Question

Crack By Work remaining CON and Act provided t o chan g ed to view ing in supermarchedoutes. The channel showed that features, and advertisements. The length of the program was based on the assumption that the population meantimea shopper stands in a supermarket checkoutlines 3 minutes. Asmole of actual w ing times will be used to test this motion and determine whether actual meaning differs from this standard a. Formulate the hypotheses for this application HE b. A sample of 120 shoes showed a sample mean 2 demais pv to 4 decimale) caa0.01 what is your conclusion? d. Computea confidence interval for the population mean. Does support your condition Select your Styr i ntheterval Check My Worker

Answer 1

Solution: The provided sample mean is X = 8.7 and the known population standard deviation is o = 3.7, and the sample size is n = 120. (1) Null and Aiternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ 8.3 Ha: u # 8.3 This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.01, and the critical value for a two-tailed test is ze = 2.58. The rejection region for this two-tailed test is R = {2: 2|> 2.58} (3) Test Statistics The z-statistic is computed as follows: 8.7 – 8.3 X- H0 = 1.184 %3D %3D 3.7/V120 (4) Decision about the null hypothesis Since it is observed that 2 = 1.184 < %e = 2.58, it is then concluded that the nu hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.2363, and since p = 0.2363 0.01, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is different than 8.3, at the 0.01 significance level. Confidence Interval The 99% confidence interval is 7.83 <H< < 9.57.