A market research company was interested in assessing how much money a single adult typically spent on groceries in one week.Participants were 772 single adults and the mean weekly grocery expenditure was found to be $69.70 with a standard deviation of $2.80.The 99% confidence interval for the mean weekly grocery expenditures for single adults would be _____.
A)[63.19,76.21]
B)[62.49,76.91]
C)[69.65,69.75]
D)[69.47,69.93]
E)[69.44,69.96]

Question

Quizplus · Accepted Answer

The 99% confidence interval is calculated using the formula: mean ± (z-score * (standard deviation/√n)). For 99% confidence, the z-score is approximately 2.576. Plugging in the values: 69.70 ± (2.576 * (2.80/√772)) gives the interval [69.44, 69.96].

A Market Research Company Was Interested in Assessing How Much