Question

A long history of testing water samples in a certain lake has shown that the level of a certain pollutant is approximately normally distributed with standard deviation 5 mg/L. What is the minimum number of samples required to estimate today’s level to with in 0.5 mg/L with 91% confidence?

Answer 1

Solution :

Given that,

standard deviation=  $\sigma$ =5

Margin of error = E = 0.5

At 91% confidence level the z is,

$\alpha$ = 1 - 91%

$\alpha$ = 1 - 0.91 = 0.09

$\alpha$/2 = 0.045

Z_$\alpha$/2 = 1.695

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

n = ( 1.695* 5 / 0.5)²

n =287.3025

Sample size = n =287