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? Group of answer choices

Answer 1

Answer: 461

Step-by-step explanation:

Formula to find the sample size is given by :-

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

Given : Margin of error : E=  3 pounds

Standard deviation: $\sigma= 25\ pounds$

Critical value for 99% confidence interval (Two-tailed ): $z_{\alpha/2}=2.576$

Required sample size = $n=(\dfrac{(2.576)\cdot 25}{3})^2$

$=460.817777778\approx461$

Hence, the required sample size = 461