Chapter 11, Problem 22P

22. The previous problem demonstrates that removing individual differences can substantially reduce variance and lower the standard error. However, this benefit only occurs if the individual differences are consistent across treatment conditions. In problem 21, for example, the participants with the highest scores in the more-sleep condition also had the highest scores in the less-sleep condition. Similarly, participants with the lowest scores in the first condition also had the lowest scores in the second condition. To construct the following data, we started with the scores in problem 21 and scrambled the scores in treatment 1 to eliminate the Consistency of the individual differences.

Number of Academic Problems
Student	Following Nights

with Above

Average Sleep Following Nights

with Below

Average Sleep A 10 13 B 8 14 C 5 13 D 5 5 E 4 9 F 10 6 G 11 6 II 3 6

a. Treat the data as if the scores are from an independent-measures study using two separate samples, each with $n=8$ participants. Compute the pooled variance, the estimated standard error for the mean difference, and the independent-measures t statistic. Using a two-tailed test with $\alpha =.05$ , is there a significant difference between the two sets of scores? Note: The scores in each treatment are the same as in Problem 21. Nothing has changed.

b. Now assume that the data are from a repeated-measures study using the same sample of $n=8$ participants in both treatment conditions. Compute the variance for the sample of difference scores, the estimated standard error for the mean difference and the repeated-measures K statistic. Using a two-tailed test with $\alpha =.05$ , is there a significant difference between the two sets of scores? (You should find that removing the individual differences with a repeated-measures t no longer reduces the variance because there are no consistent individual differences.)

To determine

Is there a significant difference between the two sets of scores by using an independent-measures design and then by using repeated measures design.

Answer to Problem 22P

Solution:

1. Using an independent-measures design:

   STEP 1: State the hypotheses. In symbols, the null and alternative hypotheses are:

   $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_a:{\mu }_1\ne {\mu }_2\end{array}$ STEP 2: Locate the critical region. The critical value of t values for the critical region are $t=\pm 2.145$ .

   STEP 3: Compute the test statistic. The test statistic is $t=-1.15$ .

   STEP 4: Make a decision about the null hypothesis. We cannot conclude that there is a significant difference between the two sets of scores by using an independent-measures design.

2. Using repeated measures design:
STEP 1: State the hypotheses. In symbols, the null and alternative hypotheses are:

$\begin{array}{l}{H}_0:{\mu }_D=0\\ H_a:{\mu }_D\ne 0\end{array}$ STEP 2: Locate the critical region. The critical value of t values for the critical region are $t=\pm 2.365$ .

STEP 3: Compute the test statistic. The test statistic is $t=2.94$ .

STEP 4: Make a decision about the null hypothesis. We can conclude that there is a significant difference between the two sets of scores by using repeated measures design.

Explanation of Solution

We are given to use a two-tailed test. That's why alternative hypothesis has $\ne$ sign.

The critical value of t values for the critical region are $t=\pm 2.145$. So, reject null hypothesis if $t<-2.145$ or $t>2.145$ .

The critical value of t values for the critical region are $t=\pm 2.365$. So, reject null hypothesis if $t<-2.365$ or $t>2.365$ .

Given:

The table that summarizes the scores for a sample of $n=8$ participants and $\alpha =0.05$ .

Formula used:

$df=n_1+n_2-2$ $S_P^2=\frac{S{S}_1+S{S}_2}{n_1+n_2-2}$ $SE=\sqrt{\frac{S_P^2}{n_1}+\frac{S_P^2}{n_2}}$ $t=\frac{M_1-M_2}{SE}$

$\begin{array}{l}df=n-1\\ S_D=\sqrt{\frac{SS}{n-1}}\\ t=\frac{M_D-{\mu }_D}{S_D/\sqrt{n}}\end{array}$

Calculation:

1. Using an independent-measures design:
STEP 1: State the hypotheses. The null hypothesis states that there is no difference between the two sets of scores. In symbols:

$H_0:{\mu }_1={\mu }_2$

The alternative hypothesis states that there is a significant difference between the two sets of scores. In symbols:

$H_a:{\mu }_1\ne {\mu }_2$

STEP 2: Locate the critical region. Degree of freedom is:

$\begin{array}{c}df=n_1+n_2-2\\ =8+8-2\\ =14\end{array}$ From the t distribution table, for a two-tailed test with $\alpha =0.05$ for $df=14$, the critical value of t values for the critical region are $t=\pm 2.145$ .

STEP 3: Compute the test statistic. The data are as follows:

Participant	Above Average Sleep $\left(X_1\right)$	Below Average Sleep $\left(X_2\right)$	${\left(X_1-M_1\right)}^2$	${\left(X_2-M_2\right)}^2$
A	10	13	9	16
B	8	14	1	25
C	5	13	4	16
D	5	5	4	16
E	4	9	9	0
F	10	6	9	9
G	11	6	16	9
H	3	6	16	9
	${\displaystyle \sum X_1}=56$	${\displaystyle \sum X_2}=72$	$S{S}_1=\text{68}$	$S{S}_2=100$

The mean of the above average sleep group is:

$\begin{array}{c}{M}_1=\frac{{\displaystyle \sum X_1}}{n_1}\\ =\frac{56}{8}\\ =7\end{array}$

The mean of the above below sleep group is:

$\begin{array}{c}{M}_2=\frac{{\displaystyle \sum X_2}}{n_2}\\ =\frac{72}{8}\\ =9\end{array}$

The pooled variance is:

$\begin{array}{c}{S}_P^2=\frac{S{S}_1+S{S}_2}{n_1+n_2-2}\\ =\frac{68+100}{8+8-2}\\ =12\end{array}$

The standard error for the mean difference is:

$\begin{array}{c}SE=\sqrt{\frac{S_P^2}{n_1}+\frac{S_P^2}{n_2}}\\ =\sqrt{\frac{12}{8}+\frac{12}{8}}\\ =\sqrt{3}\\ =1.732\end{array}$

The test statistic is:

$\begin{array}{c}t=\frac{M_1-M_2}{SE}\\ =\frac{7-9}{1.732}\\ =-1.15\end{array}$

STEP 4: Make a decision about the null hypothesis. Since test statistic does not fall outside the critical region, fail to reject the null hypothesis. We cannot conclude that there is a significant difference between the two sets of scores by using an independent-measures design.

2. Using repeated measures design:
STEP 1: State the hypotheses. The null hypothesis states that there is no difference between the two sets of scores. In symbols:

$H_0:{\mu }_D=0$

The alternative hypothesis states that there is a significant difference between the two sets of scores. In symbols:

$H_a:{\mu }_D\ne 0$

STEP 2: Locate the critical region. Degree of freedom is:

$\begin{array}{c}df=n-1\\ =8-1\\ =7\end{array}$ From the t distribution table, for a two-tailed test with $\alpha =0.05$ for $df=7$, the critical value of t values for the critical region are $t=\pm 2.365$ .

STEP 3: Compute the test statistic. The data are as follows:

Participant	Above Average Sleep $\left(X_1\right)$	Below Average Sleep $\left(X_2\right)$	Difference $D=X_2-X_1$	${\left(D-M_D\right)}^2$
A	10	13	3	1
B	8	14	6	16
C	5	13	8	36
D	5	5	0	4
E	4	9	5	9
F	10	6	-4	36
G	11	6	-5	49
H	3	6	3	1
			${\displaystyle \sum D}=\text{16}$	$SS=152$

$\begin{array}{c}{M}_D=\frac{{\displaystyle \sum D}}{n}\\ =\frac{16}{8}\\ =2\end{array}$

The standard deviation of the difference between the two sets of scores is:

$\begin{array}{c}{S}_D=\sqrt{\frac{SS}{n-1}}\\ =\sqrt{\frac{152}{8-1}}\\ =\sqrt{21.7143}\\ =4.66\end{array}$

The test statistic is:

$\begin{array}{c}t=\frac{M_D-{\mu }_D}{S_D/\sqrt{n}}\\ =\frac{2-0}{4.66/\sqrt{8}}\\ =1.21\end{array}$

STEP 4: Make a decision about the null hypothesis. Since test statistic does not fall outside the critical region, fail to reject the null hypothesis. We cannot conclude that there is a significant difference between the two sets of scores by using repeated measures design.

Conclusion:

1. We cannot conclude that there is a significant difference between the two sets of scores by using an independent-measures design.
2. We cannot conclude that there is a significant difference between the two sets of scores by using repeated measures design.