Question

The FDA regulates that fresh Albacore tuna fish that is consumed is allowed to contain 0.82 ppm of mercury or less. A laboratory is estimating the amount of mercury in tuna fish for a new company and needs to have a margin of error within 0.023 ppm of mercury with 97% confidence. Assume the population standard deviation is 0.143 ppm of mercury. What sample size is needed?

Answer 1

Solution :

Given that,

standard deviation =   $\sigma$ =0.143

Margin of error = E = 0.023

At 97% confidence level the z is,

$\alpha$ = 1 - 97%

$\alpha$ = 1 - 0.97 = 0.03

$\alpha$ /2 = 0.03

Z $\alpha$ /2 = 2.170  

sample size = n = [Z $\alpha$ /2* $\sigma$ / E] 2

n = (2.170* 0.143/0.023)2

n =182

Sample size = n =182