Question

A student believes that the average grade on the statistics final examination was 87. A sample of 36 final examinations was taken. The average grade in the sample was 83.96 with a standard deviation of 12. The student wants to test whether the average is different from 87 at 90% level of confidence. Compute the p-value for this test.

Answer 1

Answer:

The p-value for this test is 0.1375.

Step-by-step explanation:

A student believes that the average grade on the statistics final examination was 87. The student wants to test whether the average is different from 87 at 90% level of confidence.

At the null hypothesis, we test if the average is 87, that is:

$H_0: \mu = 87$

At the alternate hypothesis, we test if the average is different from 87, that is:

$H_a: \mu \neq 87$

The test statistic is:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

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

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $s$ is the standard deviation of the sample and n is the size of the sample.

87 is tested at the null hypothesis:

This means that $\mu = 87$

The average grade in the sample was 83.96 with a standard deviation of 12. Sample of 36

This means that $X = 83.96, s = 12, n = 36$

Value of the test-statistic:

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

$t = \frac{83.96 - 87}{\frac{12}{\sqrt{36}}}$

$t = -1.52$

Pvalue:

Probability of the sample mean differing of 87, which means that we have a two-tailed test, with t = -1.52 and 36 - 1 = 35 degrees of freedom.

With the help of a calculator, the pvalue is of 0.1375.

The p-value for this test is 0.1375.