A study has indicated that the sample size necessary to estimate the average electricity use by residential customers of a large western utility company is 900 customers. Assuming that the margin of error associated with the estimate will be ±30 watts and the confidence level is stated to be 90 percent, what was the value for the population standard deviation?
A) 265 watts
B) Approximately 547.1 watts
C) About 490 watts
D) Can't be determined without knowing the size of the population.

Question

Quizplus · Accepted Answer

The sample size formula for estimating a mean with a specific margin of error at a given confidence level is $n = \left(\frac{Z \cdot \sigma}{E}\right)^2$, where $n$ is the sample size, $Z$ is the Z-score corresponding to the confidence level, $\sigma$ is the population standard deviation, and $E$ is the margin of error. Given $n = 900$, $E = 30$ watts, and a 90% confidence level (which corresponds to a Z-score of approximately 1.645 for a two-tailed test), we can rearrange the formula to solve for $\sigma$: $\sigma = \frac{E \cdot \sqrt{n}}{Z}$. Plugging in the given values, we find that $\sigma$ is approximately 547.1 watts.

A Study Has Indicated That the Sample Size Necessary to Estimate