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 l

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 2 gallons and the mean is 16 gallons per day. If they are using a 95% level of confidence, how large of a sample is required to estimate the mean usage of water

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Thành Công 1 week 2021-07-22T09:20:19+00:00 1 Answers 1 views 0

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    2021-07-22T09:22:14+00:00

    Answer:

    A sample of 784 is required to estimate the mean usage of water.

    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 - 0.95}{2} = 0.025

    Now, we have to find z in the Z-table as such z has a p-value of 1 - \alpha.

    That is z with a pvalue of 1 - 0.025 = 0.975, so Z = 1.96.

    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.

    The standard deviation is 2 gallons

    This means that \sigma = 2

    They would like the estimate to have a maximum error of 0.14 gallons. How large of a sample is required to estimate the mean usage of water?

    This is n for which M = 0.14. So

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

    0.14 = 1.96\frac{2}{\sqrt{n}}

    0.14\sqrt{n} = 1.96*2

    \sqrt{n} = \frac{1.96*2}{0.14}

    (\sqrt{n})^2 = (\frac{1.96*2}{0.14})^2

    n = 784

    A sample of 784 is required to estimate the mean usage of water.

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