1

1. An experiment to study the effect of two training programs is to be conducted. A
completely randomized design is to be used. An effect size of .30 is of interest and theresearcher wants power to be at least .5. The researcher plans to use   .05 and a onetailed test. How many subjects in total should the researcher plan to use? Answer thequestion again if a randomized block design is used and the correlation is expected to be.75.
The following are the specifications for the first part of the questions:

DesignCRD

Number of levels

2ES MAX.3

Power.5

Alpha.05

Test

Directional

We can use G*Power for Means: Difference between two independent means (two groups).

2

3

For a design with only between-subjects factors the Total sample size is the total number ofsubjects required for the design. Thus the total number of subjects required is 122.

4

For the second part of the question the design changes to a randomized block deign. We can use
G*Power for Means; Difference between two dependent means (matched pairs).
The following shows the screen that appears after pressing Determine The arrows point to my inputs.

56For a design with only within-subjects factors the Total sample size is the number of scores in
one cell of the design. This is a diagram of the design, showing 17 scores in each level
Treatment
A1
A2
17 17
Because the design is a randomized block design, we have subjects nested in levels and we
require subjects 34 subjects.

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D.2
2. An experiment is to be conducted to study the effects of two factors in acompletely within-subjects design. Factor A has four levels and factor B has two levels.
Blocked random assignment is to be used. Factor A is the primary factor of interest andthe researcher is interested in detecting an effect size of .25 with power of .7. The
correlation between the scores for any pair of cell variables is expected to be .65 and  isexpected to be .90. The researcher believes the mean configuration will be similar to themaximum range configuration. If the experimenter plans to use   .05 , how many
subjects should there be in total?
The following are the specifications
Design4  2 , completely within-subjects

Number of levels in factor of interest

4Target Power.70Alpha.05

Epsilon.90

ES MAX.25

Correlation.65

I used this program to calculate f. Note that es_max is .25, levels is 4 and M is 2. A value for M
other than M  1 is only required in a design that has no between-subject factors and more than
one within-subject factors.
data;
format f 10.7;
input es_max levels M;
remainder= MOD(levels, 2);
if remainder=0 then do;
mean_configuration='maximum variability';f=(sqrt(M))*es_max/2;
output;end;
if remainder^=0 then do;
mean_configuration='maximum variability';
f=sqrt(M*levels**2-1)*es_max/(2*levels);
output;end;
mean_configuration='two extreme';
f=sqrt(M*2*(levels-2))*es_max/levels;

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output;mean_configuration='one extreme';
f=sqrt(M*(levels-1))*es_max/levels;
output;mean_configuration='equally spaced';
f=sqrt(M*(levels+1)/(12*(levels-1)))*es_max;
output;mean_configuration='minimum variability';
f=sqrt(M/(2*levels))*es_max;
output;datalines;.25 4 2
proc print;
var mean_configuration levels es_max f;
run;
Obs

mean_configurationlevelses_max1
2
3
45maximum variability
two extreme
one extreme
equally spaced
minimum variability

44
4
4
4

0.25
0.25
0.25
0.25
0.25

The value of f to use is 0.1767767

f0.1767767
0.1767767
0.1530931
0.1317616
0.1250000

910For a design with only within-subjects factors the Total sample size is the number of scores in
one cell of the design. Following is a diagram of the design, showing 28 scores in each cell.

11Factor B
Factor A
A1B128

B228

A228

28

A32828A428

28

The design is randomized block so we need 28 subjects in each of eight cells. The total sample
size is 8  28  224.

123. An experiment is to be conducted to study the effects of two factors in a design in which
Factor A is between-subjects and has three levels; Factor B is within-subjects and has twolevels. For Factor B subjects are used as their own controls. The correlation between thescores in any two cells is expected to be .50. Epsilon is 1.0 because the within-subjects
factor only has two levels.
A. The researcher is most interested in the A main effect and expects the spacing of
the marginal means to be similar to an equally spaced configuration. An effect size of .40 isof interest. How many subjects should there be in each level of A if the target power is .6
and   .05 ? Answer the question again assuming the researcher wants to be very
cautious in planning the study.
The following are the specifications
Design3  2 split-plot; the factor with 2 levels iswithin-subjects

Number of levels in factor of interest

3Target Power.6

Alpha.05

ES MAX.40

Correlation.50

We can use G*Power for ANOVA: Repeated measures, between factors.

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I used this program to calculate f. Note that es_max is .40, levels is 3 and M is 1.
data;format f 10.7;
input es_max levels M;
remainder= MOD(levels, 2);
if remainder=0 then do;
mean_configuration='maximum variability';f=(sqrt(M))*es_max/2;
output;end;
if remainder^=0 then do;
mean_configuration='maximum variability';
f=sqrt(M*levels**2-1)*es_max/(2*levels);
output;end;
mean_configuration='two extreme';
f=sqrt(M*2*(levels-2))*es_max/levels;
output;mean_configuration='one extreme';
f=sqrt(M*(levels-1))*es_max/levels;
output;mean_configuration='equally spaced';
f=sqrt(M*(levels+1)/(12*(levels-1)))*es_max;
output;mean_configuration='minimum variability';
f=sqrt(M/(2*levels))*es_max;
output;datalines;.40 3 1
proc print;
var mean_configuration levels es_max f;
run;

Obs

mean_configurationlevelses_max12345

maximum variability
two extreme
one extreme
equally spaced
minimum variability

333
3
3

0.4
0.4
0.4
0.4
0.4

The value of f to use is 0.1632993

f0.1885618
0.1885618
0.1885618
0.1632993
0.1632993

141516For a design with one–between-subject factor and one within-subjects factors the Total sample
size is the number of subjects in one level of the between-subjects factor multiplied by thenumber of levels of the factor. That is 180  60  3 . Thus we need 60 scores in each cell of thedesign.
This is a diagram of the design, showing 60 scores in each level
Factor B
Factor A
A1B160B260

A26060

A36060The design is 3  2 with repeated measures on the second factor. Thus we need a total of 180subjects.
The most conservative configuration is minimum range. For J  3 equally spaced andminimum range are equivalent, so we would need a total of 180 subjects for this design.

17B. The researcher is most interested in the B main effect. An effect size of .25 is ofinterest. How many subjects should there be in total if the target power is .80 and   .05 ?
The following are the specifications

Design3  2 split-plot; the factor with 2 levels is
within-subjects

Number of levels in factor of interest

2Target Power.80

Alpha.05

ES MAX.25

Correlation.50

Epsilon1.0

We can use G*Power for ANOVA: Repeated measures, within factors.
I used this program to calculate f. Note that es_max is .25, levels is 2 and M is 1.
data;format f 10.7;
input es_max levels M;
remainder= MOD(levels, 2);
if remainder=0 then do;
mean_configuration='maximum variability';f=(sqrt(M))*es_max/2;
output;end;
if remainder^=0 then do;
mean_configuration='maximum variability';
f=sqrt(M*levels**2-1)*es_max/(2*levels);
output;end;
mean_configuration='two extreme';
f=sqrt(M*2*(levels-2))*es_max/levels;
output;mean_configuration='one extreme';
f=sqrt(M*(levels-1))*es_max/levels;
output;mean_configuration='equally spaced';
f=sqrt(M*(levels+1)/(12*(levels-1)))*es_max;
output;18
mean_configuration='minimum variability';
f=sqrt(M/(2*levels))*es_max;
output;datalines;.25 2 1
proc print;
var mean_configuration levels es_max f;
run;

123
45maximum variability
two extreme
one extreme
equally spaced
minimum variability

22222

0.25
0.25
0.25
0.25
0.25

0.1250000
0.0000000
0.1250000
0.1250000
0.1250000

Because there are two levels to the factor of interest the two-extreme means configuration is notapplicable and all the other configurations have the same value for f. So f should be .125.

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For a design with one–between-subject factor and one within-subjects factors the Total sample
size is the number of subjects in one level of the between-subjects factor multiplied by thenumber of levels of the factor. That is 129  43  3 . Thus we need 43 scores in each cell of thedesign.

21This is a diagram of the design, showing 43 scores in each level
Factor B
Factor A
A1

B143

B243

A24343

A34343The design is 3  2 with repeated measures on the second factor. Thus we need a total of 129
subjects.