Question

A tire company advertises that its tires will last, on average, 60,000 miles. You are interested in determining if the average life of the company's tires is less than what was advertised. One hundred of the company's tires were randomly selected, and the average life of those tires was 58,500 miles with a standard deviation of 10,000 miles. (alpha=0.05) The parameter represents the population mean life of all the company's tires.
a) State the null and alternative hypotheses.

b) What is the sample?

c) Calculate the standard score.

d) What type of test was performed?

e) The p-value is...

f) What is the decision?

g) What is the conclusion?

Answer 1

Answer:

We conclude that the average life of the company's tires is more than or equal to what was advertised.

Step-by-step explanation:

We are given that a tire company advertises that its tires will last, on average, 60,000 miles.

One hundred of the company's tires were randomly selected, and the average life of those tires was 58,500 miles with a standard deviation of 10,000 miles.

Let $\mu$ = average life of the company's tires.

(a) Null Hypothesis, $H_0$ : $\mu$ $\geq$ 60,000 miles   {means that the average life of the company's tires is more than or equal to what was advertised}

Alternate Hypothesis, $H_A$ : $\mu$ < 60,000 miles   {means that the average life of the company's tires is less than what was advertised}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

                       T.S.  = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample average life of tires = 58,500 miles

             s = sample standard deviation = 10,000 miles

             n = sample of company's tires = 100

So, test statistics  =  $\frac{58,500-60,000}{\frac{10,000}{\sqrt{100} } }$   ~ $t_9_9$

                               =  -1.50

Hence, the value of test statistics is -1.50.

Also, P-value is given by the following formula;

         P-value = P( $t_9_9$ < -1.50) = 0.072 or 7.2%

Now at 5% significance level, the t table gives critical value of -1.98 at 99 degree of freedom for left-tailed test. Since our test statistics is more than the critical value of t as -1.50 > -1.98, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region and due to this we fail to reject the null hypothesis.

Therefore, we conclude that the average life of the company's tires is more than or equal to what was advertised.