Question

1. Perform a repeated-measures ANOVA using the following data. The data represent ten subjects’ measures repeated over five different trials.

Trial 1	Trial 2	Trial 3	Trial 4	Trial 5
1	2	4	5	6
1	1	3	5	6
1	2	5	7	4
2	1	4	6	7
3	3	5	8	8
2	2	4	7	8
1	3	4	6	7
0	2	5	7	8
3	3	6	8	9
2	2	4	7	8

1. Complete the following ANOVA table. You may use SPSS (attach output) but record your answers here.

Source	SS	df	MS	F	Critical Value
Subjects (S)
Trials (T)
Error
Total

2. State in words the relevant null and alternate hypotheses being examined. Make sure you phrase them as accurately as possible.

3. Write an APA-style summary of your interpretation of the results in part (a) including the appropriate statistics (F test, effect size, power, etc.) to support your report. You need to tell the story behind the results obtained from the analysis.

4. Suppose the data for the five trials were not repeated but rather were collected from five independent groups. Rerun the analysis as a one-way ANOVA and summarize your results in the below table. Compare the results of this table to those in part (a) focusing on the SS, df, MS, and F statistic.

Source	SS	df	MS	F	Critical Value
Trials (T)
Error
Total

5. Between the repeated-measures design and one-way ANOVA, which one is more powerful in finding significant differences under the assumption that the null hypothesis is true? Explain how you arrived at your conclusion.

Answer 1

a) ANOVA Table

Source	DF	SS	MS	F	P	Critical Value
Subject	9	28.32	3.1467	5.48	0.000	2.1526
Trial	4	252.52	63.1300	109.90	0.000	2.6335
Error	36	20.68	0.5744
Total	49	301.52

b)

Subject:

Null Hypothesis: The mean of data at 10 subjects are the same.

Alternative Hypothesis: At least one subject has a significant mean difference.

Trial

Null Hypothesis: The mean of data at 5 trials are the same.

Alternative Hypothesis: At least one trail has a significant mean difference.

c) Subject:

The estimated F-statistic for treatment subject is 5.48 and the corresponding critical value at the 0.05 significance level is 2.1526. Here, the test statistic is more than the critical value. Hence, reject the null hypothesis and conclude that at least one subject has a significant mean difference from the other remaining subjects at the 0.05 significance level.

Trail:

The estimated F-statistic for treatment subject is 109.90 and the corresponding critical value at the 0.05 significance level is 2.6335. Here, the test statistic is more than the critical value. Hence, reject the null hypothesis and conclude that at least one trail has a significant mean difference from the other remaining trails at the 0.05 significance level.

d)

Source	DF	SS	MS	F	P	Critical Value
Trial	4	252.52	63.13	57.98	0.000	2.5787
Error	45	49	1.09
Total	49	301.52

Comparing the above two ANOVA tables, we observe that the SS of Trail and SS of the total do not change whereas the SS of Error for the Error is increased in the second ANOVA. Similar to the SS, the df is changed only on error. Further, the F statistic value for the Train is smaller in the one-way ANOVA table.

e) Between the repeated-measures design and one-way ANOVA, the repeated-measures design is more powerful in finding significant differences under the assumption that the null hypothesis is true. Because the SS od error is smaller in the repeated-measures design together with significant Subject effect.