1EDF 6481
Quantitative Research Methods in Education
You will need pages 235 to 251 of the Target Material and a t table and an F table.
1. The problem calls for using the Welsch test to conduct planned pairwise comparisons for adesign with one between-subjects factor. From page 235 of the Target Material, the correct
formula for t istY j  Y j11MS S / A    n j n j To illustrate substitutions consider groups 1 and 2:t

4  16
1 1
13.5   6 6 5.66

The following is the table of t statistics for conducting the Welsch procedure:GroupGroupMean

2654314781012162467584103121
16

--

-1.41
--

-1.89
- .47
--

-2.83
-1.41
- .94
--

-3.77
-2.36
-1.89
- .94
--

-5.66
-4.24
-3.77
-2.83
-1.89
--

q

2

2


The degrees of freedom for MS S / A are N  J  36  6  30 . From page 241 of the Target
material, the critical values areStretc
hSize65432qq±4.33±4.11±4.11±3.96±3.59±3.06±2.91±2.91±2.80±2.542The following table shows the critical values and the significant results. A single asterisk means
the t was significant by comparison to the critical value. A double asterisk means the t wasdeclared significant by implication.

GroupGroup

Mean

24

6
7

584
10312

116q 2±3.06
±2.91
±2.91
±2.80
±2.54

-1.89
-3.77*-5.66**
24--2.83
-1.41
-1.41
67--4.24**-2.36- .47-1.89
58--3.77**- .94
410--2.83*- .94
312--1.89
116-Note. Stretch size 2 (SS2) in bold, SS3 in italics, SS4 in bold italic, SS5 in shadow, SS6 inshadow italics. The critical value for a t statistic has the same font as the t statistic.

We can conclude that the means for treatment 2, 4, 5 and 6 are significantly different from themean for treatment 1 and the means for treatments 2 and 3 are significantly different.

A convenient way to report the results is to use the superscript method:

Group
2
6
5
4
3
1

3

Mean
4a7a,b8 a,b10a,b12b,c16cNote. Means that share at least one common superscript are not significantly different

For example the means 4 and 7 are not reported as non-significantly different because they sharethe “a” superscript. The means 4 and 16 are reported as significantly different because they do
not share a common superscript.
In the preceding table the rows are order by the values of the means. Following is the table withthe rows ordered by the numbers that identify the treatments.

Group1
2
3
4
5
6

Mean
16c4a12b,c10a,b8 a,b7a,b4

2. The problem calls for using the Shaffer-Holm test to conduct planned pairwise comparisonsfor a design with one between-subjects factor. The same t statistics are used. The values of C aretaken from Table 1 on page 240 of the Target materials. For C  1 , the critical values are fromthe Bonferroni table (pages 243-247). For C  1 , the critical values are from the standard t table.
Of course N  J remains 30.
Since H 0 is rejected we can proceed to determine significance of the t statistics. The t statistics
and critical values are presented in the following table.
StepTCt fw / 2, C , N  JPair123456789101112131415-5.66*-4.24*-3.77*-3.77*-2.83-2.83-2.36-1.89-1.89-1.89-1.41-1.41- .97
- .97
- .47

101010101010
777644321±3.0298±3.0298±3.0298±3.0298±3.0298±3.0298±2.8872±2.8872±2.8872±2.8247±2.6574±2.6574±2.5357±2.3596±2.0423

2 vs. 1
6 vs. 1
2 vs. 3
5 vs. 1
4 vs.1

We can see that the t statistics are significant up to step 4. The step 5 t is non-significant.
Therefore all the t statistics at lower stages are declared non-significant without further testing.

We can conclude that the means for treatment 2, 5, and 6 are significantly different from themean for treatment 1 and the means for treatments 2 and 3 are significantly different.

Groups 4 and 1 are not significantly different by the Shaffer-Holm test, but were significantlydifferent using the Welsch procedure. This is consistent with the power advantage for Welsch.

5
The program to obtain the critical values would bedata;
input alphafw df c;
prob =1-alphafw2*c);cval = tinv(prob,df);
cards;.05 30 10.05 30 7.05 30 6.05 30 4.05 30 3.05 30 2.05 30 1
proc print;
run; quit;

The results are
Obsalphafw

dfcprob

cval

123
4
5
6
7

0.05
0.05
0.05
0.05
0.05
0.05
0.05

30
30
30
30
30
30
30

107
64
3
2
1

0.99750
0.99643
0.99583
0.99375
0.99167
0.98750
0.97500

3.02980
2.887212.82470
2.657362.53574
2.35956
2.04227

3. After inspecting the means:GroupMean

24

6758410312116I chose to test a contrast comparing the three largest means and the three smallest means. You
may have selected a different hypothesis. The hypothesis tested is
H 0 : 1   2  3   4  5  6  0
From page 235 of the Target material, the correct formula for t ist

w1Y1  w2Y2    wJ YJ
 w12 w22wJ2 
MS S / A  nJ 
 n1 n26The substitutions aret

16  4  12  10  8  712 12 12 12 12 12 13.5    
66 666 
6 5.17

From page 237 the critical value is the Scheffé critical value:  J  1 F , J 1, N  J =  6  1 F.05.61,366=  5(2.53) =  3.56

We can reject the null hypothesis and conclude that the sum of the three largest means is
significantly different than the sum of the three smallest means.
4. The design has one between-subjects factor and the problem calls for a trend analysis for.
From page 235 of the Target material, the correct formula for t is
w1Y1  w2Y2    wJ YJt
 w2 w2w2 MS S / A  1  2    J nJ  n1 n2I will use the hypothesis of a cubic trend to illustrate the analysis. From page 249 in the Target
materials, which contains the orthogonal polynomial contrast weights, the hypothesis is
H 0 : 11  3 2  33  1 4
The substitutions are6.49 + 3  4.82  - 3  4.25  + 3.8

 .52
 12 32 32 12 
1.41 + ++ 888
 8
The t statistic for the linear trend is -4.58 and that for the quadratic trend is 1.44. From page 239of the Target materials, if we are going to incorporate the omnibus test in the procedure, theproper test is the Shaffer-Holm procedure. Otherwise the Bonferroni-Holm procedure is used. Iwill use the Shaffer-Holm procedure. The omnibus F is F (3, 28)  7.8 , which is significant.
The following is the table for conducting the remaining steps of the Shaffer-Holm procedure.
t=_________________________________

StagetC7t fw / 2, C , N  J_________________________________
1 -4.58 2±2.3685
2 1.44 2±2.3685
3 -0.52 1±2.0484_________________________________The results indicate that only the first t is significant. Consequently we conclude there is a linear
trend, but no non-linear trends. Together these results imply a linear relationship between
number of prior practice problems and solution time.
5. The problem requires planned pairwise comparisons for a design with one within-subjects
factor. From page 239, either the Shaffer-Holm or Bonferoni-Holm can be used. I choose theShaffer-Holm. The test of the omnibus H 0 yields F (1.67, 21.69)  17.5, p  .05 so weproceed to calculate t statistics. From page 235 the following formula is used to calculate tstatistics:
Ct
SC2nTo calculate the t statistics, you can run the following program:data;
input face 1-3 cc 5-7 news 9-11 sc 13-15;
facecc = face - cc;
facenews = face - news;
facesc = face - sc;
ccnews = cc - news;
ccsc = cc - sc;
newssc = news - sc;
cards;
3.1 3.4 1.7 1.8
1.3 0.6 0.7 0.5
2.1 1.7 1.2 0.7
1.5 0.9 0.6 0.4
0.9 0.6 0.9 0.8
1.6 1.8 0.6 0.8
1.8 1.4 0.8 0.6
1.4 1.2 0.7 0.5
2.7 2.3 1.2 1.1

8
1.5 1.2 0.7 0.6
1.4 0.9 1.0 0.5
1.6 1.5 0.9 1.0
1.3 1.5 1.4 1.6
1.3 0.9 1.2 1.4
proc means n mean var t prt maxdec=3;
var facecc facenews facesc ccnews ccsc newssc;

The results for the pairwise contrast variables areVariableNMeanVarianceTProb>|T|---------------------------------------------------------------FACECC
14
0.257
0.093
3.148
0.0077FACENEWS 14
0.707
0.230
5.518
0.0001FACESC
14
0.800
0.314
5.343
0.0001CCNEWS
14
0.450
0.344
2.870
0.0131
CCSC
14
0.543
0.333
3.518
0.0038NEWSSC
14
0.093
0.055
1.487
0.1607----------------------------------------------------------------I will use the comparison of faces and concentric circles to illustrate the calculations:t.257

 3.15.093
14
To use these results in the Shaffer-Holm procedure, complete the following table. (TheComparison column is not required.) The degrees of freedom for the t statistics follow from thedegrees of freedom for SC2 . These are n - 1. The values of C are taken from Table 1 on page 240of the Target materials. For C  1 , the critical values are from the Bonferroni table. For C  1 ,the critical values are from the standard t table.
_________________________________________________

Step tC t  fw/2, C, n - 1

Comparison_________________________________________________
1 5.51 3 ±2.7459 Faces-News
2 5.34 3 ±2.7459 Faces-Simple
3 3.51 3 ±2.7459 Concentric-Simple
4 3.14 3 ±2.7459 Faces-Concentric
5 2.86 2 ±2.5326 Concentric-News
6 1.48 1 ±2.1604 News-Simple_________________________________________________The results indicate that all pairs of stimuli except newspaper and simple circle result insignificantly different gaze times.

6. Post hoc comparison for a design with one within-subjects factor. Based on an inspection of
the sample means (not reported in this document) I decided to test
H 0 :1/ 2(  face   conc )  1/ 2(  news   simp )  0or equivalently
H 0 :  face   conc   news   simp  0
From page 235 the following formula is used to calculate t statistics:Ct
SC2
nTo calculate the t, you can run the following program:data;
input face 1-3 cc 5-7 news 9-11 sc 13-15;
comp = face + cc - news - sc;
cards;
3.1 3.4 1.7 1.8
1.3 0.6 0.7 0.5
2.1 1.7 1.2 0.7
1.5 0.9 0.6 0.4
0.9 0.6 0.9 0.8
1.6 1.8 0.6 0.8
1.8 1.4 0.8 0.6
1.4 1.2 0.7 0.5
2.7 2.3 1.2 1.1
1.5 1.2 0.7 0.6
1.4 0.9 1.0 0.5
1.6 1.5 0.9 1.0
1.3 1.5 1.4 1.6
1.3 0.9 1.2 1.4
proc means n mean var t prt maxdec=3;
var comp;

The program calculates the t; the following calculation is used910t

1.250

 4.52
1.073
14
From page 237 of the Target material, the critical value is(n  1)( P  1)
F , P  1, n  p  1(n  P  1)

(14  1)(4  1)
F.05,14 4  1
14  4  1

(13)(3)
3.59   3.57
11

Consequently we can conclude that average gaze time for faces and concentric circles is different
than reaction time for newspaper and simple circles.

1. From page 235 the Target Materials the required formula for a between-subjects design istY j  Y j
11
MS S / A    n j n j The required statistics are the means for the five groups and MS S / A . These can be calculated byusing the following program:data;
input group 1 score 3-4;
cards;
the data go hereproc sort;
by group;
proc means;by group;
proc glm;class group;
model score=group/ss3;

11



2. From page 235 in the Target Materials the required formula for a within-subjects design ist

CSC2nThe required statistics can be obtained by using the following program
data;
input y1 1-2 y2 4-5 y3 7-8 y4 10-11;
linear = -3*y1 - y2 + y3 + 3*y4;
quad = 1*y1 - y2 - y3 + y4;
cubic = -1*y1 + 3*y2 - 3*y3 + y4;
cards;
the data go hereproc means n mean var t prt maxdec=3;
var linear quad cubic;