Question

“Dawit Lube” is a regional chain that provides a standard service for performing oil changes and basic checkups on automobiles. The chain has a standard that says the average time per car for this service should be 12.5 minutes. There is considerable variability in times due to differences in layout of engines, degree of time pressure from other jobs, and many other sources. As a result, the management of the chain decided to study if the standard is being met and randomly selected 28 oil change and basic checkup services within a month and carefully recorded the time they took in minutes. The resulting data are reported in the following table.

11.2	8.1	16.5	18	10.4	12.1	13.2	8.2	13.3	35.1	24.1	12.1	7.5	12.3
8.8	15.0	9.3	9.0	10.1	14.1	17.3	13.7	12.8	12.8	19.8	22.2	8.3	28.3

a) Using the given data, please construct the 90% and the 95% confidence intervals for the mean time it takes to complete oil change and basic checkup services at Dawit Lube. Clearly show all the necessary steps and interpret your result. Based on your results would you conclude that the chain is achieving its stated standard? Please justify your answer.

b) What is the probability that the mean time for oil change and basic checkup services at Dawit Lube could be less than the lower limit of the 90% confidence interval? Also explain which confidence interval is wider and why. What do you think will happen to the confidence intervals if the sample size is increased to 50? Please justify your answers.

c) Using the given data and at 10% level of significance, please test whether the chain is achieving its stated standard for the average time for oil change and basic checkup services. Show the necessary steps and explain your conclusion. Is your test result consistent with the conclusion you arrived at based on the 90% confidence interval you constructed under question (a) above? Please explain.

Answer 1

From the given data we have
xbar = 14.4143
s = 6.5248

a)
for 90% CI, t-value = 1.7033
CI = (14.4143 -/+ 1.7033*6.5248)
= (12.314 , 16.5146 )

for 95% CI, t-value = 2.0518
CI = (14.4143 -/+ 2.0518*6.5248)
= (11.8842 , 16.9444 )

b)
P(X < 12.314) = 0.05

The 95% CI is wider because the t-value are greater than 90% CI

Increasing the sample size decrease the width of CI

c)
As 12.5 is included in the CI, we fail to reject H0
Hence the average time is 12.5 minutes