Question

17. A random sample of n = 139 individuals results in X4 = 41 successes. An independent sample of ny = 152 individuals results in xy =61 successes. Does this represent sufficient evidence to conclude that p, <, at the a= 0.05 level of significance? Click here to view the standard normal distribution table (page 1.' Click here to view the standard normal distribution table (page 2). Click here to view the table of critical l-values. Click here to view the chi-square critical values table." The given situation about a (1) Write the hypotheses for the test. Ho: (2) Hy: (3) Calculate the test statistic (4) (Round to two decimal places as needed.) Identify the critical region. Select the correct choice below and fill in all answer boxes within your choice. (Type an integer or decimal rounded to two decimal places as needed.) O A. test statistic > B. test statistic or test statistic OC. test statistics What is the conclusion? (5) the null hypothesis and conclude there (6) sufficient evidence that the parameter of interest of population 1 is (7) the parameter of interest of population 2 at the e=0.05 level of significance 1: Standard Normal Distribution Table (page 1)

Answer 1

It is given that, h = 139 X, 541 ng = 152 1 X = 61 thew, P, = X 41 139 -0.294964 a 0.2950 61 152 no P = Xg te = 0.401315789 = 0.4013 The given Situation in test of proportion, p. Hypothesis can be helsitten as: Hos P, = Pa H,: Р 4 Р., Test Steetistic : zo Pi-Pe PC1-(n. + totese , PEX,+ X, nitha 41+61 139 +152 = 0.35051546 291 NOW 0.3505 Zo 11 0.2950 - 0.4013 10.3505 x 0.6495*(1 + i) -0.1063 -1.89 838 -1.90 0.055995 Here i Corto cofa male = Zq=20005 -1.65 test Steelestie < Corto col rolee. Hence, we rejeet Ho. Thebefore, we conclude there is sufficient evidence that the Pupametés of interest. et por primetion 1 in less than the posamotes of interest of populalion &.