Question

Construct a 96% confidence interval if a sampling distribution has a mean of 20, a standard deviation of 5, and a size of 100.

Answer 1

Using the t-distribution, the 96% confidence interval is given as follows:

(18.96, 21.04).

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 96% confidence interval, with 100 - 1 = 99 df, is t = 2.0812.

The parameters are given as follows:

$\overline{x} = 20, s = 5, n = 100$

Hence the bounds of the interval are:

$\overline{x} - t\frac{s}{\sqrt{n}} = 20 - 2.0812\frac{5}{\sqrt{100}} = 18.96$

$\overline{x} + t\frac{s}{\sqrt{n}} = 20 + 2.0812\frac{5}{\sqrt{100}} = 21.04$

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

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