A research company desires to know the mean consumption of meat per week among males over age 48 . They believe that the meat consumption has a mean of 3.6 pounds, and want to construct a $80 \%$ confidence interval with a maximum error of 0.08 pounds. Assuming a standard deviation of 0.9 pounds, what is the minimum number of males over age 48 they must inctude in their sample? Round your answer up to the next integer.

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92023 Hawkes Learning

Step 1 ：A research company wants to determine the average amount of meat consumed per week by men over the age of 48. They hypothesize that the mean consumption is 3.6 pounds. They want to construct an 80% confidence interval with a maximum error of 0.08 pounds. The standard deviation is assumed to be 0.9 pounds. The task is to find out the minimum number of men over 48 years old that they need to include in their sample.

Step 2 ：The Z-score for an 80% confidence level is 1.28.

Step 3 ：The formula for calculating the sample size is \(n = \left(\frac{Z \cdot \sigma}{E}\right)^2\), where \(Z\) is the Z-score, \(\sigma\) is the standard deviation, and \(E\) is the maximum error.

Step 4 ：Substituting the given values into the formula, we get \(n = \left(\frac{1.28 \cdot 0.9}{0.08}\right)^2\).

Step 5 ：Calculating the above expression, we get \(n = 208\). However, since we can't have a fraction of a person, we round up to the next whole number.

Step 6 ：Thus, the minimum number of men over 48 years old that they need to include in their sample is \(\boxed{208}\).