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 0.12 gallons. A previous study found that for an average family the variance is 5.29 gallons and the mean is 17 gallons per day. If they are using a $$ 85 \% $$ 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.

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UpStudy AI · Accepted Answer

Let the following be defined:

- Margin of error: $$ E = 0.12 $$ gallons.
- Known variance: $$ \sigma^2 = 5.29 $$, so standard deviation is $$ \sigma = \sqrt{5.29} = 2.3 $$ gallons.
- Confidence level: $$ 85\% $$, which implies a two‐sided confidence interval with $$ \alpha = 1 - 0.85 = 0.15 $$ and $$ \alpha/2 = 0.075 $$.

The corresponding critical value $$ z_{\alpha/2} $$ is obtained from the standard normal distribution by finding the value $$ z $$ such that

$$
P\left(Z \leq z\right) = 1 - \frac{\alpha}{2} = 0.925.
$$

This gives approximately

$$
z_{0.075} \approx 1.44.
$$

The formula for the required sample size to estimate a mean is

$$
n = \left(\frac{z_{\alpha/2}\sigma}{E}\right)^2.
$$

Substitute the values:

$$
n = \left(\frac{1.44 \times 2.3}{0.12}\right)^2.
$$

First, calculate the product in the numerator:

$$
1.44 \times 2.3 = 3.312.
$$

Then, form the fraction:

$$
\frac{3.312}{0.12} = 27.6.
$$

Now, square the result:

$$
n = (27.6)^2 = 761.76.
$$

Since the sample size must be an integer and to ensure the error is no greater than $$ 0.12 $$ gallons, we round up:

$$
n = 762.
$$

Thus, the required sample size is $$ 762 $$.
A sample size of 762 households is needed to estimate the mean water usage with a maximum error of 0.12 gallons at an 85% confidence level.

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