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NATIONAL BUREAU OF STANDARDS REPORT

1565

CONFIDENCE AND TOLERANCE INTERVALS

FOR THE NORMAL DISTRIBUTION

DV

Franc Pros chan

U. S. DEPARTMENT OF COMMERCE

NATIONAL BUREAU OF STANDARDS
U. S. DEPARTMENT OF COMMERCE

Charles Sawyer, Secretary

NATIONAL BUREAU OF STANDARDS

A. V. As tin. Acting Director

THE NATIONAL BUREAU OF STANDARDS

The scope of activities of the National &>reau of Standards is suggested in the following listing of the divisions and sections engaged in technical work. In general, each section is en- gaged in specialised research, development, and engineering in the field indicated by its

title. A brief description of the activities, and of the resultant reports and publications,

appears on the inside of the back cover of this report.

s. ELECTRICITY. Resistance Measurements. Inductance and Capacitance. Electrical Instruments. Magnetic Measurements. Electrochemistry.

2. OPTICS AMD METROLOGY. Photometry and Colorimetry. Optical Instruments. Photographic

Technology. Length. Gage.

3 . HEAT AMD POWER. Teaperature Measurements . Thermodynamics . Cryogenics. Engines and

Lubrication. Engine Fuels.

ATOMIC AND RADIATION PHYSICS. Spectroscopy. Radioroe try. Mass Spectrometry. Physical Electronics. Electron Physics. Atomic Physics. Neutron Measurements. Nuclear Physics. Radioactivity. X-Rays. Betatron. Nucleonic Instrumentation. Radiological Equipment. Atomic Energy Commission Instruments Branch.

5« CHEMISTRY, Organic Coatings. Surface Chemistry. Organic Chemistry. Analytical Chemistry. Inorganic Chemistry. Electrodeposition. Gas Chemistry. Physical Chemistry. Thermochemistry. Spectrochemistry. Pure Substances.

6® MECHANICS. Sound. Mechanical Instruments. Aerodynamics. Engineering Mechanics. Hydraulics. Mass. 'Capaci ty. Density, end Fluid Meters. 7. ORGANIC km FIBROUS MATERIALS. Rubber. Textiles. Paper. Leather. Testing and

Specifications. Organic Plastics. Dental Research.

3 . METALLURGY. Thermal Metallurgy. Chemical Metallurgy. Mechanical Metallurgy . Corrosion.

9* MINERAL PRODUCTS. Porcelain and Pottery. Glass. Refractories. Enameled Metals. Building Stoae. Concreting Materials. Constitution and Microstnicture. Chemistry of Mineral

Products.

10. BUILDING TECHNOLOGY. Structural Engineering. Fire Protection. Heating and Air Con- ditioning. Exterior and Interior Coverings. Codes and Specifications.

11. APPLIED MATHEMATICS. Numerical Analysis. Confutation. Statistical Engineering. Ma- chine Developeeat.

82. ELECTRONICS. Engineering Electronics. Electron Ihbes. Electronic Computers. Elecfaraaie IutmieatatioEa.

13 . OjRDNAMCE DEVELOPMENT. Mechanical Research and Development. Electromechanical

Fares. Technical Service®. Missile Fusing Research. Missile Fusing Development. Pro- jectile Fuses. Ordnance Coefonem ts . Ordnance Tests. Ordnance Research.

RADIO PROPAGATION. Upper Ataosphere Research. Ionospheric Research. Regular Propagation Services. Frequency Utilisation Research. Tropospheric Propagation Research. High

Frequency Standards. Microwave Standards.

85. M8SSSLE DEVELOPMENT. Missile Engineering. Missile Dynamics. Missile Intelligence.

Missile Instrumentation. Technical Services. Combustion.
NBS PROJECT

3011 - 60-0002

3 April 1952 NBS REPORT

1565

CONFIDENCE AND TOLERANCE INTERVALS

FOR THE NORMAL DISTRIBUTION

by

Frank Pros chan

Approved for public release by the

This report i. issued Director of the National Institute of

in any form, either in Standards and Technology (NIST)

fro. the Office of the 0 n October 9, 2015

ion, reprinting, or reproduction

emission in writing is obtained

Washington 25, D. C.

FOREWORD

This is the text of an incited address, "On intervals of

the- form x * ks, n presented bj F« Proschan of the Statistical

Engineering Laboratory (Section 3 of Division 11, Applied

Hatheimtiss ) at the 111th Annual Meeting of the American

Statistical Association, Bheton, Massachusetts * SB December

* 1951 » It will appear in published form at a later date In th®

Journal of the, American Statistical Association ^ Confidence and tolerance intervals for the normal

distribution are presented for the various cases of known

and unknown mean and -standard deviation « Practical illuai tration and Interpretation of these intervals are given®

Tables are presented permitting a comparison among th© intervals o Finally 8 the relationship between the two types of

intervals is described c

Jo H« Curtiss

Chief, National Applied

Mathematics Laboratories

A « Y* Astin

Acting Director

Rational Bureau ©X' Standards
. II- I III1IIIs • 
CONFIDENCE AND TOLERANCE INTERVALS

FOR THE NORMAL DISTRIBUTION

by

Frank' Proschan

I, Introduction 0 Di scissions of the theory of errors

will sometimes state that the mean + the probable error will

include 50 percent of future observations (assumed normally

distributed) , This* of course* is true only if the mean and

- » the probable error of the population Itself are used® TJnfor=*

tunately, in most practical problems, one or both of those

may not be known® Experimenters who use the sample mean + the

\ samel e probable error with th© expectation that this Interval

will contain 50 percent of future observations may be seriously

deluding themselves®

However it Is possS.ble to construct intervals of the tvpe

x ± ks (1 sa sample mean* s * sample standard deviation} which

will * on the average s Include 50 percent or the population©

From this, one Is led to a more general consideration of such

Intervals* and to the uses to which they can be put®

2* Summary. « All populations discussed in this paper are

normal unless otherwise specified* Let>u* o refer to the popular!-,

mean and standard deviation respectively*

Any one of four possible situations may exists (a) m9 o

both known; (b)iu unknown, O known; {c} ax known, o unknown $

(d)Ai* c both unknown

Let m represent either >u or xj let s*d 0 represent either
f

1

I

I

1

I

I

I

I

I

I

I
ffl 2 6 Pro schan

o or -So

r? ben two Important typos of assertions, my be made

about internals of the form

m ? k s.d* (X)

A 0 Oonf id enc 8 X nt erval <, The probability is / that

the Interval (1) contains the population mean {or alternately*

the second sample mean) e

Bo Tolerance Interval * In repeated samples* the

proportion, p5 of the population contained in (1) is

Bol) a* on the average,

B*2) P, or more, * of the time*

In this paper, a. comparison is made among the values of k

aporopriate to the respective cases obtained from various com®

binations ©f A and B with (a), Cb), (o), and (d)„ Practical

Illustrations and interpretations are given of these cases*

In addition, details of a proof are given of a result by

Wilks (1941) for the case B*l» These details are given because

they are suggestive of & general method applicable in such

problems® Also a table is present ad of values of k for combination

B.l(d) where E(p) ss a(as«50, *75, *90, 0 95 g £>99, *999) and sample

size n as 2(1)50,40,60,120,® 0 Finally the relationship between confidence intervals and

tolerance intervals is discussed*

5* Confidence Intervals 0 A cherist makes n determinations

of the Iron content of a solution* '.'.hat Interval shall he

select so that he can assert with 50 percent confidence that the

M true'* value .-to lies within that interval? The distribution of
-,v ... •• •

• — • - - ' - -

1

I

I

'

LsolSo^S b

)

:

I

I

S

-m

i

ao&a&aMmoo to! lo e@sj.tsv Jo- badass

I

9

I

tbmsy'iOslb ?;J aljav^tfrJ at nsi^loS

„ 8

I
jo 3.3 ev Proschan

observations is normal with mean-giK'

3 o_l _ F o r th e Popula ti on Mean , o Known a First s consider

the case where he know® 0 * (The determination is of a routine

f° r‘ which a great many sets of previous observations are

available* from which o is calculated )

In this ease

S *

*3745 0 (s)

will contain the w truew value (population mean) 50 percent of

the time 3 This may be seen from the following diagrams

Figure 2,

Suppose AB b the intervals £ Then,, sine© I Is

Tn

normally distributed with mean M8 standard deviation —£— 3 the

ysprobability is P 50 that 1 will be in ax + <j „ Notice hew-

“ yn“

6745

ev er *. that for the interval ju * 0 to contain x is exactly

vn

6745 equivalent to the interval CDe 1 £ —-— o * containing jq. Hence*

yn

the probability is a 50 that 1 -f o will contain ax 0
.’.it/ n$ : s.r •; ">, -I i|

a

a

a

a

T% 'v-!-.-. -•rv^W*i*s . , -i.

• a

a

:w.m%&£b z-: Jvrz.:Jz\ Vj.rl;. mo-.. \ -:tcf t; ^ . a

a

• i

’ : .

•' r 3* o*. '

a

>

a

>

.

'

i

i

i
4 Proschan

'^r

values of 33 for n s 2(1)30# 4O o 60,120 # cd » are

i/a

presented in table 1 9 column L

To generalize* when the confidence coefficient is ^(instead

of #50}, the confidence interval for the population mean Is

L.

ra

Jj, to £ i —pp— o

V'n

(3)

where

“l-r

1«*3 VS"

dt at ^ (4)

3o2 For the Population Mean, q Unknoivru Consider

ncuj the case where the only information about a is in the present

sample. 'hen the interval

X

,50,n«i.

0 (5)

Vn

(where t

nc^

is the Student^t value for n-l degrees of freedom

which is exceeded, in absolute value, with probability „50) will.

50 percent of the time, contain the population mean*

The following diagram demonstrates this*

Figure 2

r ... .... ?_ . rn i nan , ft

A . _j

— 'S?

C- h- P

- /> M o
rf?::6 :.'e Cr:'-

v 1 iS(K.r,fIo'a 9 »

•'

• ••

‘ 1.^; •• !;'> V. j ;V7

: -:.o

A

.. . r

1 .L: v-O; : '..V ' J - i»V: ' -

:

V-' • Iffjftr ,-MtH

a^ebl'BXioC •' *ilW<

©jtf£ dI eX 6 '-/c'Js %£&c - ,v .
<*> O «s> Prog chan

Lay off ABt jvx + g -nd CEs x £ —

8 e -y a

Notice that. when x lies in i\B 0 >a must of necessity lie in 03D$

and when I does not lie in AS, ax must fall outside of CD* But

the probability of

M

is, SO a n<s3l <S3> & X 6,

^

a 50»n»l

fT~ Vn (6)

^ ^

cs> /jj

is a 50 since is distributed as Student 5 s to Hence the

i/sr

probability that

is *50© values of s for n s? 2(X.)«50 S 40

,

60 * 120*00 0W 5,

are presented in table 1, column 2 « Comparison of and kg

Shows kg>k^ # but as n—*oo, kg—«fk^e

To generalize, when the confidence coefficient is $ (instead

of ©50), .the confidence internal becomes

AX

t& t

,n<»X ® X«»y »n®l X jfU 4* »w«r »»« ^ ( 8 )

5o5 Confidence Interval for Second Sample Mean®

Suppose the chemist who made the iron determinations wishes to

% sot up a confidence interval , not for the true mean* but for the

o

moan xgS of a second sample of observations* Suppose as in

paragraph 5a 2 , o I 3 unknown®

Let us now call the mean of the first sample and. the
. 1

arid ®;>n& E ..3 a

' rj •& • $ js - ba dm- J: is a.'-. C. ~ ~

.

**y*«»^ -':^r •’ .' ;r. - ,
6 Proschan

size of the first sample n^ E we may set up the statistic

( 9 )

The numerator is a normally distributed variable, while the

denominator Is an Independent estimate, of the standard deviation

of the numerator a Hence the ratio* t. Is distributed as Student

fca It follows that the Interval

*1* t 50 .+ L.

nl ng

Si (10)

will constitute a 50 percent confidence Interval for Sg, (ll

That does tJ is mean? It simply means that If pairs of

samples of size and ns respectively*, with means x^l an<^ Xgj respectively (I » • • • 0 oo ) , are dram repeatedly 9 then

for 50 percent of these pairs will lie in

*11 ®1 (11)

It does not mean that if one first sample of size with mean

x

!

1 s drawn, to be followed by the drawing of a great many

"second" samples of size with means x0

.^(I »? 1 9 2 9 **» 9 <d ),

that for 50 percent of the "second" samples xgi will lie In (11)

When ng s rip the coefficient of Bj In (11) becomes

fT

'^3 a fc *50,nx"ly nf ( 12 )

Values of for a 2(1 5 30* 40^ 60^120^00 are giv en In table

1* column 3*, for purposes of comparison. Note that k5

«-;/2k2 , simply.
s

-

'

,

' •

!

.

'

..

‘

. sv
7 Proschan

To generalize (10), If the confidence coefficient is

(instead of o50},(lQ) becomes

. iL + t. i - , Y ,n^l ( 13 )

4s T olerance Intervals » In paragraph 3, an Interval (X)

was formed to contain the population mean (with a certain

probabili ty) 0 Suppose, now, we are interested in setting up

the population* Such an Interval is known as a tolerance

interval

If eitherju or a is unknown, then the interval (!)_., containing

5 or 3 , is a random variable • Henc© the proportion, p, contained

In (1) will he a random variables

may be used as the tolerance Intervals The proportion p contained

in (14) is constant, and the appropriate value for specified p

may be obtained from a table of normal areas® Thus for p = e 50 s k » *6745 (listed in table 1, column 4, for purposes of comparison) 0 4»1.S a Unknown * Unfortunately In most

practical problems and o are nob known* Hence $ and s must

an interval ( 1 5 which will contain a certain proportion,, p a of

4sl Expected value of p a In 4»1 we determine k- so

that in repeated sampling F;(p) ss a, a constant® In 4*2 w©

determine k so that In a large series of samples from normal

universes a certain proportion § of the Intervals (1) will

include p or more of the universe®

4*1*1 o Known® In this case

ax +• k a (14)
' . ' . I

toe : Tjstvjno*) :i;o '8(- -iO'l1' mxlco . j ©Ic>J - rrl fo©tfs H; 3-1-VS
^ Pros chan

Ow u 3 od. 3 How shall os uc '5 ersnn© Is so t aa t in s i&rg© s8Pi@s of*

sample s i toth g, normaj. imiv©rs©* the ayeru 3 p contained in

% t ^s l (1 338 ) will be a?

4 solution was given by .Vi Iks, In [3] without giving

details of the proofs (For an independent derivation see append!

0 ) Stated explicitlys 1st

P 2S

, />2+kB

,, ,8, 2 4 | B"t(x»a} /o'5

^nWWft'CS a T/B>? C

J

(15)

jmei

Then

E(p) n

CO os

ssr P ds dx

-J «co J 0

n

8

t

ds

f-o~~ Y*r (n«l) 1 \jg^j U*^x)n/S

(16)

far-***** t s k fn^T

where fC^s) is the .joint distribution of JE and $§

': C ac* s

n-X

"Tf~ n^2

Vn ( n«>l ) S

*" «4'[n(S^/a) :i + (n«l)s 2

]

? n*i % <> 1*5 (17)

Xn other words* the tolerance limits which will include,,

on the average {for repeated samples)- a proportion $ a g of the

normal universe are

- . mx

2 ~X»a,n“X ^ n (18)

where neai is the value of t for which the integral in (16)

is equal to a Q Hence

kz __~ t fea

H**,!**1 T n

( 19 )
.

s

I

I

I

I

*

I
9 Proschan

Values of k for n m 2(1)5Q 9 40,60,X20,cd , and for a »

cSO $ *75, o90 s o 95 9 0 99 e 0 999 are given In taxi© 2 9 l his table

should be of use both to the experimenter and to the quality

controller'® Table 2 will si ppl eraent the values of k given in

[ 3] e An example of the use of table 2 is given:

BXMPX35g An Industrial quality control engineer

measures the voltages of a random sample of 30 batteries from

his production lino, (Production is in statistical control*,

and the successive battery voltages may be assumed to be random

values from a normal universe*) from the sample mean voltage*

x S3 7*52, and the sample standard deviation of voltages* s = o 90 p he wishes to estimate tolerance limits that trills on the average,

contain 95 percent of the population of batteries., What shall

these tolerance limits bet

The tolerance lira tg will be of the form x g ks* To find

k,he enters table 2 with n *s 30, The value of k is given as

2,079c Hence the tolerance limits ares

7o52 £ 3«079( 5,90) « 7,52 ± 1*8? m 5,65 to 9 C 39 9 Notice that 3c gg

s 2*079 is larger than the value 1,96 that

would be used if ya and o were available.

For purposes of comparison, values of k for n s 2(1)30,

o vV

40, 60*120,co , are included, in table 1, column 5,

One Sided Tolerance Limits , Suppose now the problem is

to find the value of such that, on the average, the proportion

of' the normal population less than £ + k 5 s is a specified value a 0
:-Jb . ishx . ... % -xu . -

3*3 • ' U . BIIIIBIIII1IIIB

II
10 Proschan «o>

In other words, if

<P® S

X4k 1 s ©

i'£ dx

*oo

(20)

find the value of k® such that

E{p® ) & & e From the previous proof, it

Hence to get the answer from table

desired value is

( 21 ) follows that the answer now

( 22 )

distributions

Then the

(23)

the Studenfc*»t

2 9 find _

2a«*l

k ’ 3 k3a-l

A similar result holds if the proportion of the normal

population greater than 1 k*s Is to be a specified value a

on the average,

EXAMPLE s A pilot run of 40 electron tubes is made. For

each tub#* a certain critical characteristic, x, is measured? for

the sample x m 12*25* s « ,68® From past experience, it is

known, that x is noixially distributed, What is the value of L

such that 99 percent of the population of tubes will, on the

average* have a value less than L?

Vie may write

L E3 1 -5- k9 s (24)

Then according to (23)

csa XX Proschan

U’ k

£( „99)-l “ k,98 of fcat,le 2- Table 2 yields k * 2© 455* Hence

1 L ss X2o25 + 2.455U68) s 13*92 6

4 »lo5 ax Unknown, q Known * In. this case an

of the form

1 £ kc?

must be usecU

Using the same method as in the proof above., the

result may be derived?

If the expected value S(p) of the proportion., p,

normal universe contained in (25) is to be a* then '

. k Ll»a

where L. is the nor-ml curve, (N(Q,1)), deviate such that the

JL^S,

area between + L- is. &»

•» wsg; For purposes of comparison, k of (28) is given in table 1 0 column 6, for a » o 50, and a 2(1)30,40,60,120,^ e 4*1 ©4 Ai Known, q Unknown « In this case the

interval

£ ks (27)

must be used©

Again using the same method as in' the proof above, the

appropriate value for k for (27) to include, on the average, a

is given by

interval

(25)

following

of the

(E6)

k aa t

'!«a on--X ( 28 ) where fc~ n

- is the value of t for which the integral in (16)
kMBBBB
® IS «f» Proschan

is equal to a 4 For purposes of comparison, values of k of (28) are given

in table i # 'column 7, for a » a 5G and n « 2(1)30940,60,120,® „ Statement , About Tolerance Interval 0 A number of papers have been written on the problem of confidence

statements for tolerance .intervals* [2] ,[o] ,[6] ,[7] ,[0]

The

problem may he illustrated as follows;

bffeqivn • Suppose the battery engineer

of 4.1,8 asked th© following questions vVhat value of k shall

1 take sc that I can b© 95 percent confident that I cs> + ks will

Include at least 80 percent of my population of batteries?

[3] contains extensive tables of k such that n

3Ln a large

series of samples for normal universes a certain proportion 'i

of- tbe Intervals x ± ks will include p or more of the universe?

/ ia c ailed the 9 confidence coefficient 9 since it is a measure

of the confidence with which we may assert that a given tolerance

range includes at least F of the universe” * [3] In these tables

f s *75, *30, o95, *99, O 999o

Knownx o Tfnknov/n* Consider the case

where Jb Is known. Then an interval of the form (27) can be set

up to Include at least P of the population with confidence y as

follows

Lot us take specific values of P « o 80 and y« 0 95 for

Illustrative purposes* .7© note first that P is monotonia

increasing with s (and with s

w

) 0 Hence, when takes on its
I —t.v-viMI!' . - : : 'T? :i d '

•; 0 ' . S ' HIIII1 * V
13 Fro s chan

v ax lie exceeded 95 percent of the t ilr*© (call i. t- 6^l5§), F will tak©

on its value exceeded 95 percent of the fcimee But

S' 9

v o

o95 c rXi a

n«l

Then the appropriate value of k to use in (2?) is

(29

L 20/-/x?05.n-l/(n-1)

* f *.9S,n-l

Values of k for P> $ ® o50 for n ^ 2(X}&>, 40*60,120,00

are given in table 1, column 8, for purposes of comparison.

For general P, if Llw? is defined as in (26), then t appropriate value of k to use in (27) is

L

(*>:

(31)

v

,s # 5 n«'l / (n«»Xf

__ u TJri.kn own » o Fhowru In this case s interval

(35) nmst be used* Let us solve for k when P « ,80, 0 95 to

illustrate the reasoning..

Jj

v7e first note that 95 percent of the l? s lie inyu + —a”?, o n in other words, 95 percent of the x £ tec? intervals have their

I# A

7 centers inside^ + -»-TM o » . Bov/ find such that

‘ n 9 ^ 4- a 4- k0

c?

L

a 03

ytt + G - kca

n

t~ju) ^ ®~ 2c^

"

88 a SO ( 32 )

:’hen 95 percent of the £ 4 k c intervals will contain P >,*80 9

f namely those Intervals for which I lies In^u * —i2iL ©5

14- Proschan

It follows that the interval

( 33}

will contain oSQ or more of the population, <,95 of the time*

where Is defined as in (26)*

3 a Relationship Between Confidence Intervale and Tolerance

Interval

e There is a very interesting relationship between

confidence intervals and tolerance Intervals that can he illus~

t rated by the following examples

Suppose as in paragraph £.3 we wanted to find a confidence

interval for the mean of a second samples. But now let = l a In other words, we will now be finding a confidence interval for

a single future observation* According to the result In para®

graph 3,3, our answer la

where a Is the confidence coefficients.

•That does (35) mean? One way of looking at it is that If

repeatedly a sample of size Is first drawn and then a second

sample of one item is drawn, then a of the time the single item

for n s 2(1 >30,40,60,120,co « For general P,df , k Is found from

(34)
I

I

•7

'

1

1

1

1

1

1

1

1

.... T

- --• ,

•

'

'

;
15 Pros chan

will lie in the corresponding interval {,35 5 o' But a little

thought .shows that this is exactly equivalent to stating that- in

repeated samples of size n^ y the average proportion, P, of the

population contained in (55) is a 0 In other words, confidence

limits with confidence coefficient a for a second sample of sis©

1 are identical with tolerance limits t Vat will include a pro^

portion, a, on the average* This is confirmed by the fact that

(-35) is identical with (18) „ The above is an illustration of a theorem by Paulson [5]$

n If confidence limits o • *nXri ) e °^xr^

on a probability level » € q

are determined for g, a function of

a future sample of k observations, (with distribution (g)],

and p = V 2 fKg)dg, t: enE(p) ss<

0

o w

The proof is now given, because it is short and instructive?

^Let 4p(g)dg and pf{ TT^ a TJg ^dU1 nXJg denote the distribution

of g and TJ^ ,-Ug respectively* Then by the definition of expected

value

.co joct r f)Vg 1

ip Cs5dg ] 0(V1

,u2 )dU1dU2

(56)

--V.'

E{p) f^ «*00 I/f«00

This triple integral is however exactly the probability that g

will lie between IT, and U - which by the nature of confidence

(So limits must equal

In the illustration given above, g corresponds to the value

Of the single future observation, and k s l 0 Similarly we can check the results of paragraphs 4olo3 and

4*1 u’4 by the use of Paulson’s theor&mo
-

'• .v-: -v - ; . .
Pros chan

References

[X] G„ A* Baker , The probability that the mean of a second Mmolejviill . differ from the mean of a first sample bv lass

ascertain multiple of the standard deviation o? thsT

5* £j^§l,sample g Annals of Math* Stat,« Vol* Vi'; IoT47~D©c7 1935e [2] A, H. Bowker, Computation of factors for tolerance limits

distribution when the sample is larrre* Annals

of Hath, 8tat 7, ^oi e XVII, N0.2, June 194S7~~“

[3] Eisenhart, Hastay, and Wallis* Techniques of Statistical

Analysis , McGraw-Hill, 1947, Chapter 2 & -

[4] A* II. Mood, Introduction to the Theory of Statistics,,

McGraw-Hill, 1950, Chapter 11 T

~~~~ ~ '

Paulson, A note on tolerance limits, annals of Math.

State , Vol. XIV, No *1, March 1943*

[6] A, Wald, Settinr: of -tolerance limits when the sample is

lar^e. Annals of Math. 3tat., Vol, XIII, Ho, 4, Dec, 1942,

!?] A* Wald and J *> Wolfowltz. Tolerance limits for a normal

distribution . Annals of Math, Stat., Vol, XVII, Ho, 2,

” June 1946 s

[8] S, 3* Wilks, Determination of sample sizes for settinr;

tolerance limits . Annals of Math® Stat»« Vol . XII, Uo.l, March 3.941,

*

l9] Wo Allen Wallis, Tolerance intervals for linear regression ^ Second Berkeley Symposium on Mathematical Statistics ancT”‘

Probability, edited by Jerzy Heyman, 1951®
*
APPENDIX Pros chan

Hathemafcloal Pro of o f JJlSjL ° The details of the proof

{independently derived by I* R© Savage of the Statistical

Engineering Laboratory * National Bureau of Standards) of (16)

are given, since the method is a suggestive ones

By an appropriate linear transformation, the problem maybe reduced to that of finding

Blp)

jx> ^x-fks

_ ^ ©“^ dfc an^e +(n«*l) (36)

L' 0 0 ~cq u S>ks

where is a constant free of k u In the following* Cf

ss constant

free of

The conditions for differentiating 'under the integral

holds Hence we have

.co ptt2 gVoE *«» _ nVfc>

’* 6k

30

0 a/ «iU *

p

re f

|( i-Jfcs

)

s

4.se^|( I«ks

)

2

j ^ftySe»jtnX®+(n«l)a2

aids (57)

atS C

1

-

5

- 0.

v 0 if «oo

a? * co

f*

^ 0 v <&>

4(Vrar S+;-TM)^n»l+k?^~j)s£

,n=.la

sae is/n-Ki *

r

«TB'

is ,2 K/n^l S- ”r}B+(n-l+k2^-)«2

an-ldxag (gg)

-va+:

Let u ^%/n+l i +

ks first integral and

V

vn*± x

ks in the second o Then
nMIHIaaaaaiBaa
Pro 3 chan

A'OO „„00

6F; r

6k. 1

0

r tin »w( n~A flrc' ^ ) j»2 ® a

x

© 8in ***

rT?I

8

dta

hl+j

’ 0 vji^OD

+ C

1

gd c©

p

r , . 2

j

“+u _S2_ a

11"1

3“i(a=l+ks Jj^s8

du ds ' 11+3

0 ^<®0D

( 39 )

or

6e „ 6k

53 'J2

,00

( n^l A«^(:n«*l + k a e

v Q

2 J*: n2

n+l

X. 0t y s J(n®l + k

2 ““Is2 « Then

-6e p 5«" 52 ok *

©

A n -I n X tt-1 n

2^ y^ e“^/{ n«l + k

2 £y ( 41 )

r g { n=»X +

n

Hence

Nov/ let

so that

n

E(p) s 0 3

k

4c

dk

C.n-1 + k

2

t s k ( 44 )

E'Cp) » c

dt

4

(n«*l + t2

) n9T,»

( 45 )

Pro s chan

n

n

'3 dt/( 1 +t'^/( n«>X ) } ^ ( 46 }

\J

But 'She integrand is the well known Sfcv.dent«*t density emotion*

How when k = go 9 E(p) » 1, Hence C r must he identical with the

o » constant of the student«*t distributions ( 16 ) follows

/
1

I
TABLE 1

Factors for confidence and tolerance intervals

Sample

Size

n k

x

k2 k3 k4 . kk

s

k6 k

7 8 k9

2 .477 O 707 1.000 .674 1.225 ©825 XoOGO XoOOQ ,754

3 o389 . 471 .666 .674 .942 .779 *816 .810 .727

4 .557 .382 .541 .674 .855 .754 ©765 ©759 ©714

5 o 302 .531 .469 .674 . 812 .739 ©741 ,736 .706

8 © 27 5 .297 .420 .674 .©785- .729 .727 ©723 .700

7 o 25 5 .271 .384 .674 .768 .721 .718 ©714 ©697

8 *238 .251 .356 .674 .754 .715

' .711 ©708 .694

9 o225 .235 .-333 .674 .744 • 711 .706 ,704 .692

10 o 21

.222 . ol4 .674 ©737 *707 ©705 ,701 o690

11 .203 *211 ©299 cS74 ©731 .704 ©700 ©698 .688

12 o!95 ©201 ©285 .674 .725 ©702 ^697 ©698 .687

IS ©187 ©195 .273 ©674 .781 o7G0 ©695 ©694 ©686

14 .180 ©185 .262 .674 .718 .698 .694 ,692 ©686

15 .174 .179 .253 .674 .715 .697 ©692 ©691 ©685

16 el 69 ol73 .244 .674 ©712 .695 © 691 ©690 ©684

17 ©164 .167 .237 .674 .710 .694 ©690 ,689 ©684

• IS . .159 .162 .230 .674 ©708 ©693 *689 *688 .683

19 0 155 ol58 .223 .674 ©706 ©692 ©888 ©687 © 683

20 ©151 .* 154 .218 .674 .705 ©691 .688 ,687 ©682

21 o!47 .150 .212 .674 ©703 .690 .687 ,686 .682

22 ©144 o!46 .207 .674 ©701 .690 .686 ©685 ©681

23 .141 oX45 .202 .674 .701 .689 0686 .685 .681

24 ©138 .140 .198 ©674 .,699 .688 .685 ©684 © 681

25 o!35 .137 .194 .674 ©699 .688 o 685 ,684 ©681

26 ©152 e 134 .190 .674 .697 .687 .684 ©684 .680

27 ©ISO *132 *186 .674 .697 *687 *685 © 685 o 880

28

* ©127 .129 .183 .674 .896 .686 ©684 © 683 © 680

29 • « ju25 .127 *179 .674 .695 *686 .683 ©685 .680

30 .123 .125 .176 .674 *694 .685 ©683 ©682 ©680

40 .107 .108 .152 .674 *689 *683 oSSX *680 ©678

60 .087 .088 .124 .674 .685 .680 ,679 .678 ,677

120 .062 *062 .087 .674 ©680 .677 .677 ,676 .676

oo 0 0 0 .674 *674 .674 .674 ©674 ,674

For

xplanat;ion 3 © 1 3.2 3.3 4 0 1 9 1 4.1.2 4.1.3 4 ©1 © 4 4 © 2 o 2 4 & 2 o o

ee para.graph
J

’ d

.OB;: . i;

,

! «' vis o
TABLE II

0 ample

Size

Factors for tolerance interval a <>

n %5G ]r ,75k

.90

k

0 95 k

.9B k

.99

k

o999

2 1,225 2.957 7,733 15.562 -38 .973 77 a 964 779 © 699

5 » 9 42 1 ,852 3® 372 4.989 8*042 Ho 460 36,486

4 ® 355 la 593. 3,631 3.553 5.077 6,550 14.469

5 *8X2 1.473 Sy 335 3.041 4.105 5,043 9.0 432

6 .785 1,405 2, 176 2.777 3,635 4.355 7 c 409

7 .768 1 a 561 3 c 077 2,618 3,360 3 ® 953 6.370

8 .754 x o 330 2,010 2.508 3.180 5.711 5.733

9 o744 1 e 307 1,961 . 2 . 4ol 3.053 Oo53S 5,514

10 .737 1 6 290 1,922 2,372 2,859 3,409 5.014

11 ,731 1.276 1.893 2 c 327 2.887 3,310 4,791

12 .725 1,264 1.869 2.291 2.829 3o 23d 4.618

15 e 7 21 1,255 2,8.49 2 c 261 2,782' 3.170 4,481

14 .718 1.246 If833 2,.236 2.743 5*118 4 0 oGQ

3-5 ,715 1.239 1,819 2,215 2 a 710 3.075 4.27.6

16 ,712 1,234 1 ,807 2,197 2.688 3.038 4.198

1? • 710. 1.228 1,797 2.161 8.658 3,006 4.131

18 ,708 1,224 1,788 2.168 2 . 637 2,977 4.074

19 .706 1,220 1,779 2,156 2.618 2.953 4.024

20 c705 1 o216 1,772 2 n JL 45 8,602 2 . 958 3 c 879

21 ,705 1 ,213 1,766 2.135 2,587 2.912 3,941

22 ,701 1*210 1.760 2.127 2.575 2.895 5.905

23 ,701 1 o 207 1,754 2.119 2.562 2.830 3.874

24 .699 1.205 1,749 2,112 8,552 2,865 O.B45

25 = 699 1 <> 202 1,745 2.105 2.541. 2,852. 3.819

26 ,69V 1.200 1.741 2.099 2;532 2.840 3 ©796

27 .697 X .198 1,737 2.094 2,524 2,830 3.775

28 ,696 1 e 197 1.733 2.088 8,517 2 a 820 3,755

29 « 695 1,195 1 9 7 50 2.083 2,509*' 2.810 3.737

30 ©694 1,193 1,7:37 2,079 2 p 503 2 0 802 3.719

40 « 689 X cl82 1 *706 2.047 2 3 455 2 « 7 4X 3.602

SO ,685 1.171 1 a 636 2,017 2 0 411 2.684 3.492

*.20 ,630 1,161 1 .865 1.988 2.368 2,628 5.388

CD • 674 1,150 1.645 1.950 2.326 2,576 3,291

+ . -( 3S>Ul£

Let P * )

1 e 2e^ dx. The -value of k,

t given i If 0

table Is such that E(p) a a In repeated sampling,, (See par, 4,1,2)

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