bank branch located in a commercial district of a city has

- bank branch located in a commercial district of a city has developed an improved process for serving customers during the noon-to-1:00 p.m. lunch period.The waiting time (dened as the time the customer enters the line until he or she reaches the teller window) of all customers during this hour is recorded over a period of 1 week. A random sample of 16 customers is selected.

Variable      N N*   Mean SE Mean StDev Minimum     Q1 Median     Q3
waiting time 16    0    4.408    0.414     1.655    0.380       3.285   4.635 5.962

a) Construct a 90% condence interval for the population mean of waiting time.

b) Give an interpretation for the 90% condence interval you calculated in part (b).

c) At the 0.01 level of signicance, is there evidence that the population mean waiting time is less than 6 minutes?

i.State the hypotheses.

ii.Find the test statistic.

iii.Find the p-value.

iv.Provide an interpretation of the p-value within the context of the problem.

Provide an interpretation of the p-value within the context of the problem.

v.Based on the p-value and the level of signicance make a decision about Ho

vi.Based on the decision write a conclusion in terms of the problem

d) Create a normal probability plot of the data using MINTAB, and use it to check whether the assumption of normality seems reasonable for these data.

e)Are the conditions required to construct the condence interval and the hypothesis test met? List them and comment on whether they hold or not. (See part(D).)

f)As a customer walks into the branch oce during the lunch hour, she asks the branch manager how long she can expect to wait. The branch manager replies,\\Typically, not longer than 6 minutes.\" On the basis of the results of (c), evaluate this statement.

g) Using MINTAB, conduct the hypotheses test and compare with your answer in (c).

Please make the answers clear.

Solution

Here we use the t- test for statistical inference as the sample size is less than 30.
a.
the t- value at 90% confidence interval(CI) at 15(16-1)degrees of freedom is 1.753
Now with the given SE of the Mean
CI= Mean +/- t- value(SE of Mean )
=4.408+/- 1.753(0.414)
=4.408+/-0.725742
= 5.133472, 3.682258
b.
This CI implies that if the actual waiting time is between 5.133472 minutes and 3.682258, then our hypothesis with the taken sample that Mean waiting time is 4.408 minutes holds true.
c.
At 0.01 significance level if we wish to test the hypothesis
i)H0 : Mean ?6
And
H1 : Mean <6 i.e. mean waiting time is less than 6
Then the tabulated t- defining the critical region is 2.6 at 15 degrees of freedom.
With the given data from the generated sample, the t-value for 6
ii)T6 = 6-4.408/ 0.414
= 3.3854
This is > 2.6 and thus falls in the acceptance region.
iii)p-value=0.999

iv)From the t- table the p- value is the area corresponding to the Calculated t which here is 3.85. The area to the left of 3.85 is 0.001 and that to the right is 0.999. Since this shows the probability of the acceptance of the null hypothesis, the H0 would be accepted as this probability is greater than 0.99 the assumed significance level.

v and vi) The Null hypothesis is thus accepted and the alternative hypothesis is rejected. There is thus evidence that the mean waiting time is greater than equal to 6 minutes.