Question

A research company desires to know the mean consumption of meat per week among males over age 25. They believe that the meat consumption has a mean of 3.8 pounds, and want to construct a 85% confidence interval with a maximum error of 0.06 pounds. Assuming a standard deviation of 1.3 pounds, what is the minimum number of males over age 25 they must include in their sample? Round your answer up to the next integer.

Answer 1

Answer:  974

Step-by-step explanation:

Given : Significance level : $\alpha: 1-0.85=0.15$

Using the standard normal distribution table for z,

Critical value for significance level of $\alpha:0.15$ : $z_{\alpha/2}= 1.44$

Sample mean : $\overline{x}= 3.8$

Standard deviation : $\sigma= 1.3$

Margin of error : E=0.06

The formula to find the minimum sample size is given by :-

$n=(\dfrac{z_{\alpha/2}\ \sigma}{E})^2$

i.e .$n=(\dfrac{(1.44)(1.3)}{0.06})^2=973.44\approx974$

Hence, the minimum number of males over age 25 they must include in their sample = 974