Question

Dean Halverson recently read that full-time college students study 20 hours each week. She decides to do a study at her university to see if there is evidence that students study an average of less than 20 hours each week. A random sample of 33 students were asked to keep a diary of their activities over a period of several weeks. It was found that he average number of hours that the 33 students studied each week was 17. 7 hours. The sample standard deviation of 3. 9 hours. Find the p -value. The p -value should be rounded to 4-decimal places

Answer 1

Using the t-distribution, as we have the standard deviation for the sample, it is found that the p-value is of 0.0009.

What are the hypotheses tested?

At the null hypothesis, it is tested if the students study 20 hours, that is:

$H_0: \mu = 20$.

At the alternative hypothesis, it is tested if they study less than 20 hours, that is:

$H_1: \mu < 20$.

What is the test statistic?

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.

• $\mu$ is the value tested at the null hypothesis.

• s is the standard deviation of the sample.

• n is the sample size.

In this problem, the values of the parameters are given as:

$\overline{x} = 17.7, \mu = 20, s = 3.9, n = 33$.

Hence, the value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{17.7 - 20}{\frac{3.9}{\sqrt{33}}}$

$t = -3.39$

What is the p-value?

Using a t-distribution calculator, considering a left-tailed test, as we are testing if the mean is less than value, the p-value is of 0.0009.

More can be learned about the t-distribution at https://brainly.com/question/13873630