Question

A nationally branded manufacturer of tires wishes to review its warranty for its all-terrain radials. The warranty is for 40,000 miles. The distribution of tire thread life is normally distributed with a standard deviation of 6,125 miles. The manufacturer claims that the tire actually lasts more than 40,000 miles. A sample of 55 tires revealed that the mean number of miles is 41,655 miles.

a. Using a 0.02 significance level, determine the five-step hypothesis for the claim (show all steps).

Answer 1

Here standard deviation is given that is known , so we need to use one sample Z test for testing the population mean.

Claim:  the tire actually lasts more than 40,000 miles.

From the above claim the null hypothesis ( H₀ ) and the alternative hypothesis ( H_a ) are as follows:

Let's write the given information.

Sample size = n = 55

sample mean = = 41655

level od significance = = 0.02

Standard deviation = = 6125

Let's use minitab

The command is Stat>>>Basic Statistics >>1 sample Z...

Click on summarized data

Sample size = 55

Mean = 41655

Standard deviation = 6125

then click oon "Perform hypothesis test

Hypothesized mean = 40000 (from null hypothesis)

then click on Option

Confidence level = (1.00 - )*100 = 98.0

Alternative "greater than"

then click on OK

again click on OK

We get the following output

Decision rule: 1) If p-value <= level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.023 > 0.02 so we used 2nd rule.

That is we fail to reject null hypothesis

Conclusion: At 5% level of significance there are not sufficient evidence to support the claim as the tire actually lasts more than 40,000 miles.