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 0.12 gallons. A previous study found that for an average family the variance is 5.29 gallons and the mean is 17 gallons per day. If they are using a 85% level of confidence, how large of a sample is required to estimate the mean usage of water

Answer 1

`n=((1.440(2.3))/(0.12))^2 =761.76 \approx 762`

So the answer for this case would be n=762 rounded up to the nearest integer

Explanation:

Information given

`\bar X = 17` represent the mean

`\mu` population mean (variable of interest)

`\sigma= √(5.29)= 2.3` represent the standard deviation

Solution to the problem

The margin of error is given by this formula:

`ME=z_(\alpha/2)(s)/(√(n))` (a)

And on this case we have that ME =0.12 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=((z_(\alpha/2) \sigma)/(ME))^2` (b)

The confidence level is 85%, the significance level would be
`\alpha=1-0.85 = 0.15` and
`\alpha/2 =0.075` the critical value for this case would be
`z_(\alpha/2)=1.440`, replacing into formula (b) we got:

`n=((1.440(2.3))/(0.12))^2 =761.76 \approx 762`

So the answer for this case would be n=762 rounded up to the nearest integer