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 variance is 1.69 gallons and the mean is 15.9 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? Round your answer up to the next integer.

Answer 1

Answer:

A sample of 179 is needed.

Step-by-step explanation:

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

$\alpha = \frac{1 - 0.85}{2} = 0.075$

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 $1 - 0.075 = 0.925$, so Z = 1.44.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

A previous study found that for an average family the variance is 1.69 gallon?

This means that $\sigma = \sqrt{1.69} = 1.3$

If they are using a 85% level of confidence, how large of a sample is required to estimate the mean usage of water?

A sample of n is needed, and n is found for M = 0.14. So

$M = z\frac{\sigma}{\sqrt{n}}$

$0.14 = 1.44\frac{1.3}{\sqrt{n}}$

$0.14\sqrt{n} = 1.44*1.3$

$\sqrt{n} = \frac{1.44*1.3}{0.14}$

$(\sqrt{n})^2 = (\frac{1.44*1.3}{0.14})^2$

$n = 178.8$

Rounding up

A sample of 179 is needed.