Answer to Question #246687 in Statistics and Probability for Chris

Question

Polokwane Butchery supplies Vienna sausages in the entire of KZN province. An Inspector from

the CCSA has been skeptical of the mass of the Vienna sausage packs after receiving numerous

complaints. He wantsto investigate the marked massshown on Vienna sausages. A pilotstudy showed

a mean of 11.8kg per pack and a variance of 0.49kg. How many packs should the Inspector sample in

order to be 99% confident that the sample mean will differ by at most 0.2kg?

Answer 1

$ME=z_{0.005}\frac{\sigma}{\sqrt{n}}$

$n=(\frac{z_{0.005}\sigma}{ME})^2=(\frac{2.576*\sqrt{0.49}}{0.2})^2=82.$