Question

Problem I: Suppose you are the manager of Speedy Oil Change which claims that it will change the oil in customers’ cars in less than 30 minutes on average. Further suppose that several complaints have been filed from customers stating that their oil change took longer than 30 minutes. Upper-level management at Speedy Oil Change headquarters has requested that you investigate the complaints. To begin your investigation, you randomly audit 36 oil changes performed by Speedy Oil Change and record the time each customer waited for the oil change. The number of minutes to complete each of the 36 oil changes is reported below.

42	29	19	11	10	27
41	27	22	26	28	32
17	15	25	35	32	22
13	31	17	30	33	25
18	24	28	21	40	19
34	30	14	23	22	10

(data similar to http://org.elon.edu/econ/sac/Descriptive.htm)

In the questions below, you will test if the mean is significantly less than 30 minutes at a significance level of 0.01.

6. Using the confidence interval you created, calculate the range of values that answers the question “how much less than 30 minutes is our mean oil change?”

7. What other factors might you need to consider in order to address the customer complaints?

Answer 1

Hypothesis test

The single sample t test is used to compare the sample mean to hypothetical population mean

The null hypothesis and the alternate hypothesis are,

The t statistic is,

From the data values,

The P-value is obtained using the t distribution table for t = -3.672 and degree of freedom = N - 1 = 36- 1 = 35.

Since the P-value is less than 0.01 at 1% significance level. The null hypothesis is rejected. Hence the mean is significantly less than 30 minutes at a significance level of 0.01.

Confidence interval

The 99% confidence interval for means is obtained using the formula,

From the data values,

The t critical value for 99% confidence interval and degree of freedom = n -1 = 36 - 1 = 35 is obtained from t distribution table.

Mean oil change is 30 - 20.904 = 9.096 minutes less.