The Retail Advertising and Marketing Association would like to estimate the average amount of money that a person spends for Mother's Day with 99% confidence interval and a margin of error within plus or minus $6. Assuming the standard deviation for spending on Mother's Day is $36, the required sample size is ________.
A) 210
B) 239
C) 284
D) 313

Question

Quizplus · Accepted Answer

To calculate the required sample size for estimating the average amount of money spent on Mother's Day with a specific margin of error and confidence level, we use the formula for sample size in estimating a mean: $n = \left(\frac{Z\sigma}{E}\right)^2$, where $n$ is the sample size, $Z$ is the Z-score corresponding to the desired confidence level, $\sigma$ is the population standard deviation, and $E$ is the margin of error. For a 99% confidence interval, the Z-score is approximately 2.576. Given $\sigma = $36$ and $E = $6$, substituting these values into the formula gives $n = \left(\frac{2.576 \times 36}{6}\right)^2$, which calculates to approximately 239. Therefore, the required sample size is 239.

The Retail Advertising and Marketing Association Would Like to Estimate