In an effort to estimate the mean dollars spent per visit by customers of a food store, the manager has selected a random sample of 100 cash register receipts. The mean of these was $45.67 with a sample standard deviation equal to $12.30. Assuming that he wants to develop a 90 percent confidence interval estimate, the upper limit of the confidence interval estimate is:
A) about $2.02
B) approximately $65.90
C) about $47.69
D) None of the above

Question

Quizplus · Accepted Answer

To calculate the upper limit of a 90% confidence interval for the mean, we use the formula: $\bar{x} + Z \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $Z$ is the Z-score corresponding to the desired confidence level. For a 90% confidence interval, the Z-score is approximately 1.645. Plugging in the given values: $45.67 + 1.645 \frac{12.30}{\sqrt{100}} = 45.67 + 1.645 \times 1.23 = 45.67 + 2.02 = 47.69$.

In an Effort to Estimate the Mean Dollars Spent Per