Here we conduct Z-test for two proportions as :
Hence,
there is not enough evidence to conclude that Female are
more likely to experience higher than-normal levels of OCD
indicators at 10% level of significance.
For sample 1, we have that the sample size is Ni-92, the number of favorable cases is X1-24, 24 = 0.2609 so then the sample proportion is pl For sample 2, we have that the sample size is N2 -56, the number of favorable cases is X2 - 11, so then the sample proportion is P2 = 0.1964 The value of the pooled proportion is computed as p = TT 24-11 = 0.2365 Also, the given significance level is α = 0.1 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: 92+56 This corresponds to a right-tailed test, for which a z-test for two population proportions needs to be conducted. 2) Rejection Region Based on the information provided, the significance level is 0.1, and the critical value for a right- tailed test is zc-1.28. The rejection region for this right-tailed test is R z: z1.28)
(3) Test Statistics The z-statistic is computed as follows: P1P2 0.2609 - 0.1964 0.895 VP(1 - p)(1/ni + 1/n2) V0.2365 - (1-0.2365) (1/92 +1/56) (4)_Decision about the null hypothesis Since it is observed that z rejected. 0.895 S zc-1.28, it is then concluded that the null hypothesis is not Using the P-value approach: The p-value is p 0.1855, and since p 0.1855 2 0.1, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion pi is greater than p2, at the 0.1 significance level