Question

To test the relationship between gender and ratings of a promiscuous partner, a group of men and women was given a vignette describing a person of the opposite sex who was in a dating relationship with one, two, or three partners. Participants rated how positively they felt about the individual described in the vignette, with higher ratings indicating more positive feelings.

Source of Variation	SS	df	MS	F
Gender	10
Promiscuity
Gender × Promiscuity	146
Error	570	114
Total	816

(a) Complete the F-table and make a decision to retain or reject the null hypothesis for each hypothesis test. (Assume experimentwise alpha equal to 0.05.)

Source of Variation	SS	df	MS	F
Gender	10
Promiscuity
Gender × Promiscuity	146
Error	570	114
Total	816

Answer 1

Source of Variation	SS	df	MS	F
Gender	10	1	10	2
Promiscuity	90	2	45	9
Gender × Promiscuity	146	2	73	14.6
Error	570	114	5
Total	816	119

First we will calculate all df values.

df Gender = Number of levels of Gender - 1 = 2 - 1 = 1

df Promiscuity = Number of levels of Promiscuity - 1 = 3 - 1 = 2

df Gender x Promiscuity = df Gender x df Promiscuity = 1 * 2 = 2

df Total = df Gender + df Promiscuity + df Gender x Promiscuity + df Error

= 1 + 2 + 2 + 114 = 119

SS Promiscuity = SS Total - (SS Gender + SS Gender x Promiscuity + SS Error)

= 816 - (10 + 146 + 570)

= 90

MS Gender = SS Gender / df Gender = 10 / 1 = 10

MS Promiscuity = SS Promiscuity / df Promiscuity = 90 /2 = 45

MS Gender × Promiscuity = SS Gender × Promiscuity / df Gender × Promiscuity = 146 / 2 = 73

MS Error = SS Error / df Error = 570 / 114 = 5

F Gender = MS Gender / MS Error = 10 / 5 = 2

F Promiscuity = MS Promiscuity / MS Error = 45 / 5 = 9

F Gender × Promiscuity = MS Gender × Promiscuity / MS Error = 73 / 5 = 14.6

Critical value of F at 0.05 significance level and df = df Gender, df Error = 1, 9 is 5.12

Since the observed F (2) is less than the critical value, we fail to reject the null hypothesis H0 and conclude that there is no significant main effect of gender on ratings.

Critical value of F at 0.05 significance level and df = df Gender, df Error = 2, 9 is 4.26

Since the observed F (9) is greater than the critical value, we reject the null hypothesis H0 and conclude that there is significant main effect of Promiscuity on ratings.

Since the observed F (14.6) is greater than the critical value, we reject the null hypothesis H0 and conclude that there is significant interaction effect of Gender and Promiscuity on ratings.