Chapter 13, Problem 9P

One of the primary advantages of a repealed-measures design, compared to an independent-measures design, is that it reduces the overall variability by removing variance caused by individual differences. The following data are from a research study comparing three treatment conditions.

1. a. Assume that the data are from an independent- measures study using three separate samples, each with n = 6 participants. Ignore the column of P totals and use an independent-measures ANOVA with σ = .05 to test the significance of the mean differences.
2. b. Now assume that the data are from a repealed- measures study using the same sample of n = 6 participants in all three treatment conditions. Use a repealed-measures ANOVA with α = .05 to test the significance of the mean differences.
3. c. Explain why the two analyses lead to different conclusions.

To determine

To test: The significance of the mean difference using independent measure ANOVA with $\alpha =0.05$.

Answer to Problem 9P

The test of significance of the mean difference using independent measure ANOVA with $\alpha =0.05$ shows that there is no significant mean difference among the treatments.

Explanation of Solution

Given info:

The data is provided in the question.

Treatment 1	Treatment 2	Treatment 3	$\begin{array}{l}N=18\\ G=108\\ \sum X^2=800\end{array}$
$M=4$	$M=6$	$M=8$
$T=24$	$T=36$	$T=48$
$SS=42$	$SS=28$	$SS=34$

Calculation:

Step 1:

The null and alternative hypotheses are:

Null hypothesis:

  $H_0:{\mu }_1={\mu }_2={\mu }_3$,

Where ${\mu }_1\text{, }{\mu }_2\text{, }{\mu }_3$ are the means of the 3 treatments.

Alternate hypothesis:

$H_a:$ At least one treatment mean is different.

Step 2:

Compute the degrees of freedom (df) for the between treatment effects, within treatment effects and total and corresponding sum of squares (SS).

Now, it is known that, for a given sample size, the degrees of freedom (df) is:

$df=\left(\text{number of units}\right)-1$.

Thus,

$\begin{array}{c}d{f}_{\text{total}}=N-1\\ =18-1\\ =17\end{array}$

The number of treatments, $k=3$. Thus,

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

As within treatments degrees of freedom, $d{f}_{\text{within}}=d{f}_{\text{total}}-d{f}_{\text{between}\text{treatment}}$, thus,

$\begin{array}{c}d{f}_{\text{within}}=d{f}_{\text{total}}-d{f}_{\text{between}\text{treatment}}\\ =17-2\\ =15\end{array}$

With $\alpha =0.05$ and $df=\left(2,15\right)$ critical value from table is $CV=3.68$

Step 3:

Compute F-ratio.

The total sum of squares is:

$S{S}_{\text{total}}={\displaystyle \sum X^2-\frac{G^2}{N}}$.

The within treatment sum of squares is:

$S{S}_{\text{within}\text{treatment}}={\displaystyle \sum S{S}_{\text{inside}\text{each}\text{treatment}}}$.

The within treatment sum of squares is:

$S{S}_{\text{between}}={\displaystyle \sum \frac{T^2}{n}-\frac{G^2}{N}}$.

Thus, here,

$\begin{array}{c}S{S}_{total}={\displaystyle \sum X^2-\frac{G^2}{N}}\\ =800-\frac{{108}^2}{18}\\ =152\end{array}$

$\begin{array}{c}S{S}_{within}={\displaystyle \sum S{S}_{insideeachtreatment}}\\ =42+28+34\\ =104\end{array}$

$\begin{array}{c}S{S}_{between}=S{S}_{total}-S{S}_{within}\\ =152-104\\ =48\end{array}$

Now,

$\begin{array}{c}M{S}_{between}=\frac{S{S}_{between}}{d{f}_{between}}\\ =\frac{48}{2}\\ =24\end{array}$

Similarly,

$\begin{array}{c}M{S}_{within}=\frac{S{S}_{within}}{d{f}_{within}}\\ =\frac{104}{15}\\ =6.933\end{array}$

Finally F-ratio formula is:

$\begin{array}{c}F\text{-}ratio=\frac{M{S}_{between}}{M{S}_{within}}\\ =\frac{24}{6.933}\\ =3.46\end{array}$

Step 4:

Decision rule:

If $F\text{-value}\ge \alpha$, then reject the null hypothesis $H_0$.

If $F\text{-value<}\alpha$, then fail to reject the null hypothesis $H_0$.

Since $F\text{-}ratio\left(=3.46\right)>\text{critical value}\left(=3.68\right)$, no evidence is there to reject the null hypothesis.

Based on the result we fail to reject null and conclude, there is no significance difference among the three treatment.

To determine

The significance of mean difference using repeated-measures ANOVA with $\alpha =0.05$

Answer to Problem 9P

Using repeated-measures ANOVA with $\alpha =0.05$, one may reject null hypothesis and conclude significant mean difference exists.

Explanation of Solution

Calculation:

Step 1:

The null and alternative hypotheses are:

Null hypothesis:

  $H_0:{\mu }_1={\mu }_2={\mu }_3$,

Where ${\mu }_1\text{, }{\mu }_2\text{, }{\mu }_3$ are the means of the 3 treatments.

Alternate hypothesis:

$H_a:$ At least one treatment mean is different.

Step 2:

First stage:

Compute the degrees of freedom (df) for the between treatment effects, within treatment effects and total and corresponding sum of squares (SS).

$d{f}_{within}=15$ and $d{f}_{between}=2$, $S{S}_{total}=152$, $S{S}_{within}=104$, $S{S}_{between}=48$.

Second stage:

Compute the values of between subject sum of squares and sum of squares due to error.

The between subject sum of squares is:

$S{S}_{\text{between}\text{subject}}={\displaystyle \sum \frac{p^2}{k}-\frac{G^2}{N}}$.

The sum of squares due to error is:

$S{S}_{\text{error}}=S{S}_{\text{within}}-S{S}_{\text{between}\text{subject}}$.

Thus,

$\begin{array}{l}S{S}_{betweensubject}={\displaystyle \sum \frac{p^2}{k}-\frac{G^2}{N}}\\ =\frac{{27}^2}{3}+\frac{{24}^2}{3}+\frac{{21}^2}{3}+\frac{{12}^2}{3}+\frac{9^2}{3}+\frac{{15}^2}{3}-\frac{{108}^2}{18}\\ =84\end{array}$

$\begin{array}{c}S{S}_{error}=S{S}_{within}-S{S}_{betweensubject}\\ =104-84\\ =20\end{array}$

$\begin{array}{c}d{f}_{error}=d{f}_{within}-d{f}_{betweensubject}\\ =15-5\\ =10\end{array}$

Here, $\alpha =0.05$ and $df=\left(2,10\right),CV=4.10$

Thus,

$\begin{array}{l}M{S}_{between}=\frac{S{S}_{between}}{d{f}_{between}}\\ =24\end{array}$

$\begin{array}{l}M{S}_{error}=\frac{S{S}_{error}}{d{f}_{error}}\\ =2\end{array}$

Step 3:

$\begin{array}{l}F\text{-}ratio=\frac{M{S}_{between}}{M{S}_{error}}\\ =\frac{24}{2}=12\end{array}$

Step 4:

Decision rule:

If $F\text{-value}\ge \alpha$, then reject the null hypothesis $H_0$.

If $F\text{-value<}\alpha$, then fail to reject the null hypothesis $H_0$.

Since $F\text{-}ratio\left(=12\right)>CV\left(=4.10\right)$ reject $H_0$.

Based on result reject null and conclude significance mean difference exists.

To determine

To find: Why the results of analysis differ in part a and part b.

Explanation of Solution

In part a, in independent-measures ANOVA, the individual differences is not present while in part b the individual differences exists. As a result, the individual differences have been eliminated in part b as these effects impact the results.

Thus, the results of analysis differ in part a and part b.