Question

Test the claim that the proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.01 significance level. The null and alternative hypothesis would be: HPM = PF HOPM = PF HUM = Up HPM = PF H:MM = MF HIM = MF HP <PF HPM > PF HUM > HF HPM + PF HUM uf HUM<Hp The test is: left-tailed right-tailed two-tailed Based on a sample of 40 men, 40% owned cats Based on a sample of 40 women, 65% owned cats The test statistic is: (to 2 decimals) The p-value is: (to 4 decimals) Based on this we Fail to reject the null hypothesis Reject the null hypothesis

Answer 1

Solution Given data 2 = 0.01 X,: The number of own Cats by men (40 X 0.4=16) Xg: The number of own cats by women (40X0.65= 26). ^ = 0.4 P2 = 40 P2 = 0.65 P= X, 4X2 mit ng 16+26 40+40 P= 42 P = 0.525 hypothesis Now to state the null and alternative Nutt hypothesis Ho: P = PF Alternative hypothesis Hi: PM & PF The test is two-tailed test

statistic Now to find the test We know that The formula of test Test statistic , z = , Z is given by statistic Pi-la V P(-7) (thi + ) 0.4-0.65 z = 10.595 (1-0.525) ! ! ( 40 tt 40 - 0.25 Z = 0.525 X 0.475 0.025 + 0.025) - 0.25 N 0.525 X0.475 X 0.05 -0.25 N 0.0125 -0.25 N = 0.1118 za - 2.236 Test statistic , Z = -2.24

Now to find the p-value p-value = 0.0251 Decision p-value > L 0.0251 7o.ol Since the p-value is greater than significance level. Therefore we fail to reject the null hypothesis (Ho) Hence the "option=1" is the correct answer.