Question

A plan for an executive travelers' club has been developed by an airline on the premise that 5% of its current customers would qualify for membership. A random sample of 500 customers yielded 40 who would qualify. (a) Using this data, test at level 0.01 the null hypothesis that the company's premise is correct against the alternative that it is not correct. State the appropriate null and alternative hypotheses. Ho: P = 0.05 Ha: p = 0.05 Ho: P = 0.05 Ha: p < 0.05 O Ho: p = 0.05 Ha: p = 0.05 Ho: p = 0.05 Ha: p > 0.05 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = 3.078 P-value = 0.0021 State the conclusion in the problem context. Reject the null hypothesis. There is sufficient evidence to conclude that the company's premise is incorrect. Do not reject the null hypothesis. There is sufficient evidence to conclude that the company's premise is incorrect. Do not reject the null hypothesis. There is not sufficient evidence to conclude that the company's premise is incorrect. Reject the null hypothesis. There is not sufficient evidence to conclude that the company's premise is incorrect. (b) What is the probability that when the test of part (a) is used, the company's premise will be judged correct when in fact 10% of all current customers qualify. (Round your answer to four decimal places.) 0.1360 X

Answer 1

Answers Given that n= 500. x=40 customers a) Hypothesis, Ho: P=0.05 Hoop & 0.05. option Ċ is correct x=0.01 500 . & calculfate - 0.08 Test Statstic PP JPGP)/n 0.08 -0.05 Jo.08(1-0.05) 500 003 0.05(0.95 500 0.03 0.03 0.0475 500 COOGES SOO 0.03 0.009747 pvalue 3.077870u - 3.078 = 20/2>3.07) - 1-Plzs 3:01) =)

- 1-0.9989 = 0.0011 6 It is Two tailed test so = 2x0.0011 0022 Reject the hull hypothesis There is sufficient evidence to conclude that the company's premise incorrect- - option À is correct. p=0.10 10.08-0.10 + 70008 | 00 (0.95 Blopho) = 0 2008-010-201 / 2009 y 500. (voor FOD (0.9) -0.05 & 2.58 (0.0097 -0.05 - 2158 x 0.0097 √0.00018 T0100018 -05-005+0.025026 ) - 01-0.05 -00250267 0.0136166079 0.0134164079 591 010134164079) - 01 -0.029974 7-01-0.075026 0.0136169017 = 0[-1.86)-01-5.60] - 861.86]-o = 0.03199 500 0018