Question

Test the claim about the population mean, , at the given level of significance using the given sample statistics. Claim: u = 30; a = 0.09; 6 = 3.02. Sample statistics: x = 28.8, n=62 Identify the null and alternative hypotheses. Choose the correct answer below. O A. Ho: u = 30 Hu> 30 0 C. Ho: < 30 Hau= 30 E. Ho: u = 30 H: #30 OB. Ho:u#30 Ha: u = 30 OD. Ho: > 30 Ha: H = 30 OF. Ho: u = 30 H: < 30 Calculate the standardized test statistic. The standardized test statistic is . (Round to two decimal places as needed.)

Answer 1

Solution : - Given that, n = 62 X = 28.8 6 3.02 The null and alternative hypothesis aze, Hoil = 30 Hand & 30 The test statistic Z, is z - Xu - 6/462 28.8 -30 3.02% [62 -3.13

a the null and alternative hypotheses are Ho u < 860 Hau > 860 Hese, u= 69 x = 880,0 = 120 the test statistic 2 z X-1 oldu TS. 880-860 120/ /69 - 1.38 > Sign in Ha =) Right tailed test. The caitical value - The Siguificance level de 0.05 : Za 1.64 1. Caitical value is 1.64 The Rejection region is 2:27 1.64 Here, 2. 1.38 < z fail to Reject Ho. = 1.64

& hypothesis aa. The null and alternative - Ho uc 1270 - Ha u z 1270 Here in= 300, x= 1285, o. 206.24 The test statistic z is, x- el 1285 - 1270 olu . 206.24/√300 a 1.26 * The null and alteonative hypothesis ader Ho: Ha u=7000 u 7000 I Hese, n = 33, À = 6700, o .416 The test Statistic z 2 . X-M is, 6700 - 7000 416/ V33 - 4.14

Given, test Statistic z = 1.83 - 2 = 0.03. Right tailed test - P-value = P(Z > 1.83] a 1-P(Z 2 1.83) 1 0.9664 - Using z tob.. - 0.0336 Since, p-value 0.0336 > x = 0,03 4 fail to Reject to