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To estimate the average cost of a food-shopping event, Delcore, Inc. randomly sampled 100 shoppers and found a sample mean of $72. Assuming a population standard deviation of $5, a 99% confidence interval for average cost for a food-shopping event is _______.
Correct Answer: A
For a 99% confidence interval we find z(0.005), the cut-off for the top 0.5% of the normal distribution. Looking up 0.995 in the middle of the table the reading to the row/column values, we get 2 .575. Working with the formula for E we get E = 1.29. So, the 99% confidence interval is $72 - 1.29 m $ 72 + 1.29 or $70.71 m $73.29.