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The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day. They would like the estimate to have a maximum error of 0.14 gallons. A previous study found that for an average family the standard deviation is 2 gallons and the mean is 16 gallons per day. If they are using a 95% level of confidence, how large of a sample is required to estimate the mean usage of water

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Answer:

A sample of 784 is required to estimate the mean usage of water.

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 - 0.95)/(2) = 0.025

Now, we have to find z in the Z-table as such z has a p-value of
1 - \alpha.

That is z with a pvalue of
1 - 0.025 = 0.975, so Z = 1.96.

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.

The standard deviation is 2 gallons

This means that
\sigma = 2

They would like the estimate to have a maximum error of 0.14 gallons. How large of a sample is required to estimate the mean usage of water?

This is n for which M = 0.14. So


M = z(\sigma)/(√(n))


0.14 = 1.96(2)/(√(n))


0.14√(n) = 1.96*2


√(n) = (1.96*2)/(0.14)


(√(n))^2 = ((1.96*2)/(0.14))^2


n = 784

A sample of 784 is required to estimate the mean usage of water.

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