Question

ou 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 96%, how many citizens should be included in your sample

Answer 1

Answer:

$(\frac{4\sigma}{2.056})^2$, rounded up, if needed, citizens should be included in the sample, in which $\sigma$ is the standard deviation of the population.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.96}{2} = 0.02$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$.

That is z with a pvalue of $1 - 0.02 = 0.98$, so Z = 2.056.

Now, find the margin of error M as such

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

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Within 4 years of the actual mean

We have to find n for which M = 4. So

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

$4 = 2.056\frac{\sigma}{\sqrt{n}}$

$2.056\sqrt{n} = 4\sigma$

$\sqrt{n} = \frac{4\sigma}{2.056}$

$(\sqrt{n})^2 = (\frac{4\sigma}{2.056})^2$

$n = (\frac{4\sigma}{2.056})^2$

$(\frac{4\sigma}{2.056})^2$, rounded up, if needed, citizens should be included in the sample, in which $\sigma$ is the standard deviation of the population.