Question

The data below consists of the test scores of 32 students. Construct a 99% confidence interval for thepopulation mean.

Answer 1

Using the t-distribution, it is found that the 99% confidence interval for the population mean is: (100.61, 111.07).

We suppose that:

• The sample mean is of $\overline{x} = 105.84$.
• The sample standard deviation is of $s = 14.27$.
• We have the standard deviation for the sample, thus, the t-distribution is used to solve this question.

The first step is finding the number of degrees of freedom, which is the sample size subtracted by 1, thus $df = 32 - 1 = 31$.

Then, we find the critical value for a 99% confidence interval with 31 df, which, looking at the t-table, is t = 2.0395.

The margin of error is:

$M = t\frac{s}{\sqrt{n}} = 2.0395\frac{14.27}{\sqrt{31}} = 5.23$

The confidence interval is:

$\overline{x} \pm M$

Then

$\overline{x} - M = 105.84 - 5.23 = 100.61$

$\overline{x} + M = 105.84 + 5.23 = 111.07$

The 99% confidence interval for the population mean is (100.61, 111.07).

A similar problem is given at brainly.com/question/14885491