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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?

1 Answer

6 votes

Answer: 461

Explanation:

Formula to find the sample size is given by :-


n=((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=(((2.576)\cdot (25))/(3))^2\approx460.817777778\approx461

Hence, the required minimum sample size = 461

User Markbarton
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