Chapter 13, Problem 11P

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

1. a. Use a repealed-measures ANOVA with α – 05 to determine whether there are significant differences among the treatments and compute η’ to measure the size of the treatment effect.
2. b. Double the number of scores in each treatment by simply repeating the original scores in each treatment a second time. For example, the n = 8 scores in Treatment I become 1.4, 2.1, 1.4, 2.1. Note that this will not change the treatment means but it will double SS_{between treatments}, SS_{between subjects}, and the SS value for each treatment. For the new data, use u repeated-measures ANOVA with α - 05 to determine whether there are significant differences among the treatments and compute η’ to measure the Size of treatment effect.
3. c. Describe how doubling the sample size affected the value of the F-radio and the value of η’.

   

To determine

There are significant differences among the treatments.

To compute: The value of ${\eta }^2$.

Answer to Problem 11P

There are significant differences between the treatments. The value of ${\eta }^2$ is 0.848.

Explanation of Solution

Given info:

Following data are from a repeated-measure study comparing three treatments conditions.

	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	$P=12$	$\sum X^2=378$
	$M=2$	$M=6$	$M=7$
	$T=8$	$T=24$	$T=28$
	$SS=6$	$SS=10$	$SS=6$

Calculation:

The hypotheses are given below:

Null Hypothesis: There are no differences among treatments.

Alternate Hypothesis: There are significant differences among the treatments.

Let k and n represents the total numbers of treatments and participants. Then,

$\begin{array}{l}k=3\\ n=4\end{array}$

Let, N represents total numbers of observations. Then,

$\begin{array}{c}N=k\times n\\ =3\times 4\\ =12\end{array}$

$S{S}_{total}$ is given as:

$\begin{array}{c}S{S}_{total}=\sum X^2-\frac{G^2}{N}\\ =378-\frac{{\left(60\right)}^2}{12}\\ =378-300\\ =78\end{array}$

$S{S}_{withintreatments}$ is given as:

$\begin{array}{c}S{S}_{withintreatments}=\sum SS\\ =6+10+6\\ =22\end{array}$

$S{S}_{betweensubjects}$ is given as:

$\begin{array}{c}S{S}_{betweensubjects}=\sum \frac{P^2}{k}-\frac{G^2}{N}\\ =\frac{{\left(12\right)}^2}{3}+\frac{{\left(18\right)}^2}{3}+\frac{{\left(18\right)}^2}{3}+\frac{{\left(12\right)}^2}{3}-\frac{{\left(60\right)}^2}{12}\\ =48+108+108+48-300\\ =12\end{array}$

$S{S}_{betweentreatments}$ is given as:

$\begin{array}{c}S{S}_{betweentreatments}=\sum \frac{T^2}{n}-\frac{G^2}{N}\\ =\frac{{\left(8\right)}^2}{4}+\frac{{\left(24\right)}^2}{4}+\frac{{\left(28\right)}^2}{4}-\frac{{\left(60\right)}^2}{12}\\ =16+144+196-300\\ =356-300\\ =56\end{array}$

$S{S}_{error}$ is given as:

$\begin{array}{c}S{S}_{error}=S{S}_{withintreatments}-S{S}_{betweensubjects}\\ =22-12\\ =10\end{array}$

Degrees of freedom corresponding to $M{S}_{betweentreatments}$ are:

$\begin{array}{c}d{f}_{betweentreatments}=k-1\\ =3-1\\ =2\end{array}$

Degrees of freedoms corresponding to $d{f}_{error}$ are:

$\begin{array}{c}d{f}_{error}=\left(k-1\right)\times \left(n-1\right)\\ =\left(3-1\right)\times \left(4-1\right)\\ =2\times 3\\ =6\end{array}$

F-ratio is given as:

$\begin{array}{c}F=\frac{\frac{S{S}_{betweentreatments}}{d{f}_{betweentreatments}}}{\frac{S{S}_{error}}{d{f}_{error}}}\\ =\frac{\frac{56}{2}}{\frac{10}{6}}\\ =\frac{28}{1.67}\\ =16.77\end{array}$

From the table $B.4$ of appendix B in the textbook, critical value corresponding to $df=2,6$ is 5.14.

Since, F-ratio is greater than the critical value; therefore, reject the null hypothesis and conclude that there are significant differences between the treatments.

${\eta }^2$ to measure the size of the treatment effect is given as:

$\begin{array}{c}{\eta }^2=\frac{S{S}_{betweentreatments}}{S{S}_{error}+S{S}_{betweentreatments}}\\ =\frac{56}{10+56}\\ =\frac{56}{66}\\ =0.848\end{array}$

Conclusion:

There are significant differences between the treatments. The value of ${\eta }^2$ is 0.848.

To determine

There are significant differences among the treatments for the new data.

To compute: The value of ${\eta }^2$.

Answer to Problem 11P

There are significant differences between the treatments. The value of ${\eta }^2$ is 0.848.

Explanation of Solution

Calculation:

The new data is obtained by doubling the number of scores in each treatment by simply repeating the original scores in each treatment a second time.

The hypotheses are given below:

Null Hypothesis: There are no differences among treatments.

Alternate Hypothesis: There are significant differences among the treatments.

Let k1 and n1 represents the total numbers of treatments and participants of the new data. Then,

$\begin{array}{l}k1=3\\ n1=8\end{array}$

Let, N1 represents total numbers of observations of new data. Then,

$\begin{array}{c}N1=k1\times n1\\ =3\times 8\\ =24\end{array}$

$S{S}_{total}$ of new data is given as:

$\begin{array}{c}S{S}_{total}1=2\times S{S}_{total}\\ =2\times 78\\ =156\end{array}$

$S{S}_{withintreatments}$ for the new data is given as:

$\begin{array}{c}S{S}_{withintreatments}1=2\times S{S}_{withintreatments}\\ =2\times 22\\ =44\end{array}$

$S{S}_{betweensubjects}$ for the new data is given as:

$\begin{array}{c}S{S}_{betweensubjects}1=2\times S{S}_{betweensubjects}\\ =2\times 12\\ 24\end{array}$

$S{S}_{betweentreatments}$ for new data is given as:

$\begin{array}{c}S{S}_{betweentreatments}1=2\times S{S}_{betweentreatments}\\ =2\times 56\\ =112\end{array}$

$S{S}_{error}$ for the new data is given as:

$\begin{array}{c}S{S}_{error}1=2\times S{S}_{error}\\ =2\times 10\\ =20\end{array}$

Degrees of freedom corresponding to $M{S}_{betweentreatments}$ for the new data are:

$\begin{array}{c}d{f}_{betweentreatments}1=k1-1\\ =3-1\\ =2\end{array}$

Degrees of freedoms corresponding to $d{f}_{error}$ for the new data are:

$\begin{array}{c}d{f}_{error}1=\left(k-1\right)\times \left(n1-1\right)\\ =\left(3-1\right)\times \left(8-1\right)\\ =2\times 7\\ =14\end{array}$

F-ratio is given as:

$\begin{array}{c}F1=\frac{\frac{S{S}_{betweentreatments}1}{d{f}_{betweentreatments}1}}{\frac{S{S}_{error}1}{d{f}_{error}1}}\\ =\frac{\frac{112}{2}}{\frac{20}{14}}\\ =\frac{56}{1.43}\\ =39.16\end{array}$

From the table $B.4$ of appendix B in the textbook, critical value corresponding to $df=2,14$ is 3.74.

Since, F-ratio is greater than the critical value; therefore, reject the null hypothesis and conclude that there are significant differences between the treatments for the new data.

${\eta }^2$ to measure the size of the treatment effect for the new data is given as:

$\begin{array}{c}{\eta }^{2}1=\frac{S{S}_{betweentreatments}1}{S{S}_{error}1+S{S}_{betweentreatments}1}\\ =\frac{112}{20+112}\\ =\frac{112}{132}\\ =0.848\end{array}$

Conclusion:

There are significant differences between the treatments. The value of ${\eta }^2$ is 0.848.

To determine

The affect on F-ratio and ${\eta }^2$ by doubling sample size.

Answer to Problem 11P

The F-ratio got increased and there is no effect on ${\eta }^2$.

Explanation of Solution

Given info:

Sample size has been doubled.

Calculation:

From part a. the F-ratio is $F=16.77$.

After doubling the sample size, from part b. the F-ratio is $F=39.16$.

But value of ${\eta }^2$ is same after doubling the sample size that is ${\eta }^2=0.848$.

Conclusion:

The F-ratio got increased and there is no effect on ${\eta }^2$ by doubling the sample size.