Question

Many medical professionals believe that eating too much red meat increases the risk of heart disease and cancer (WebMD website, March 12, 2014).
Suppose you would like to conduct a survey to determine the yearly consumption of beef by a typical American and want to use 3 pounds as the desired margin of error for a 99% confidence interval of the population mean amount of beef consumed annually.
Use 25 pounds as a planning value for the population standard deviation.

How large a sample should be taken?

Answer 1

Answer: 461

Step-by-step explanation:

Formula to find the sample size is given by :-

$n=(\dfrac{z_{\alpha/2}\cdot \sigma}{E})^2$ , where n is the sample size , $\sigma$ is the population standard deviation and $z_{\alpha/2$ is the two tailed test value of z for significance level ($\alpha$).

Given : $\sigma=25\text{ pounds}$

Margin of error : 3 pounds

Confidence level = 99%

Significance level : $\alpha=1-0.99=0.01$

Critical value : $z_{\alpha/2}=z_{0.005}=2.576$

Then, the required minimum sample size would be :-

$n=(\dfrac{(2.576)\cdot (25)}{3})^2\approx460.817777778\approx461$

Hence, the required minimum sample size = 461