Question

We want to obtain a sample to estimate a population mean. Based on previous evidence, researchers believe the population standard deviation is approximately
σ
=
24.4
. We would like to be 99.5% confident that the estimate is within 0.5 of the true population mean. How large of a sample size is required?

n
=

Answer 1

Using the z-distribution, it is found that a sample size of n = 18,805 is required.

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:

$z = 2.81, \sigma = 24.4, M = 0.5$.

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

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

$0.5 = 2.81\frac{24.4}{\sqrt{n}}$

$0.5\sqrt{n} = 24.4 \times 2.81$

$\sqrt{n} = 48.8 \times 2.81$

$(\sqrt{n})^2 = (48.8 \times 2.81)^2$

n = 18,805.

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

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