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19 March, 01:04

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 0.1 gallons. A previous study found that for an average family the variance is 5.76 gallons and the mean is 17.9 gallons per day. If they are using a 99% level of confidence, how large of a sample is required to estimate the mean usage of water?

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  1. 19 March, 02:31
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    19,652

    Step-by-step explanation:

    Error (E) = (t*sd) / √n

    √n = (t*sd) / E

    sd = √variance = √5.76 = 2.4, E = 0.1

    Assuming the average family is a family of 4

    n = 4, degree of freedom = n-1 = 4-1 = 3, t-value corresponding to 3 degrees of freedom and 99% confidence level is 5.841

    √n = 5.841*2.4/0.1 = 140.184

    √n = 140.184

    n = 140.184^2 = 19651.55

    n = 19,652
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