A bank located in a commercial district of a city has developed an improved proc

Question

A bank located in a commercial district of a city has developed an improved process for serving customers during the noon to 1 pm peak lunch period. The waiting time (as defined as the period the cutomer enters the line until he is served) of all customers during this hous is recorded over a period of 1 week. A random sample of 15 customers is selected, and the results are as follows:

4.25, 5.55, 3,05, 5,15 , 4,75, 2,35, 3,55, 3,20, 4,50, 6,10, 0,40, 5,15, 6,45, 6,20, 3,80

A. Find sample mean and standard deviation

B.Find standard error of the estimator i.e standard deviation of the sample mean

C. Set up 95% confidence interval for the estimating the average waiting during peak time

D. at the 0.01 level of significance, is there evidence that the waiting time is less tha 5 minutes

e. find the p-value for part (D)

f. what assuption must hold in order to perfom the test in part (d)?

Explanation / Answer

a) from above mean =4.297 and std deviation =1.633

b) std error of the estimator =0.422

C)for 14 degree of freedom and 95% CI, t=2.1448

hence confidence interval =mean +/- t*std error =3.392 ; 5.201

D) for test stat t=(X-mean)/std error =(4.297-5)/0.422=-1.66588

abd for 0.01 level and 1 tail test critical value t=-2.6245

therefore we do not have sufficient evidence that the waiting time is less tha 5 minutes at 0.01 level.

e) p value for above test stat =0.059

f) we made the assumption that population is normally distributed and sample is not biased

X 4.250 5.550 3.050 5.150 4.750 2.350 3.550 3.200 4.500 6.100 0.400 5.150 6.450 6.200 3.800 mean(X) 4.297 std deviation(S) 1.633 std error =S/(n)1/2 0.422

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