Question

Conduct the following test at the a = 0.05 level of significance by determining (a) the null and alternative hypotheses, (b) the test statistic, and (c) the P-value. Assume that the samples were obtained independently using simple random sampling. Test whether pe p2. Sample data are xy = 28, n, = 255, x2 = 38, and nz = 301 (a) Determine the null and alternative hypotheses. Choose the correct answer below. O A. Ho: P." P2 versus H: PP2 OC. Ho: P = P2 versus He: #P2 OB. Ho: P = 0 versus Hy: P = 0 OD. Hy: P = P2 versus H: P4 > P2 (b) The test statistic Zo is (Round to two decimal places as needed.) (c) The P-value is .(Round to three decimal places as needed.) Test the null hypothesis. Choose the correct conclusion below. O A. Roject the null hypothesis because there is not sufficient evidence to conclude that p, <p2 OB. Do not reject the null hypothesis because there is sufficient evidence to conclude that pq > P2 OC. Reject the nut hypothesis because there is sufficient evidence to condude that P, P2 OD. Do not reject the nut hypothesis because there is not sufficient evidence to conclude that P, MD2 FO Click to select your answer(s) 2

Answer 1

For sample 1, we have that the sample size is N1 = 255, the number of favorable cases is X1 = 28, so then the sample proportion is ê1 = 1 25% = 0.1098 = For sample 2, we have that the sample size is N2 = 301, the number of favorable cases is X2 = 38, so then the sample proportion is P2 38 = 0.1262 N2 301 X1+X2 28+38 The value of the pooled proportion is computed as p = = 0.1187 Ni+N2 255+301 Also, the given significance level is a = 0.05. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: P1 = P2 Ha: p = P2 This corresponds to a two-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 a = 0.05, and the critical value for a two- tailed test is zc 1.96. = The rejection region for this two-tailed test is R = {z : 12 > 1.96} (3) Test Statistics


The z-statistic is computed as follows: 2= = P1 – 2 VP(1 - 7)(1/nı +1/n2) 0.1098 – 0.1262 10.1187 · (1 – 0.1187)(1/255 + 1/301) -0.597 (4) Decision about the null hypothesis Since it is observed that |2= 0.597 < ze = 1.96, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.5503, and since p= 0.5503 > 0.05, 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 pı is different than p2, at the 0.05 significance level. Confidence Interval The 95% confidence interval for pi – P2 is: -0.07 <P1 – P2 < 0.037.

Hence
option D is correct: We donot reject H₀ as we dont have sifficient evidence to conclude that p₁ is not equal to p₂