Question

A marketing research company desires to know the mean consumption of meat per week among males over age 26. They believe that the meat consumption has a mean of 3.7 pounds, and want to construct a 80% confidence interval with a maximum error of 0.09 pounds. Assuming a standard deviation of 1.4 pounds, what is the minimum number of males over age 26 they must include in their sample

Answer 1

Answer:

80% confidence interval with Maximum error is

(3.61, 3.79)

Step-by-step explanation:

Explanation

Given the mean of the sample ( x⁻) = 3.7

The maximum error  = 0.09

80% confidence interval is determined by

$(x^{-} - Z_{0.80} \frac{S.D}{\sqrt{n} } , x^{-} +Z_{0.80} \frac{S.D}{\sqrt{n} })$

( x⁻ - M.E , x⁻ + M.E)

( 3.7 - 0.09 , 3.7 +0.09)

(3.61 , 3.79)

Final answer:-

80% confidence interval with Maximum error is

(3.61, 3.79)