Final answer:
To estimate the mean water usage with a 0.1 gallon margin of error and 99% confidence, the water works commission should sample approximately 3832 households, based on a known population variance of 5.76 gallons and a mean of 17.9 gallons per day.
Step-by-step explanation:
To estimate the mean household usage of water by the residents of a small town with a maximum error of 0.1 gallons at a 99% level of confidence, we must determine the appropriate sample size. Given the variance of 5.76 gallons and a previously studied mean of 17.9 gallons per day, we can use the formula for the sample size of a mean when the population standard deviation is known:
n = (Z^2 * σ^2) / E^2
Where:
n is the sample size
Z is the Z-value corresponding to the desired level of confidence (from Z-table)
σ is the population standard deviation (the square root of the variance)
E is the maximum error of the estimate
For a 99% confidence level, the Z-value is approximately 2.576. The population standard deviation σ is the square root of the variance, which is √5.76. The maximum error E is 0.1 gallons.
Thus, n = (2.576^2 * √5.76^2) / 0.1^2 = (6.64 * 5.76) / 0.01 = 38.3104 / 0.01 = 3831.04. Therefore, the water works commission would need to sample approximately 3832 households (since we round up to the next whole number).