To test the relationship between gender and ratings of a promiscuous partner, a
ID: 3297388 • Letter: T
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.
(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.)
State the decision for the main effect of gender.
Retain the null hypothesis.Reject the null hypothesis.
State the decision for the main effect of promiscuity.
Retain the null hypothesis.Reject the null hypothesis.
State the decision for the interaction effect.
Retain the null hypothesis.Reject the null hypothesis.
(b) Based on the results you obtained, what is the next step?
Compute simple main effect tests for the significant interaction.Compute pairwise comparisons for the promiscuity factor. No further analysis is needed, because none of the effects are significant.Compute pairwise comparisons for the gender factor.
Source of Variation SS df MS F Gender 5 Promiscuity Gender × Promiscuity 152 Error 570 114 Total 867Explanation / Answer
a. Note there are two levels of factor gender (A)-men and women and three levels of factor promiscuity (B)-one, two and three. Compute and extra column P to come to conclusion regarding rejection of null hypotheses.
Per rejection rule based on p value, reject null hypothesis if p value is less than alpha=0.05. Hence, main effect gender is not significant (p value>0.05). Therefore, retain the null hypothesis. The main effect promiscuity is significant (p value<0.05). Therefore, reject null hypothesis. The interaction effect is significant (p value<0.05). Therefore, reject null hypothesis.
b. Since, interaction effect and promiscuity main effect are significant, conduct pairwise comparison for promiscuity factor. First, third and fourth options are inevitably wrong.
Source of variation SS df MS F P Gender 5 1[(a-1)=2-1] 5 [SSA/(a-1)=5/1] 1[MSA/MSE=5/5] 0.3194[p value at F(1,114)=1] Promisquity 140[867-(5+152+570)] 2[(b-1)=3-1] 70[SSB/(b-1)=140/2] 14[MSB/MSE=70/5] 0.0000[p value at F(2, 114)=14] Gender*Promisquity 152 2 [(a-1)(b-1)=1*2] 76[SS(AB)/(a-1)(b-1)=152/2] 15.2[MS(AB)/MSE=76/5] 0.0000[F(2,114)=15.2] Error 570 114 5[SSE/ab(n-1)=570/114] Total 867 119Related Questions
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