1. 35 pt. (Assumptions: 3 pt. Hypotheses: 6 pt. Means: 2pt. Variances:3pt. Test

Question

1. 35 pt. (Assumptions: 3 pt. Hypotheses: 6 pt. Means: 2pt. Variances:3pt. Test Statistics: 10 pt. Critical Value: 6 pt. Conclusion: 5 pt.) In a test of the hypothesis that men are more aggressive than females, males and females were videotaped while interacting in stressful scenarios and the number of aggressive behaviors by each sex was noted. Using the following number of aggressive behaviors in these interactions, test the null hypothesis that there are no sex differences in the prevalence of aggressive behaviors (=0.01) (Hint: This question is about Chapter 7) Males Females 6 4 7 6 6 4

Explanation / Answer

Here the mean and standard deviation table is given below :

Alternative Hypothesis : Ha : There is significant difference in aggressive behavior between men and women, in which men are tend to be more aggressive then women.

Null Hypothesis : H0: There is no difference in aggressive behavior between men and women.

Here , variance can be considered equal so we can use t- test for two sample mean for hypothesis testing.  

Pooled Variance Sp2= (s12+ s22)  /4 = (2.50 + 2.30)/2 = 2.4

Test statistic :

t = (xmen - xwomen)/ sqrt [sp (1/n1+ 1/n2) ]

t = (5 - 4.4)/ sqrt [2.4 * (2/5)]

t = 0.6/ 0.98

t = 0.612

Here for dF = 5 + 5-2 = 8 and alpha = 0.01

tcr=t0.01,8 = 2.8965

so here t < tcritical so we shall reject the null hypothesis and can conclude that Males and females are equally aggressive in stressful situations.

Males Females 6 3 4 6 7 6 3 3 5 4 Mean 5 4.4 Variance 2.50 2.30

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