Question

9 athletes from a high school weightlifting team were randomly selected to begin a new resistance weight program. After 8 weeks, the difference in how much weight they could lift (after - before) was recorded below in kg 4, 13.5, 15, 6, 12, 14, 19.5, 7.5, 13, Assume the population distribution is approximately normal. Find a 95% confidence interval for the mean strength gain of all students from this school if they joined the program.

Answer 1

Using the t-distribution, as we have the standard deviation for the sample, it is found that the 95% confidence interval for the mean strength gain of all students from this school if they joined the program is (7.85, 15.37).

What is a t-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm t\frac{s}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• t is the critical value.

• n is the sample size.

• s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 9 - 1 = 8 df, is t = 2.306.

The other parameters are given as follows, with the help of a calculator.

$\overline{x} = 11.61, s = 4.89, n = 9$

Hence:

$\overline{x} - t\frac{s}{\sqrt{n}} = 11.61 - 2.306\frac{4.89}{\sqrt{9}} = 7.85$

$\overline{x} + t\frac{s}{\sqrt{n}} = 11.61 + 2.306\frac{4.89}{\sqrt{9}} = 15.37$

The 95% confidence interval for the mean strength gain of all students from this school if they joined the program is (7.85, 15.37).

More can be learned about the t-distribution at https://brainly.com/question/16162795