Question

Someone claims that the average amount of time that a freshman at TAMU studies is 7 hours. We think it’s higher than that and decide to test, using a random sample of 49 freshmen. The sample mean is 8.5 hours with a sample variance of 4 hours. What are the test statistic and p-value in this case?

Answer 1

Test statistic t = 5.25

P-value = 0.000002 (one-tailed test)

Explanation:

This is a hypothesis test for the population mean.

The claim is that the average amount of time that a freshman at TAMU studies is significantly higher than 7 hours.

Then, the null and alternative hypothesis are:

`H_0: \mu=7\\\\H_a:\mu> 7`

The significance level is 0.05.

The sample has a size n=49.

The sample mean is M=8.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=√s^2=√4=2.

The estimated standard error of the mean is computed using the formula:

`s_M=(s)/(√(n))=(2)/(√(49))=0.29`

Then, we can calculate the t-statistic as:

`t=(M-\mu)/(s/√(n))=(8.5-7)/(0.29)=(1.5)/(0.29)=5.25`

The degrees of freedom for this sample size are:

`df=n-1=49-1=48`

This test is a right-tailed test, with 48 degrees of freedom and t=5.25, so the P-value for this test is calculated as (using a t-table):

`\text{P-value}=P(t>5.25)=0.000002`

As the P-value (0.000002) 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 the average amount of time that a freshman at TAMU studies is significantly higher than 7 hours.