Question

) A tire manufacturing company is reviewing its warranty for their rainmaker tire. The warranty is for 40,000 miles. The distribution of tire wear is normally distributed with a population standard deviation of 15,000 miles. The tire company believes that the tire actually lasts more than 40,000 miles. A sample of 49 tires revealed that the mean number of miles is 45,000 miles. Test the hypothesis with a 0.05 significance level. What is our decision? A) Accept the null hypothesis B) Reject the null hypothesis C) Reject the both hypothesis D) Reject the alternative hypothesis

Answer 1

What is our decision?

Answer: B) Reject the null hypothesis

Solution:

Here, we have to use one sample z test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H0: The tire lasts about 40,000 miles.

Alternative hypothesis: Ha: The tire actually lasts more than 40,000 miles.

H0: µ = 40000 versus Ha: µ > 40000

This is an upper or right tailed test.

The test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

From given data, we have

µ = 40000

Xbar = 45000

σ = 15000

n = 49

α = 0.05

Critical value = 1.6449

(by using z-table or excel)

Z = (45000 – 40000)/[15000/Sqrt(49)]

Z = 2.3333

P-value = 0.0098

(by using Z-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that the tire actually lasts more than 40,000 miles.