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

Answer:

A sample of 164 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:

$\alpha = \frac{1-0.8}{2} = 0.1$

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

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

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

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

$0.14 = 1.28*\frac{1.4}{\sqrt{n}}$

$0.14\sqrt{n} = 1.28*1.4$

$\sqrt{n} = \frac{1.28*1.4}{0.14}$

$(\sqrt{n})^2 = (\frac{1.28*1.4}{0.14})^{2}$

$n = 163.84$

Rounding up

A sample of 164 is needed.