Question

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

A psychologist would like to examine the effects of three 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. A separate sample of 10 students were assigned in each testing method condition. At the end of the course, the psychologist interviews each student to get a measure of the student’s overall knowledge of the material. Do these data indicate any significant differences among the three methods? Test with α = .05.

*Partial information is provided for you in the ANOVA summary table for the problem*

Source SS df MS F
Between 16
Within -
Total 70 - -

State the null and alternative hypothesis ?

b. Compute dfB, dfW, and dfT ?

c. Determine the boundary (critical) value of rejection region under alpha level equal to .05 ?

d. Compute SSW (2 pt) e. Compute MSB and MSW ?

f. Compute the F test statistic ?

g. Complete the ANOVA summary table (copy the numbers you computed to the following summary table) ?

Source SS df MS F
Between
Within -
Total - -

h. What is your statistical decision on your test of hypothesis? And your conclusion based on your test ?

i. Compute the effect size of the ANOVA analysis you just conducted.

Homework Answers

Answer #1

H0: µ1 = µ2 = µ3 : There is no significant difference among the three testing methods

H1: There is significant difference among the three testing methods

n=10

k =3 (no. of treatments)

b) dfT = 10-1=9

dfB = k-1 =3-1=2

dfW = n-k = 10-3 =7

c) Critical F (Using Excel F.INV.2T(probability, dfB, dfW)) = F.INV.2T(0.05,9,2) = 4.73

d) SSW = TSS-SSB = 70-16 = 54

e) MSB = SSB/dfB = 16/2 = 8

MSW = SSW/dfW = 54/7 = 7.71

f) F statistic = MSB/MSW = 8/7.71 = 1.04

g) ANOVA Table

Source of Variation SS df MS F
Between Groups 16.00 2 8.00 1.04
Within Groups 54.00 7 7.71
Total 70.00 9

h) Since test statistic (1.04) is less than critical value (4.73), we do not reject the null hypothesis and conclude that there is no significant difference among the three testing methods.

i) Effect size of ANOVA = SSB/TSS = 16/70 = 0.23

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