An office manager has implemented an incentive plan that she

An office manager has implemented an incentive plan that she thinks will reduce the mean time required to handle a customer complaint. The mean time for handling a complaint was 30 minutes prior to implementing the incentive plan. After the plan was in place for several months, a random sample of the records of 38 customers who had complaints revealed a mean time of 28.7 minutes with a standard deviation of 3.0 minutes.

(a) Give a point estimate of the mean time required to handle a customer complaint.

(b) What is the standard deviation of the point estimate given in (a)?

(c) Construct a 95% confidence on the mean time to handle a complaint after implementing the plan.

Interpret the confidence interval for the office manager.

(d) Is there sufficient evidence that the incentive plan has reduced the mean time to handle a complaint? (Use ? = .05.)

z =

P value=

e)

Fail to reject H₀. There is sufficient evidence to conclude that the mean time to handle a customer complaint has decreased.Reject H₀. There is not sufficient evidence to conclude that the mean time to handle a customer complaint has decreased.    Fail to reject H₀. There is not sufficient evidence to conclude that the mean time to handle a customer complaint has decreased.Reject H₀. There is sufficient evidence to conclude that the mean time to handle a customer complaint has decreased.

Solution

unvrs :

In this problem, n=38, x-bar = 28.7 minutes, s = 3.8 minutes.
a. A points estimate of the mean time required to handle a customer complaint = 28.7 minutes.

unvrs :

b. the standard deviation of the point estimate or the standard error = 3.8/sqrt(38) = 0.61644

unvrs :

c. The critical value that corresponds to confidence level of 95% and degree of freedom of 37 is 2.026.
Then, the 95% confidence interval is
28.7 +/- 2.026*0.61644 = 28.7 +/- 1.2489 = (27.451, 29.949)

unvrs :

It means the office manager is 95% confident that the mean time for handling a complaint is between 27.45 minutes and 29.949 minutes.

unvrs :

Please let me know if you have any questions related to this problem.

unvrs :

May I know how I can assist you in understanding this question better?

unvrs :

Confidence interval for a mean is calculated as point estimate +/- margin of error.
From part a, we have the point estimate of 28.7 minutes.
The margin of error = critical value * standard error.
Part b, we got the standard error as 0.61644. Part c, we have the critical value as 2.026.
That means the margin of error = 2.026*0.61644 = 1.2489.
Putting everything together, the 95% confidence interval = (28.7-1.2489, 28.7+1.2489) = (27.451, 29.949)