The average natural gas bill for a random sample of 26 homes in the 19808 zip code during the month of March was $305.30 with a sample standard deviation of $46.50. The margin of error for a 98% confidence interval around this sample mean is -----.
A)$25.20
B)$22.66
C)$18.69
D)$20.15

Question

Quizplus · Accepted Answer

To calculate the margin of error for a 98% confidence interval, we use the formula: $ME = Z * (\frac{\sigma}{\sqrt{n}})$, where $Z$ is the Z-score corresponding to the confidence level, $\sigma$ is the sample standard deviation, and $n$ is the sample size. For a 98% confidence level, the Z-score is approximately 2.33 (since the Z-score for 99% confidence level is 2.576, and for 95% confidence level is 1.96, 98% would be in between but closer to 2.576). Plugging in the values: $ME = 2.33 * (\frac{46.50}{\sqrt{26}})$, we get approximately $22.66 as the margin of error.

The Average Natural Gas Bill for a Random Sample of 26