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, which of the following is the margin of error that will be reported?
A) About ±$2.02
B) Nearly $50.20
C) $1.645
D) About $1.43

Question

Quizplus · Accepted Answer

The margin of error for a 90% confidence interval can be calculated using the formula: $ME = Z * (\sigma / \sqrt{n})$, where $Z$ is the Z-score corresponding to the desired confidence level (for 90%, $Z = 1.645$), $\sigma$ is the sample standard deviation, and $n$ is the sample size. Plugging in the values: $ME = 1.645 * (12.30 / \sqrt{100}) = 1.645 * 1.23 = 2.02$.

In an Effort to Estimate the Mean Dollars Spent Per