Question

Construct the indicated confidence interval for the population mean using the​ t-distribution. Assume the population is normally distributed.
picture of question below

Answer 1

Using the t-distribution, it is found that the confidence interval is (14.2, 14.8).

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 18 - 1 = 17 df, is t = 2.1098.

The other parameters are given by:

$\overline{x} = 14.5, s = 0.68, n = 18$.

Hence, the bounds of the interval are given by:

$\overline{x} - t\frac{s}{\sqrt{n}} = 14.5 - 2.1098\frac{0.68}{\sqrt{18}} = 14.2$

$\overline{x} + t\frac{s}{\sqrt{n}} = 14.5 + 2.1098\frac{0.68}{\sqrt{18}} = 14.8$

The confidence interval is (14.2, 14.8).

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