Question

A statistics teacher started class one day by drawing the names of 10 students out of a hat and asked them to do as many pushups as they could. The 10 randomly selected students averaged 15 pushups per person with a standard deviation of 9 pushups. Suppose the distribution of the population of number of pushups that can be done is approximately normal. The 95% confidence interval for the true mean number of pushups that can be done is _____.

Answer 1

Using the t-distribution, we have that the 95% confidence interval for the true mean number of pushups that can be done is (9, 21).

For this problem, we have the standard deviation for the sample, thus, the t-distribution is used.

• The sample mean is of 15, thus $\overline{x} = 15$.
• The sample standard deviation is of 9, thus $s = 9$.
• The sample size is of 10, thus $n = 10$.

First, we find the number of degrees of freedom, which is the one less than the sample size, thus df = 9.

Then, looking at the t-table or using a calculator, we find the critical value for a 95% confidence interval, with 9 df, thus t = 2.2622.

The margin of error is of:

$M = t\frac{s}{n}$

Then:

$M = 2.2622\frac{9}{\sqrt{10}} = 6$

The confidence interval is:

$\overline{x} \pm M$

Then

$\overline{x} - M = 15 - 6 = 9$

$\overline{x} + M = 15 + 6 = 21$

The confidence interval is (9, 21).

A similar problem is given at brainly.com/question/25157574