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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

User Sdoxsee
by
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1 Answer

4 votes

Answer:


((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.

Explanation:

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


\alpha = (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(\sigma)/(√(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(\sigma)/(√(n))


4 = 2.056(\sigma)/(√(n))


2.056√(n) = 4\sigma


√(n) = (4\sigma)/(2.056)


(√(n))^2 = ((4\sigma)/(2.056))^2


n = ((4\sigma)/(2.056))^2


((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.

User Omiz
by
8.0k points
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