Question

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 mean of 3.1 pounds and want to construct a 85% confidence interval with the maximum error of 0.06 pounds assuming a standard deviation of 0.8 pounds what is the minimum number of people over age 31 they must include in their sample? Round your answer up to the next integer

Answer 1

369

Explanation:

Given:

Mean (μ) = 3.1

Standard deviation (σ) = 0.8

Margin of error (E) = 0.06

At 85% confidence level the z is:

`\begin{gathered} \alpha=1-\text{ confidence level} \\ \\ \alpha=1-85\%=1-0.85=0.15 \\ \\ \alpha\text{/2}=(0.15)/(2)=0.075 \\ \\ \text{ The corresponding value of z according to the table is:} \\ \\ Z_{\alpha\text{/2}}=1.44 \end{gathered}`

We can determine the sample size using the following formula:

`\begin{gathered} n=\:\left(\frac{Z_{\alpha\text{/2}}\cdot\sigma}{E}\right)^2 \\ \\ \text{ We replacing:} \\ \\ n=\left((1.44\cdot0.8)/(0.06)\right)^2 \\ \\ n=368.64\cong369 \end{gathered}`

The size of the sample is 369