A psychologist would like to examine the effects of different testing methods on

Question

A psychologist would like to examine the effects of different testing methods on the final performance of college students. One group gets regular quizzes, one group gets three large exams, and the third group only gets a final exam. At the end of the course, the psychologist interviews each student to get a measure of the student’s overall knowledge of the material.

Quizzes Exams Final Only

4 1 0

6 4 2

3 5 0

7 2 2

G=36

(sigma)x^2=164

T = 20 T = 12 T=4

SS = 10 SS = 10 SS= 4

a. State the null hypothesis. (1 POINT)

b. Compute the necessary hypothesis test using a two-tailed test with = .05. Show

your calculations not just ANOVA table. (5 POINTS)

c.Make a decision about your null hypothesis. (Include the critical value you used to

base your decision.) (1 POINT)

d.Write an APA format statement summarizing your findings. (Include a measure of effect size if necessary.) (2 POINTS)

e.Explain what would need to be done next to follow up on any significant effects. (2 POINTS)

Explanation / Answer

One way ANOVA

The null and alternative hypothesis for this one way ANOVA is given as below:

Null hypothesis: H0: There is no significant difference in the averages of the scores for the quizzes, exams and final only.

Alternative hypothesis: There is a significance difference in the averages of the scores for the quizzes, exams and final only.

The ANOVA table with calculations is given as below:

ANOVA: Single Factor

SUMMARY

Groups

Count

Sum

Average

Variance

Quizzes

4

20

5

3.3333

Exams

4

12

3

3.3333

Final Only

4

4

1

1.3333

ANOVA

Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

32.0000

3 – 1 = 2

32/2 = 16.0000

16/2.6667 = 6.0150

0.0221 By using the F-table

4.2565 By using the F-table for the 0.05 level

Within Groups

24.0000

11 – 2 = 9

24/9 = 2.6667

Total

56.0000

12 – 1 = 11

Level of significance

0.05

Here, we get the p-value for this one way ANOVA test as 0.0221. We do not reject the null hypothesis if the p-value is greater than the given level of significance and we reject the null hypothesis if the p-value is less than the given level of significance or alpha value.

Here, we are given that the p-value = 0.0221 and level of significance or alpha value = 0.05, that is p-value is less than the given level of significance. So, we reject the null hypothesis that there is no significant difference in the averages of the scores for the quizzes, exams and final only. This means we conclude that there is a significance difference in the averages of the scores for the quizzes, exams and final only.

ANOVA: Single Factor

SUMMARY

Groups

Count

Sum

Average

Variance

Quizzes

4

20

5

3.3333

Exams

4

12

3

3.3333

Final Only

4

4

1

1.3333

ANOVA

Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

32.0000

3 – 1 = 2

32/2 = 16.0000

16/2.6667 = 6.0150

0.0221 By using the F-table

4.2565 By using the F-table for the 0.05 level

Within Groups

24.0000

11 – 2 = 9

24/9 = 2.6667

Total

56.0000

12 – 1 = 11

Level of significance

0.05

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