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 1.4 gallons and the mean is 15.9 gallons per day. If they are using a 80% level of confidence, how large of a sample is required to estimate the mean usage of water

Answer 1

A sample of 164 is needed.

Explanation:

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

`\alpha = (1-0.8)/(2) = 0.1`

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

So it is z with a pvalue of
`1-0.1 = 0.9`, so
`z = 1.28`

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.

How large of a sample is required to estimate the mean usage of water

We need a sample of n.

n is found when
`\sigma = 1.4, M = 0.14`. So

`M = z*(\sigma)/(√(n))`

`0.14 = 1.28*(1.4)/(√(n))`

`0.14√(n) = 1.28*1.4`

`√(n) = (1.28*1.4)/(0.14)`

`(√(n))^2 = ((1.28*1.4)/(0.14))^(2)`

`n = 163.84`

Rounding up

A sample of 164 is needed.