Step 1 :We are given that the sample mean (\(\bar{x}\)) is 15.7 gallons, the population standard deviation (\(\sigma\)) is 1.7 gallons, and the sample size (\(n\)) is 1454 families.
Step 2 :We are asked to construct a 98% confidence interval for the mean usage of water. The Z-score for a 98% confidence interval is approximately 2.33.
Step 3 :We use the formula for a confidence interval for a population mean, which is \(\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}\).
Step 4 :Substituting the given values into the formula, we get \(15.7 \pm 2.33 \frac{1.7}{\sqrt{1454}}\).
Step 5 :Calculating the margin of error, we get approximately 0.1039.
Step 6 :Subtracting and adding the margin of error from the sample mean, we get the lower and upper limits of the confidence interval, which are approximately 15.6 and 15.8 respectively.
Step 7 :Final Answer: The 98% confidence interval for the mean usage of water is \(\boxed{(15.6, 15.8)}\) gallons per day.