 10 5 2 10 5 2 11 6 3 11 6 3 12 6 

0.10 3 2 2 6 3 7 3 0.05 4 3 2 5 5 4 6 4 4 7 5 8 5 8 5 

13 

0.01 4 4 6 5 7 5 7 5 8 6 9 6 9 6 

c 9 5 2 9 5 2 10 5 2 10 5 2 11 6 3 11 6 3 12 6 3 12 6 

0.10 3 2 5 2 2 6 4 3 0.05 4 3 2 5 4 6 4 7 5 3 7 5 3 8 5 8 5 

14 

0.01 4 4 6 5 7 5 8 6 8 6 9 6 9 6 

c 9 5 2 10 5 2 10 5 2 11 6 3 11 6 3 12 6 3 12 6 3 13 7 

0.10 3 2 5 2 4 6 4 7 3 8 5 0.05 4 3 2 5 4 6 3 3 7 5 3 8 5 6 

15 

0.01 4 4 6 5 7 5 7 5 8 6 9 6 9 7 

c 10 5 2 10 5 2 11 6 3 11 6 3 12 6 3 12 6 3 13 7 3 13 7 

0.10 3 5 6 4 4 5 5 0.05 4 3 2 5 2 4 3 6 4 3 7 3 7 5 3 8 3 8 

16 

0.01 4 4 6 5 7 5 8 5 8 6 9 6 9 6 

c 10 5 2 11 6 3 11 6 3 12 6 3 12 6 3 13 7 3 13 7 3 14 7 

0.10 5 6 4 8 5 0.05 4 3 2 5 4 3 6 4 3 3 7 5 3 8 5 3 

17 

0.01 4 4 3 6 5 7 5 7 5 8 6 9 6 9 6 

c 11 6 3 11 6 3 12 6 3 12 6 3 13 7 3 13 7 3 14 7 3 14 7 

0.10 2 3 3 5 5 0.05 4 3 5 4 6 4 3 7 5 3 7 5 3 8 5 3 8 

18 

0.01 4 3 4 6 5 3 7 5 8 6 8 6 9 6 9 6 

c 11 6 3 12 6 3 12 6 3 13 7 3 13 7 3 14 7 3 14 7 3 15 8 

0.10 2 3 2 5 4 6 4 7 7 8 5 0.05 4 3 5 4 6 3 7 5 3 5 3 8 5 3 9 6 

19 

0.01 4 3 4 3 6 5 3 7 5 8 6 8 6 9 6 10 7 

c 12 6 3 12 6 3 13 7 3 13 7 3 14 7 3 14 7 3 15 8 3 15 8 

0.10 2 3 2 5 4 8 5 0.05 4 3 5 4 6 4 3 7 3 7 5 3 8 5 3 6 

m

20 

0.01 4 3 4 3 6 5 3 7 5 8 5 8 6 9 6 9 7 
EPA QA/G-9S 181 

TABLE A-19. CRITICAL VALUES FOR THE QUANTILE TEST (CONT.) 

m is the number of background samples, n is the number of site samples, and c is the number values larger than the th b1 quantile. 

If s is greater than or equal to the table value, qα, then reject the null hypothesis of no difference at the given significance level. 

n 

13 14 15 16 17 18 19 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

c 8 4 2 9 5 2 9 5 2 10 5 2 10 5 2 11 6 3 11 6 3 

0.10 8 9 10 11 0.05 9 10 11 

4 

0.01 

c 9 5 2 9 5 2 10 5 2 10 5 2 11 6 3 11 6 3 12 6 3 

0.10 0.05 9 9 10 10 11 11 12 

5 

0.01 

c 9 5 2 10 5 2 10 5 2 11 6 3 11 6 3 12 6 3 12 6 3 

0.10 8 9 9 10 10 11 11 0.05 

6 

0.01 9 10 10 11 11 12 12 

c 10 5 2 10 5 2 11 6 3 11 6 3 12 6 3 12 6 3 13 7 3 

0.10 5 5 6 6 6 7 0.05 9 9 10 10 11 11 12 

7 

0.01 10 10 11 11 12 12 13 

c 10 5 2 11 6 3 11 6 3 12 6 3 12 6 3 13 7 3 13 7 3 

0.10 5 9 10 6 6 7 0.05 9 10 6 10 6 11 11 12 7 12 

8 

0.01 10 11 11 12 12 13 13 

c 11 6 3 11 6 3 12 6 3 12 6 3 13 7 3 13 7 3 14 7 3 

0.10 9 10 11 12 0.05 9 6 6 10 6 6 11 7 7 7 

9 

0.01 10 10 11 11 12 12 13 

c 11 6 3 12 6 3 12 6 3 13 7 3 13 7 3 14 7 3 14 7 3 

0.10 0.05 9 6 10 6 10 6 11 7 11 7 12 7 12 7 

10 

0.01 10 11 11 12 12 13 13 

c 12 6 3 12 6 3 13 7 3 13 7 3 14 7 3 14 7 3 15 8 3 

0.10 9 10 6 10 11 11 12 0.05 9 6 10 6 11 7 7 12 7 13 8 

11 

0.01 10 11 11 7 12 12 13 14 

c 12 6 3 13 7 3 13 7 3 14 7 3 14 7 3 15 8 3 15 8 3 

0.10 5 6 6 6 7 7 0.05 9 10 10 11 11 7 12 12 

m

12 

0.01 10 6 11 7 11 7 12 7 12 13 8 13 8 
EPA QA/G-9S 182 

TABLE A-19. CRITICAL VALUES FOR THE QUANTILE TEST (CONT.) 

m is the number of background samples, n is the number of site samples, and c is the number values larger than the th b1 quantile. 

If s is greater than or equal to the table value, qα, then reject the null hypothesis of no difference at the given significance level. 

n 

13 14 15 16 17 18 19 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

0.50 

0.75 

0.90 

c 13 7 3 13 7 3 14 7 3 14 7 3 15 8 3 15 8 3 16 8 4 

0.10 9 9 10 6 10 6 11 11 7 12 7 0.05 6 10 6 11 11 12 7 12 13 

13 

0.01 10 7 11 7 12 7 12 7 13 8 13 8 14 8 

c 13 7 3 14 7 3 14 7 3 15 8 3 15 8 3 16 8 4 16 8 4 

0.10 3 4 12 7 4 0.05 9 6 10 6 10 6 11 7 11 7 12 7 

14 

0.01 10 7 11 7 11 7 12 8 12 8 13 8 13 8 

c 14 7 3 14 7 3 15 8 3 15 8 3 16 8 4 16 8 4 17 9 4 

0.10 9 3 9 3 10 11 4 4 12 4 0.05 10 6 10 6 11 7 11 7 12 7 12 7 13 8 

15 

0.01 11 7 11 7 12 8 12 8 13 8 13 8 14 9 

c 14 7 3 15 8 3 15 8 3 16 8 4 16 8 4 17 9 4 17 9 4 

0.10 3 6 3 10 6 4 11 4 7 4 12 4 0.05 9 6 10 7 11 7 12 7 12 8 13 8 

16 

0.01 10 7 11 7 11 8 12 8 13 8 13 9 14 9 

c 15 8 3 15 8 3 16 8 4 16 8 4 17 9 4 17 9 4 18 9 4 

0.10 9 3 6 3 10 6 6 11 7 4 7 4 12 7 4 0.05 10 6 10 11 7 4 11 7 4 12 12 8 13 8 

17 

0.01 11 7 11 7 12 8 12 8 13 8 13 9 14 9 

c 15 8 3 16 8 4 16 8 4 17 9 4 17 9 4 18 9 4 18 9 4 

0.10 9 3 10 6 11 7 4 12 7 4 0.05 6 10 6 4 11 4 11 7 4 12 7 4 12 13 8 

18 

0.01 10 7 11 7 12 7 12 8 13 8 13 8 14 9 

c 16 8 4 16 8 4 17 9 4 17 9 4 18 9 4 18 9 4 19 10 4 

0.10 9 10 11 7 12 4 0.05 10 6 4 10 6 4 11 7 4 11 7 4 12 7 4 12 4 13 8 

19 

0.01 11 7 11 7 12 8 12 8 13 8 13 8 14 9 

c 16 8 4 17 9 4 17 9 4 18 9 4 18 9 4 19 10 4 19 10 4 

0.10 9 6 10 11 12 7 12 0.05 10 6 4 10 7 4 11 7 4 11 7 4 12 7 4 13 8 4 13 8 4 

m

20 

0.01 11 7 11 8 12 8 12 8 13 8 14 9 14 9 
EPA QA/G-9S 183 February 2006 

APPENDIX B: 

REFERENCES 
EPA QA/G-9S 184 February 2006 

APPENDIX B: REFERENCES 

This appendix provides references for the topics and procedures described in this 

document. The references are broken into three groups: Primary, Basic Statistics Textbooks, and 

Secondary. This classification does not refer in any way to the subject matter content but to the 

relevance to the intended audience for this document, ease in understanding statistical concepts 

and methodologies, and accessibility to the non-statistical community. Primary references are 

those thought to be of particular benefit as hands-on material, where the degree of sophistication 

demanded by the writer seldom requires extensive training in statistics; most of these references 

should be on an environmental statistician’s bookshelf. Users of this document are encouraged 

to send recommendations on additional references to the address listed in the Foreword. 

Some sections within the chapters reference materials found in most introductory 

statistics books. This document uses Walpole and Myers (1985), Freedman, Pisani, Purves, and 

Adhakari (1991), Mendenhall (1987), and Dixon and Massey (1983). Table B-1 (at the end of 

this appendix) lists specific chapters in these books where topics contained in this guidance may 

be found. This list could be extended much further by use of other basic textbooks; this is 

acknowledged by the simple statement that further information is available from introductory 

text books. 

Some important books specific to the analysis of environmental data include: Gilbert 

(1987), an excellent all-round handbook having strength in sampling, estimation, and hot-spot 

detection; Gibbons (1994), a book specifically concentrating on the application of statistics to 

groundwater problems with emphasis on method detection limits, censored data, and the 

detection of outliers; and Madansky (1988), a slightly more theoretical volume with important 

chapters on the testing for Normality, transformations, and testing for independence. In addition, 

Ott (1995) describes modeling, probabilistic processes, and the Lognormal distribution of 

contaminants, and Berthouex and Brown (1994) provide an engineering approach to problems 

including estimation, experimental design and the fitting of models. Millard and Neerchal 

(2001) has excellent discussions of applied statistical methods using the software package SPlus. Ginevan and Splitstone (2004) contains applied examples of many of the statistical tests 

discussed in this guidance and includes some more sophisticated tests for the more statistically 

minded. Gibbons and Coleman (2001) apply sophisticated statistical theory to environmental 

issues, but the text is not intended for the casual reader. 

B.1 CHAPTER 1

Chapter 1 establishes the framework of qualitative and quantitative criteria against which 

the data that has been collected will be assessed. The most important feature of this chapter is 

the concept of the test of hypotheses framework which is described in any introductory textbook. 

 A non-technical exposition of hypothesis testing is also to be found in U.S. EPA (2000, 1994b) 

which provides guidance on planning for environmental data collection. An application of the 

DQO Process to geostatistical error management may be found in Myers (1997). 
EPA QA/G-9S 185 February 2006 

A full discussion of sampling methods with the attendant theory are to be found in 

Gilbert (1987) and a shorter discussion may be found in U.S. EPA (1989). Cochran (1966) and 

Kish (1965) also provide more advanced theoretical concepts but may require the assistance of a 

statistician for full comprehension. More sophisticated sampling designs such as composite 

sampling, adaptive sampling, and ranked set sampling, will be discussed in future Agency 

guidance. 

B.2 CHAPTER 2

Standard statistical quantities and graphical representations are discussed in most 

introductory statistics books. In addition, Berthouex & Brown (1994) and Madansky (1988) 

both contain thorough discussions on the subject. There are also several textbooks devoted 

exclusively to graphical representations, including Cleveland (1993), which may contain the 

most applicable methods for environmental data, Tufte (1983), and Chambers, Cleveland, 

Kleiner and Tukey (1983). 

Two EPA sources for temporal data that keep theoretical discussions to a minimum are 

U.S. EPA (1992a) and U.S. EPA (1992b). For a more complete discussion on temporal data, 

specifically time series analysis, see Box and Jenkins (1970), Wei (1990), or Ostrum (1978). 

These more complete references provide both theory and practice; however, the assistance of a 

statistician may be needed to adapt the methodologies for immediate use. Theoretical 

discussions of spatial data may be found in Journel and Huijbregts (1978), Cressie (1993), and 

Ripley (1981). 

B.3 CHAPTER 3

The hypothesis tests covered in this edition of the guidance are well known and straightforward; basic statistics texts cover these subjects. Besides basic statistical text books, 

Berthouex & Brown (1994), Hardin and Gilbert (1993), and U.S. EPA (1989, 1994) may be 

useful to the reader. In addition, there are some statistics books devoted specifically to 

hypothesis testing, for example, see Lehmann (1991). These books may be too theoretical for 

most practitioners, and their application to environmental situations may not be obvious. 

The statement in this document that the sign test requires approximately 1.225 times as 

many observations as the Wilcoxon rank sum test to achieve a given power at a given 

significance level is attributable to Lehmann (1975). 

B.4 CHAPTER 4

This chapter is essentially a compendium of statistical tests drawn mostly from the 

primary references and basic statistics textbooks. Gilbert (1987) and Madansky (1988) have an 

excellent collection of techniques and U.S. EPA (1992a) contains techniques specific to water 

problems. 
EPA QA/G-9S 186 February 2006 

For Normality (Section 4.2), Madansky (1988) has an excellent discussion on tests as 

does Shapiro (1986). For trend testing (Section 4.3), Gilbert (1987) has an excellent discussion 

on statistical tests and U.S. EPA (1992) provides adjustments for trends and seasonality in the 

calculation of descriptive statistics. 

There are several very good textbooks devoted to the treatment of outliers (Section 4.4). 

Two authoritative texts are Barnett and Lewis (1978) and Hawkins (1980). Additional 

information is also to be found in Beckman and Cook (1983) and Tietjen and Moore (1972). 

Tests for dispersion (Section 4.5) are described in the basic textbooks and examples are 

to be found in U.S. EPA (1992a). Transformation of data (Section 4.6) is a sensitive topic and 

thorough discussions may be found in Gilbert (1987), and Dixon and Massey (1983). Equally 

sensitive is the analysis of data where some values are recorded as non-detected (Section 4.7); 

Gibbons (1994) and U.S. EPA (1992a) have relevant discussions and examples. 

B.5 CHAPTER 5

Chapter 5 discusses some of the philosophical issues related to hypothesis testing which 

may help in understanding and communicating the test results. Although there are no specific 

references for this chapter, many topics (e.g., the use of p-values) are discussed in introductory 

textbooks. Future editions of this guidance will be expanded by incorporating practical 

experiences from the environmental community into this chapter. 

B.6 LIST OF REFERENCES

B.6.1 Primary References

Berthouex, P.M., and L.C. Brown, 1994. Statistics for Environmental Engineers. Lewis, Boca 

Raton, FL. 

Gibbons, R.D. and D. E. Coleman, 2001. Statistical Methods for Detection and Quantification 

of Environmental Contamination. John Wiley, New York, NY. 

Gilbert, R.O., 1987. Statistical Methods for Environmental Pollution Monitoring. John Wiley, 

New York, NY. 

Ginevan, M.E. and D.E. Splitstone, 2004. Statistical Tools for Environmental Quality 

Measurement. Chapman and Hall, Boca Raton, FL. 

Myers, J.C., 1997. Geostatistical Error Management, John Wiley, New York, NY. 

Ott, W.R., 1995. Environmental Statistics and Data Analysis. Lewis, Boca Raton, FL. 
EPA QA/G-9S 187 February 2006 

U.S. Environmental Protection Agency , 1992a. Addendum to Interim Final Guidance 

Document on the Statistical Analysis of Ground-Water Monitoring Data at RCRA 

Facilities. EPA/530/R-93/003. Office of Solid Waste. (NTIS: PB89-151026) 

U.S. Environmental Protection Agency, 2000. Guidance for the Data Quality Objectives 

Process (EPA QA/G4). EPA/600/R-96/055. Office of Research and Development. 

U.S. Environmental Protection Agency, 2002. Choosing a Sampling Design for Environmental 

Data Collection (EPA QA/G5S). Office of Environmental Information. 

U.S. Environmental Protection Agency, 2004. Data Quality Assessment: A Reviewer’s Guide

(Final Draft) (EPA QA/G9R). Office of Environmental Information. 

B.6.2 Basic Statistics Textbooks

Dixon, W.J., and F.J. Massey, Jr., 1983. Introduction to Statistical Analysis (Fourth Edition). 

McGraw-Hill, New York, NY. 

Freedman, D. R. Pisani, R. Purves, and A. Adhikari, 1991. Statistics. W.W. Norton & Co., 

New York, NY. 

Mendenhall, W., 1987. Introduction to Probability and Statistics (Seventh Edition). PWS-Kent, 

Boston, MA. 

Walpole, R., and R. Myers, 1985. Probability and Statistics for Engineers and Scientists (Third 

Ed.). MacMillan, New York, NY. 

B.6.3 Secondary References

Aitchison, J., 1955. On the distribution of a positive random variable having a discrete 

probability mass at the origin. Journal of American Statistical Association 50(272):901-8 

Barnett, V., and T. Lewis, 1978. Outliers in Statistical Data. John Wiley, New York, NY. 

Beckman, R.J., and R.D. Cook, 1983. Outlier......s, Technometrics 25:119-149. 

Box, G.E.P., and G.M. Jenkins, 1970. Time Series Analysis, Forecasting, and Control. HoldenDay, San Francisco, CA. 

Chambers, J.M, W.S. Cleveland, B. Kleiner, and P.A. Tukey, 1983. Graphical Methods for 

Data Analysis. Wadsworth & Brooks/Cole Publishing Co., Pacific Grove, CA. 

Chen, L., 1995. Testing the mean of skewed distributions, Journal of the American Statistical 

Association 90:767-772. 
EPA QA/G-9S 188 February 2006 

Cleveland, W.S., 1993. Visualizing Data. Hobart Press, Summit, NJ. 

Cochran, W.G., 1966. Sampling Techniques (Third Edition). John Wiley, New York, NY. 

Cohen, A.C., Jr. 1959. Simplified estimators for the normal distribution when samples are 

singly censored or truncated, Technometrics 1:217-237. 

Conover, W.J., 1980. Practical Nonparametric Statistics (Second Edition). John Wiley, New 

York, NY. 

Cressie, N., 1993. Statistics for Spatial Data. John Wiley, New York, NY. 

D'Agostino, R.B., 1971. An omnibus test of normality for moderate and large size samples, 

Biometrika 58:341-348. 

David, H.A., H.O. Hartley, and E.S. Pearson, 1954. The distribution of the ratio, in a single 

normal sample, of range to standard deviation, Biometrika 48:41-55. 

Dixon, W.J., 1953. Processing data for outliers, Biometrika 9:74-79. 

Filliben, J.J., 1975. The probability plot correlation coefficient test for normality, Technometrics

17:111-117. 

Geary, R.C., 1947. Testing for normality, Biometrika 34:209-242. 

Geary, R.C., 1935. The ratio of the mean deviation to the standard deviation as a test of 

normality, Biometrika 27:310-32. 

Gibbons, R. D., 1994. Statistical Methods for Groundwater Monitoring. John Wiley, New 

York, NY. 

Grubbs, F.E., 1969. Procedures for detecting outlying observations in samples, Technometrics

11:1-21. 

Hardin, J.W., and R.O. Gilbert, 1993. Comparing Statistical Tests for Detecting Soil 

Contamination Greater than Background, Report to U.S. Department of Energy, PNL8989, UC-630, Pacific Northwest Laboratory, Richland, WA. 

Hawkins, D.M., 1980. Identification of Outliers. Chapman and Hall, New York, NY. 

Hochberg, Y., and A. Tamhane, 1987. Multiple Comparison Procedures. John Wiley, New 

York, NY. 

Journel, A.G., and C.J. Huijbregts, 1978. Mining Geostatistics. Academic Press, London. 

Kish, L., 1965. Survey Sampling. John Wiley, New York, NY. 
EPA QA/G-9S 189 February 2006 

Kleiner, B., and J.A. Hartigan, 1981. Representing points in many dimensions by trees and 

castles, Journal of the American Statistical Association 76:260. 

Lehmann, E.L., 1991. Testing Statistical Hypotheses. Wadsworth & Brooks/Cole Publishing 

Co., Pacific Grove, CA. 

Lehmann, E.L., 1975. Nonparametrics: Statistical Methods Based on Ranks. Holden-Day, Inc., 

San Francisco, CA. 

Lilliefors, H.W., 1969. Correction to the paper "On the Kolmogorov-Smirnov test for normality 

with mean and variance unknown," Journal of the American Statistical Association 

64:1702. 

Lilliefors, H.W., 1967. On the Kolmogorov-Smirnov test for normality with mean and variance 

unknown, Journal of the American Statistical Association 64:399-402. 

Madansky, A., 1988. Prescriptions for Working Statisticians. Springer-Verlag, New York, NY. 

Millard, S.P. and N.K. Neerchal, 2001. Environmental Statistics with S-Plus. CRC Press, Boca 

Raton, FL. 

Ostrum, C.W., 1978. Time Series Analysis (Second Edition). Sage University Papers Series, 

Vol 9. Beverly Hills and London. 

Ripley, B.D., 1981. Spatial Statistics. John Wiley and Sons, Somerset, NJ. 

Rosner, B., 1975. On the detection of many outliers, Technometrics 17:221-227. 

Royston, J.P., 1982. An extension of Shapiro and Wilk's W test for normality to large samples, 

Applied Statistics 31:161-165. 

Sen, P.K., 1968a. Estimates of the regression coefficient based on Kendall's tau, Journal of the 

American Statistical Association 63:1379-1389. 

Sen, P.K., 1968b. On a class of aligned rank order tests in two-way layouts, Annals of 

Mathematical Statistics 39:1115-1124. 

Shapiro, S., 1986. Volume 3: How to Test Normality and Other Distributional Assumptions. 

American Society for Quality Control, Milwaukee, WI. 

Shapiro, S., and M.B. Wilk, 1965. An analysis of variance test for normality (complete 

samples), Biometrika 52:591-611. 

Stefansky, W., 1972. Rejecting outliers in factorial designs, Technometrics 14:469-478. 
EPA QA/G-9S 190 February 2006 

Stephens, M.A., 1974. EDF statistics for goodness-of-fit and some comparisons, Journal of the 

American Statistical Association 90:730-737. 

Tietjen, G.L., and R.M. Moore, 1972. Some Grubbs-type statistics for the detection of several 

outliers, Technometrics 14:583-597. 

Tufte, E.R., 1983. The Visual Display of Quantitative Information. Graphics Press, Cheshire, 

CN. 

Tufte, E.R., 1990. Envisioning Information. Graphics Press, Cheshire, CN. 

U. S. Environmental Protection Agency, 1989. Methods for Evaluating the Attainments of 

Cleanup Standards: Volume 1: Soils and Solid Media. EPA/230/02-89-042. Office of 

Policy, Planning, and Evaluation. (NTIS: PB89-234959)

U. S. Environmental Protection Agency, 1992b. Methods for Evaluating the Attainments of 

Cleanup Standards: Volume 2: Ground Water. EPA/230/R-92/014. Office of Policy, 

Planning, and Evaluation. (NTIS: PB94-138815) 

U. S. Environmental Protection Agency, 1994. Methods for Evaluating the Attainments of 

Cleanup Standards: Volume 3: Reference-Based Standards. EPA/230/R-94-004. Office 

of Policy, Planning, and Evaluation. (NTIS: PB94-176831) 

Walsh, J.E., 1958. Large sample nonparametric rejection of outlying observations, Annals of the 

Institute of Statistical Mathematics 10:223-232. 

Walsh, J.E., 1953. Correction to "Some nonparametric tests of whether the largest observations 

of a set are too large or too small," Annals of Mathematical Statistics 24:134-135. 

Walsh, J.E., 1950. Some nonparametric tests of whether the largest observations of a set are too 

large or too small, Annals of Mathematical Statistics 21:583-592. 

Wang, Peter C.C., 1978. Graphical Representation of Multivariate Data. Academic Press, New 

York, NY. 

Wegman, Edward J., 1990. Hyperdimensional data analysis using parallel coordinates, Journal 

of the American Statistical Association 85: 664. 

Wei, W.S., 1990. Time Series Analysis (Second Edition). Addison Wesley, Menlo Park, CA.