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	5
Promiscuity
Gender × Promiscuity	150
Error	570	114
Total	815

(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	5
Promiscuity
Gender × Promiscuity	150
Error	570	114
Total	815

Answer 1

(a)

Source of Variation	SS	df	MS	F
Gender	5	2-1=1	5/1=5	5/5=1
Promiscuity	815-5-150-570=90	3-1=2	90/2=45	45/5=9
Gender × Promiscuity	150	(2-1)*(3-1) =2	150/2=75	75/5=15
Error	570	114	570/114=5
Total	815	1+2+2+114=119

p-value corresponding gender=P(F>1|F~F1,114)=0.3194>0.05

(Use R code for finding p-value: "round(1-pf(1,1,114),4)")

Since p-value>0.05 so we fail to reject the null hypothesis, H01: Effect of gender is absent and hence effect of gender is insignificant.

p-value corresponding gender=P(F>9|F~F2,114)=0.0002<0.05

(Use R code for finding p-value: "round(1-pf(9,2,114),4)")

Since p-value<0.05 so we reject the null hypothesis, H02: Effect of Promiscuity is absent and hence effect of Promiscuity is significant.

p-value corresponding gender=P(F>15|F~F2,114)=0.0000<0.05

(Use R code for finding p-value: "round(1-pf(15,2,114),4)")

Since p-value<0.00 so we reject the null hypothesis, H03: Interaction Effect Gender ×Promiscuity is absent and hence interaction effect Gender ×Promiscuity is significant.