Question

3) A power company wished to test the null hypothesis that the mean time for processing rebates was equal to 3.5 minutes, against the alternative that it was greater. A sample of mean time of 3.8 minutes was obtained from a sample of n - 40 rebate applications. State hypotheses, compute test statistic, P-value, and state if the null hypothesis would be rejected at -.01 (Assume the population standard deviation is -1.2 minutes) H : - H: : P-value ; at a = .01: ACCEPT H. REJECT H. (Circle one) 4) For the following hypotheses the test statistics, z, have already been computed from the data. Find the P-value corresponding to each test statistic. a) H.: u = H. H: b) H u , H: <o: > He: Test statistic z Test statistic z= - 2.23 1.91 P-value = P-value

Answer 1

H. : [M] =(3-6) Hi : [M]> | 3-5] na 40 : 3.8. - 12. a 20.01 X Me X 3-8 -3.5 2 1140 = 1.58 olto - p-value = [!-PC2X158)] = 10-0570) I Right tailed) A test J. 0.0570 p-value ya=o.o ACCEPT Ho J. 4 (a) Homes Mlo vs HIMLMO P-value = P(Z = -2-23) Z=-7.23 i = 0.01287 ~ 0.0129 (left tailed) (6) Horn = Mo Vs Hi: My Mo Z = 1.91 Penalue ={1-P (2=1991)] (Right-tailed test) -10.0281] using p-value calculator