Question

You are interested in estimating the the mean age of the citizens living in your community. In order to do this, you plan on constructing a confidence interval; however, you are not sure how many citizens should be included in the sample. If you want your sample estimate to be within 4 years of the actual mean with a confidence level of 98%, how many citizens should be included in your sample? Assume that the standard deviation of the ages of all the citizens in this community is 23 years. (Use Excel to find the appropriate critical value and round it to 3 decimal places.)

Answer 1

Using the z-distribution, a sample size of 180 is needed for the estimate.

What is a z-distribution confidence interval?

The confidence interval is:

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

The margin of error is given by:

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

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.

In this problem, we have a 98% confidence level, hence$\alpha = 0.98$, z is the value of Z that has a p-value of $\frac{1+0.98}{2} = 0.99$, so the critical value is z = 2.327.

The population standard deviation is of $\sigma = 23$, and to find the sample size, we have to solve for n when M = 4.

Hence:

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

$4 = 2.327\frac{23}{\sqrt{n}}$

$4\sqrt{n} = 2.327 \times 23$

$\sqrt{n} = \frac{2.327 \times 23}{4}$

$(\sqrt{n}) = \left(\frac{2.327 \times 23}{4}\right)^2$

n = 179.03.

Rounding up, as a sample size of 179 would result in an error slightly above 4, a sample of 180 is needed.

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

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