One of the major U.S. tire makers wishes to review its warranty for their rainmaker tire. The warranty is for 40,000 miles. The tire company believes that the tire actually lasts more than 40,000 miles with a standard deviation of 15,000 miles. A sample 49 tires revealed that the mean number of miles is 45,000 miles. Test the hypothesis with a 0.05 significance level.
A) At a 5% level of significance, our decision is to reject the null hypothesis. The tires last more than 40 000 miles.
B) At a 5% level of significance, our decision is to accept the null hypothesis. The tires last 40,000 miles.
C) At a 5% level of significance, we should have used the t-test, rather than the z-test.

Question

Quizplus · Accepted Answer

We will use a one-tailed z-test since the tire company believes that the tire actually lasts more than 40,000 miles. The null hypothesis is that the mean number of miles the tire lasts is equal to or less than 40,000 miles. 
H0: μ ≤ 40,000 
Ha: μ > 40,000 
The sample size (n) is greater than 30, so we can use a z-test. The z-statistic can be calculated as:
z = (x̄ - μ) / (σ / √n)
z = (45,000 - 40,000) / (15,000 / √49)
z = 3.00
Using a z-table, the p-value is less than 0.001, which is less than the significance level of 0.05. Therefore, we reject the null hypothesis and conclude that the tires last more than 40,000 miles.

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