Question

Ads for Speedster Auto Service Centers claim that, on average, they will change a customer’s oil and get them back on the road in 15 minutes. A consumer protection agency suspects that this claim is false and the true average is more than 15 minutes. They send cars to 40 randomly selected Speedster service centers and find that the mean oil change for this sample is 16.3 minutes with a standard deviation of 4.3 minutes. They plan to conduct a hypothesis test, using the .05 significance level, to determine whether their data provides evidence to dispute Speedster’s claim.

B3: Find the P-value for the test

B4: What does the P-value tell us?

B5: Based on the results obtained in this situation, what should we conclude?

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u < 15
Alternative hypothesis: u > 15

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.67989
DF = n - 1

D.F = 39
t = (x - u) / SE

t = 1.912

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 1.912.

B3) Thus the P-value in this analysis is 0.032.

B4) The p-value tells us the the probability of getting sample mean 16.3 or extreme if we assume that the null hypothesis is true.

Interpret results. Since the P-value (0.032) is less than the significance level (0.05), we have to reject the null hypothesis.

B5) From the above test we have sufficient evidence in the favor of the claim that true average is more than 15 minutes.