Question

Conduct the following test at the a = 0.01 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 p1 #P2. Sample data are X7 = 30, n = 254, X2 = 36, and n2 = 301. (a) Determine the null and alternative hypotheses. Choose the correct answer below. O A. Ho: P1 = P2 versus H: P1>P2 O c. Ho: P1 = P2 versus H : P1 P2 OB. Ho: P1 = P2 versus Hy: P7 P2 D. Ho: P1 = 0 versus H: P1 = 0 (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. Do not reject the null hypothesis because there is not sufficient evidence to conclude that P1 *P2. OB. Do not reject the null hypothesis because there is sufficient evidence to conclude that pn >P2 O c. Reject the null hypothesis because there is sufficient evidence to conclude that P1 *P2. OD. Reject the null hypothesis because there is not sufficient evidence to conclude that p <P2.

Answer 1

a) Null and alternative hypothesis are, B. Hoo P = P2 versus H1: PiZ P2 b) Ô = Sample Prortion for ist population = x1 CO 미 → ê, = 0.1181 ; Q, = 1- Â = 0.8819. 30 254 a Ộs = Sample proportion of 2nd samples X2 12 => 2 = 36 301 = 041791Q2 = 1 - 2 0.1196 0.8804 1 Didi tn 2 P2 nitna 254 + 30 +(36 254 2 su + 301 TO) - 0.1189 30+ 36 66 SSS SSS 1 - 0.1189 = 0.8811 I - Test statistic, Zo= Pi-P2 på (1 DI π Zo=08814 20 = 0.1181 - 0.1196 00:1199x0.8811) * 254 + 1 301

T -0.00 1491 = -0.054 ZO - 0.0 2751 20= -0.054 c) p-value = p 135-6.054) + P (27, 01054) - 2. PLZE-0.054) (1:2 is symmetric. plz:-0.054) - P(27,0.054) 10 p-value = 2*0.4784 0957 p-value value = .Value = 0.960 2 * 0.48006 ( Using standard normal table). p.value tet dao.01 p-value > < we do not reject Ho ㅠ A Do not reject Ho the null hypothesis because thieve is not sufficient evidence to conclude that Pi ² P2.