Question

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

Answer 1

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.