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 0.130.13 gallons. A previous study found that for an average family the variance is 2.562.56 gallons and the mean is 16.316.3 gallons per day. If they are using a 80%80% 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:

Therefore, a sample that is required to estimate the mean usage of water must be at least 249.

Step-by-step explanation:

We are given that they would like the estimate to have a maximum error of 0.13 gallons.

A previous study found that for an average family the variance is 2.56 gallons and the mean is 16.3 gallons per day.  

As we know that; Margin of error formula is given by;

   Margin of error  =  $Z_(_\frac{\alpha}{2}_) \times \text{Standard of error}$

where, $\alpha$ = level of significance = 1 - 0.80 = 20% and  $(\frac{\alpha}{2} )$ = 10%

Standard of error =  $\frac{\sigma}{\sqrt{n} }$

where, $\sigma$ = standard deviation = $\sqrt{2.56}$ = 1.6

            n = sample size

Also, the critical value of z at 10% significance level is given in the z table as 1.282.

SO,  Margin of error  =  $Z_(_\frac{\alpha}{2}_) \times \frac{\sigma}{\sqrt{n} }$

                 0.13           =  $1.282 \times \frac{1.6}{\sqrt{n} }$

                  $\sqrt{n}$  =  $\frac{1.282 \times 1.6}{0.13}$

                    n  =  $15.8^{2}$

                    n  =  248.95  ≈ 249.

Therefore, a sample that is required to estimate the mean usage of water must be at least 249.