Question

The average wait time to get seated at a popular restaurant in the city on a Friday night is 12 minutes. Is the mean wait time different for men who wear a tie? Wait times for 14 randomly selected men who were wearing a tie are shown below. Assume that the distribution of the population is normal. 12, 11, 10, 13, 9, 13, 10, 11, 10, 10, 12, 13, 11, 9 What can be concluded at the the a = 0.05 level of significance level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: HO: Select an answer v H: ? Select an answer QE c. The test statistic? (please show your answer to 3 decimal places.) d. The p-value - (Please show your answer to 4 decimal places.) e. The p-value is f. Based on this, we should Select an answer the null hypothesis. g. Thus, the final conclusion is that ... The data suggest the populaton mean is significantly different from 12 at a -0.05, so there is statistically significant evidence to conclude that the population mean wait time for men who wear a tie is different from 12. The data suggest that the population mean wait time for men who wear a tie is not significantly different from 12 at -0.05, so there is statistically insignificant evidence to conclude that the population mean wait time for men who wear a tie is different from 12. O The data suggest the population mean is not significantly different from 12 at a = 0.05, so there is statistically insignificant evidence to conclude that the population mean wait time for men who wear a tie is equal to 12.

Answer 1

Standard Deviation, s: 1.4142135623731 Count, N: 14 Sum, Ex: 154 Mean, X: 11 Variance, s2:2 Steps 1 N S = (xi – T)? VN N-1 i=1 s2 = E(xi-x)2 N-1 - (12 - 11)2 + ... + (9-11) 14-1 26 13 = 2 S = vē = 1.4142135623731

The provided sample mean is X = 11 and the sample standard deviation is s = 1.414, and the sample size is n = 12. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: A = 12 Ha: u 7 12 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a two-tailed test is te 2.201. The rejection region for this two-tailed test is R= {t: \t] > 2.201} : (3) Test Statistics The t-statistic is computed as follows: t = Ã - μo s/Vn 11 – 12 1.414/V12 -2.45 (4) Decision about the null hypothesis Since it is observed that (t) = 2.45 > te=2.201, it is then concluded that the null hypothesis is rejected. = = 0.0323< Using the P-value approach: The p-value is p 0.0323, and since p 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is different than 12, at the 0.05 significance level.

please like ??