Chapter 13, Problem 17P

The following data are from a repeated-measures study comparing three treatment conditions.

a. Use a repeated-measures ANOVA with $\alpha =.05$ to determine whether there are significant differences among the treatments and compute ${\eta }^2$ to measure the size of the treatment effect.

b. Double the number of scores in each treatment by simply repeated the original scores in each treatment a second time. For example, the $n=8$ scores in treatment 1 become 1, 4, 2, 1, 1,4, 2, 1. Note that this will not change the treatment means but it will double $S{S}_{\text{between treatments}}$ , $S{S}_{\text{between subject}}$ , and

the SS value for each treatment. For the new data, use a repeated-measures ANOVA with $\alpha =.05$to determine whether there are significant differences among the treatments and compute ${\eta }^2$ to measure the size of the treatment effect.

c. Describe how doubling the sample size affected the value of the F-ratio and the value of ${\eta }^2$ .

		Treatments
Person	I	II	III	Person

Totals A 1 4 7
   $P=12$ B 4 8 6
   $P=18$
   $N=12$ C 2 7 9
   $P=18$
   $G=60$ D 1 5 6
   $P=12$
   ${\displaystyle \sum X^2}=378$
   $M=2$
   $M=6$
   $M=7$
   $T=8$
   $T=24$
   $T=28$
   $SS=6$
   $SS=10$
   $SS=6$

To determine

• If there is significant different between the three treatments
• The size of treatment effect
• Effect of doubling sample size

Answer to Problem 17P

Solution:

There is significant difference between the three treatments, the effect size is 0.8485, and doubling the sample size alters the F ratio but the effect size remains same.

Explanation of Solution

Information given:

Let t denote number of treatments

S denote numbers of subjects

N denote total number of subjects in all treatments

We got t = 3

S = 4

N = 12

Treatments
Person	I	II	III	Person Totals
A	1	4	7	Pa=12
B	4	8	6	Pb=18	N=12
C	2	7	9	Pc=18	G = 60
D	1	5	6	Pd=12	${\displaystyle \sum x^2=378}$
	M =2	M=6	M=7
	SS=6	SS=10	SS=6

Formula used:

$Dfbetween=t–1$

$Dfwithin=n–t$

$Dfsubject=s-1$

$Dferror=dfwithin–dfsubjects$

$Dftotal=n-1$

$SSbetween=\frac{(({\displaystyle \sum x_1{)}^2}+({\displaystyle \sum x_2{)}^2}+({\displaystyle \sum x_3{)}^2})}{s}-\left(\frac{({\displaystyle \sum x{)}^2}}{n}\right)$

$SSwithin={\displaystyle \sum x^2}-\left(\frac{({({\displaystyle \sum x_1})}^2+({\displaystyle \sum x_2{)}^2}+({\displaystyle \sum x_3{)}^2})}{s}\right)$

$SSsubjects=\frac{(P{a}^2+P{b}^2+P{c}^2+P{d}^2)}{t}-\left(\frac{{({\displaystyle \sum x})}^2}{n}\right)$

$\begin{array}{l}SSerror=SSwithin–SSsubject\\ SStotal=SSbetween+SS within\\ MSbetween=\frac{SSbetween}{dfbetween}\\ MSerror=\frac{SSerror}{dferror}\\ F=\frac{MSbetween}{MSerror}\end{array}$

Formula of effect size:

$Etasquare=\frac{SSbetween}{(SSbetween+SSerror)}$

Calculations:

Answer to part a:

$\begin{array}{l}Dfbetween=t-1=3-1=2\\ Dfwithin=n–t =12-3=9\\ Dfsubject=s-1=4-1=3\\ Dferror=dfwithin–dfsubjects=9-3=6\\ Dftotal=n-1=12–1=11\\ SSbetween=\frac{\left(64+576+784\right)}{4}–\left(\frac{3600}{12}\right)=356–300=56\\ \\ SSwithin=378-356=22\\ \\ SSsubjects=\frac{(144+324+324+144)}{3}-\frac{{(60)}^2}{12}\\ SSsubjects=312–300=12\\ \\ SSerror=within–subject=22-12=10\\ \\ Total=between+within=56+22=78\\ \\ MSbetween=\frac{56}{2}=28\\ \\ MSerror=\frac{10}{6}=1.67\\ \\ F=\frac{28}{1.67}=16.77\end{array}$

Fcritical value for df between 2 and df error 6 for alpha = 0.05 is: 5.14

Inference: Since F statistic 16.77 is greater than F critical 5.14, this implies that there is significant difference in the treatments

Effect size:

$etasquare= \frac{56}{(56+10)}=0.8485$

Answer to part b:

K = 3 since the treatments are three

$N=12X2=24$ .... [since the subjects are doubled]

$\begin{array}{l}Dfbetween=t-1=3-1=2\\ Dfwithin=n–t =\text{ 24}-3=\text{ 21}\\ Dfsubject=s-1=\text{ 8}-1=\text{ 7}\\ Dferror=dfwithin–dfsubjects=\text{ 21 }-7=14\\ Dftotal=n-1=12–1=11\\ SSbetween=\frac{\left(\text{256 }+2304+3136\right)}{8}–\frac{\text{14400}}{24}=\text{ 712 }–\text{ 600 }=\text{ 112}\\ \\ SSwithin=\text{ 756 }-712=\text{ 44}\\ \\ SSsubjects=\frac{(144+324+324+144)}{3}-\frac{{(60)}^2}{12}\\ SSsubjects=\text{ 624 }–\text{ 6}00=\text{ 24}\\ \\ SSerror=within–subject=\text{ 44 }-24=\text{ 2}0\\ \\ Total=between+within=\text{ 112}+44=\text{ 156}\\ \\ MSbetween=\frac{\text{112}}{2}=\text{ 56}\\ MSerror=\frac{\text{20 }}{14}=1.43\\ F=\frac{\text{56 }}{1.43}=\text{ 179}\text{.0210}\end{array}$

F critical, for df between = 2, and df within = 14 for alpha = 0.05 is 3.74

Inference: Since F statistic 179.0210 is greater than F critical 3.74, this implies that there is significant difference in the treatments

$Effectsize=\frac{112}{(112+20)}=\text{ 0}\text{.8485}$

Answer to part c:

The if the sample size is doubled the F ratio's value increases further, but the effect size does not alter

Conclusion:

Thus we conclude that the treatments have a significant effect, and the effect size remains same even if the sample size is doubled.

Justification:

This process of repeated measures of ANOVA is applied because there are 3 groups and we need to compare the mean of the same characteristic in the three groups. ANOVA is used to compare means of multiple groups together. It lets us know if there is significant difference between the means of 3 different treatments.