1. A researcher computes the following one-way between-subjects ANOVA table for
ID: 3392928 • Letter: 1
Question
1. A researcher computes the following one-way between-subjects ANOVA table for a study where k = 3 and n = 12. State the decision at a .05 level of significance. (Hint: Complete the table first.)
Source of Variation
SS
df
MS
F
Between groups
120
Within groups (error)
Total
780
a. Reject the null hypothesis.
b. Retain the null hypothesis.
c. There is not enough information to answer this question.
2. A researcher records depression scores for 12 patients in each of four seasons. If depression scores for the same participants are recorded in each season, then what type of statistical design is most appropriate for this study?
a. a related samples t-test
b. a one-way between-subjects ANOVA
c. a one-way within-subjects ANOVA
d. both A and C
3. A one-way within-subjects ANOVA is typically associated with ______ power than the one-way between-subjects ANOVA.
a. depleted
b. larger
c. more significant
4. The term "within-subjects" refers to
a. observing the same participants in each group
b. observing different participants in each group
c. the type of post hoc test conducted
d. the type of effect size estimate measured
5. Which of the following post hoc tests is associated with the least power to detect an effect?
a. Fisher's LSD test
b. Tukey's HSD test
c. Studentized Newman-Keuls
d. None; each post hoc test is associated with the same power to detect an effect.
6. To summarize a one-way within-subjects ANOVA, we report each of the following except,
a. the test statistic value
b. the degree of freedom
c. the critical values
d. the p value
Source of Variation
SS
df
MS
F
Between groups
120
Within groups (error)
Total
780
a. Reject the null hypothesis.
b. Retain the null hypothesis.
c. There is not enough information to answer this question.
Explanation / Answer
1.k=3 n=12
sources of variation SS df MS F
between group 120 k-1=2 120/2=60 60/73.33=0.818
within group 780-120=660 n-k=9 660/9=73.33
total 780 n-1=11
critical value is at 0.05 level of significance=F0.05,2,9=4.26
but 0.818<4.26
hence correct option b) retain the null hypothesis
2) option c) a one-way within-subjects ANOVA as the scores are recorded for each student.
3) answer is more significant
4) option a-observing the same participants in each group
6) the critical value is not reported as p value is calculated
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