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 3.8 minutes.a. (5 pts) Give a point estimate of the mean time required to handle a customer complaint.b. (5 pts) What is the standard deviation of the point estimate given in a?c. (5 pts) Construct a 95% confidence on the mean time to handle a complaint after implementing the plan. Interpret the confidence interval for the office manager.[Hint: use minitab to find the t-value for 37 degrees of freedom, point estimate refers to sample mean, standard deviation of the point estimate refers to the standard deviation of sample mean which is also called standard error.]NEED HELP WITH PART C USING THE HINT
I FOUND THE T VALUE TO BE 2.02619, IM NOT SURE HOW TO FIND THE 95% CONFIDENCE INTERVAL
Hi, Welcome to Just Answer.
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.
b. the standard deviation of the point estimate or the standard error = 3.8/sqrt(38) = 0.61644
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)
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.
Please let me know if you have any questions related to this problem.
May I know how I can assist you in understanding this question better?
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)
Please let me know if you need further explanation.