Question

How large a sample should be selected to provide a 95% confidence interval with a margin of error of 6? Assume that the population standard deviation is 40 . Round your answer to next whole number.

Answer 1

Using the z-distribution, it is found that a sample of 171 should be selected.

What is a z-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

The margin of error is:

$M = z\frac{\sigma}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• z is the critical value.

• n is the sample size.

• $\sigma$ is the standard deviation for the population.

For this problem, the parameters are:

$z = 1.96, \sigma = 40, M = 6$

Hence we solve for n to find the needed sample size.

$M = z\frac{\sigma}{\sqrt{n}}$

$6 = 1.96\frac{40}{\sqrt{n}}$

$6\sqrt{n} = 40 \times 1.96$

$\sqrt{n} = \frac{40 \times 1.96}{6}$

$(\sqrt{n})^2 = \left(\frac{40 \times 1.96}{6}\right)^2$

n = 170.7.

Rounding up, a sample of 171 should be selected.

More can be learned about the z-distribution at brainly.com/question/25890103

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