173k views
25 votes
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

User Ffff
by
8.4k points

1 Answer

6 votes

Answer:

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.

User Eirik Fuller
by
8.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.