Question

Suppose the mean wait-time for a telephone reservation agent at a large airline is 45 seconds. A manager with the airline is concerned that business may be lost due to customers having to wait too long for an agent. To address this concern, the manager develops new airline reservation policies that are intended to reduce the amount of time an agent needs to spend with each customer. A random sample of 250 customers results in a sample mean wait-time of 44.4 seconds with a standard deviation of 4.2 seconds. Using a = 0.05 level of significance. do you believe the new policies were effective in reducing wait time? Do you think the results have any practical significance? Determine the null and alternative hypotheses. Ho 45 seconds 45 seconds Calculate the test statistic (Round to two decimal places as needed.) Calculate the P-value P-value = (Round to three decimal places as needed.) State the conclusion for the test. A. Reject H, because the P-value is greater than the a=0.05 level of significance. B. Do not reject H, because the P-value is greater than the c = 0.05 level of significance. OC. Reject H, because the P-value is less than the a= 0.05 level of significance. OD. Do not reject H, because the P-value is less than the q = 0.05 level of significance. State the conclusion in context of the problem. There sufficient evidence at the a= 0.05 level of significance to conclude that the new policies were effective. Do you think the results have any practical significance? O A. Yes, because the test concluded that there is a significant difference between the two mean wait-times. Therefore, there is a practical significance OB. Yes, because while there is no significant evidence that shows the new policies were effective in lowering the mean wait-time of customers, the difference between the previous mean wait-time and the new mean wait-time is large enough to be considered important OC. No, because the test concluded that there is no significant difference between the two mean wait-times. Therefore, there is no practical significance OD. No, because while there is significant evidence that shows the new policies were effective in lowering the mean wait-time of customers, the difference between the previous mean wait-time and the new mean wait-time is not large enough to be considered important

Answer 1

x̅ = 44.4, s = 4.2, n = 250

Null and Alternative hypothesis:

Ho : µ = 45

H1 : µ < 45

Test statistic:

t = (x̅- µ)/(s/√n) = (44.4 - 45)/(4.2/√250) = -2.26

df = n-1 = 249

p-value = T.DIST(-2.26, 249, 1) = 0.012

Conclusion:

C. Reject Ho because the P-value is less than the α = 0.05 level of significance.

There is sufficient evidence.

D. No, because while there is significant evidence that shows the new policies were effective in lowering the mean​ wait-time of​ customers, the difference between the previous mean​ wait-time and the new mean​ wait-time is not large enough to be considered important.