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 1:00PM peak lunch period. The waiting time (as defined as the time the customer enters the line until he or she is served) of all customers during this hour is recorded over a period of 1 week. A random smaple of 15 customers is selected, and results are as follows:

4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50, 6.10, 0.38, 5.12, 6.46, 6.19, 3.79.

a) Set up 99% cofidence interval for estimating the average waiting time during peak hour.
b) At the 0.05 level of significance, is there evidence that the average waiting time is less tha 5 minutes?
c) Fine the p-value for part (b).
d) What assumption must hold in order to perform the test in part (b)?

Explanation / Answer

a) from above given data:

for (n-1=14) degree of freedom and  99%^ CI ; t=2.977

therefore 99% confidence interval =sample mean -/+ t*std error =3.0277 ; 5.5457

b) for test stat t=(X-mean)/std error =(4.287-5)/0.423=-1.687

for above test stat p value =0.057

as p value is higher then 0.05 level we do not have  evidence that the average waiting time is less tha 5 minutes.

c) po value =0.057

d) assumptions: distribution is normal.

sample is random

and no outliers/

X 4.21 5.55 3.02 5.13 4.77 2.34 3.54 % CI ; t= 3.2 4.5 6.1 0.38 5.12 6.46 6.19 3.79 mean(X) 4.287 std deviation(S) 1.638 std error =S/(n)1/2 0.423

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