Question

A limnologist wishes to estimate the mean phosphate content per unit volume of lake water. It is knownfrom studies in previous years that the standard deviation has a fairly stable value ofσ= 4. How manyindependent water samples must the limnologist analyze to be 90% certain that the error of estimation [halfwidth of 90% confidence interval] does not exceed 0.8 milligrams?

Answer 1

Answer: 68

Step-by-step explanation:

As per given , we have

Population standard deviation : $\sigma=4$

Critical value for  90% confidence interval =$z_{\alpha/2}=1.645$

Margin of error : E= 0.8 milligrams

The formula to find the sample size is given by :-

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

i.e. $n=(\dfrac{(1.645)(4)}{0.8})^2=(8.225)^2$

$=67.650625\approx68$

Hence, the limnologist must analyze 68 samples.