Meat Consumption
A marketing research company desires to know the mean consumption of meat per week among people over age 31. They believe that the meat consumption has a mean of 4.7 pounds, and want to construct a 80% confidence interval with a maximum error of 0.09 pounds. Assuming a variance of 0.64 pounds, what is the minimum number of people over age 31 they must include in their sample?
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1 solution
We are told in the problem that σ2 = 0.64 and E = 0.09. Since the formula uses the standard deviation instead of the variance, we can take the square root of the variance to get σ = 0.64√ = 0.8. The only thing remaining is the find the value of zα/2. Using the level of confidence, c = 0.80, we can determine that α = 1 − 0.80 = 0.20. This means that we want to find the critical value z0.10.
n= (1.28 · 0.80.09)2 ≈ 129.4538 ≈ 130