Question

Company claims that their tires outlast the tires ofCompany B by more than 10,000 miles. Data has been collected and summarized below: Test the claim at the .05 level assuming and equal Company An-16 -63,500 s- 4000 Company B n-12 K-49,500 s-6000

Answer 1

Answer:

The null hypothesis is rejected (P-value=0.0 28).

There is  enough evidence to support the claim that that Company A tires outlast the tires of Company B by more than 10,000 miles.

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that that Company A tires outlast the tires of Company B by more than 10,000 miles.

Then, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=10000\\\\H_a:\mu_1-\mu_2> 10000$

being μ1: average for Company A and μ2: average for Company B.

The significance level is 0.05.

The sample 1, of size n1=16 has a mean of 63,500 and a standard deviation of 4,000.

The sample 1, of size n1=12 has a mean of 49,500 and a standard deviation of 6,000.

The difference between sample means is Md=14,000.

$M_d=M_1-M_2=63500-49500=14000$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{4000^2}{16}+\dfrac{6000^2}{12}}\\\\\\s_{M_d}=\sqrt{1000000+3000000}=\sqrt{4000000}=2000$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{14000-10000}{2000}=\dfrac{4000}{2000}=2$

The degrees of freedom for this test are:

$df=n_1+n_2-1=16+12-2=26$

This test is a right-tailed test, with 26 degrees of freedom and t=2, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t>2)=0.028$

As the P-value (0.028) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is  enough evidence to support the claim that that Company A tires outlast the tires of Company B by more than 10,000 miles.