Question

A company wants to test a randomly selected sample of n water specimens and estimate the mean daily rate of pollution produced by a mining operation. If the company wants a 95% confidence interval estimate with a margin of error of 1.7 milligrams per liter (mg/L), how many water specimens are required in the sample? Assume prior knowledge indicates that pollution readings in water samples taken during a day are approximately normally distributed with a standard deviation equal to 7 mg/L.

The company must sample______water specimens in order to estimate the mean daily rate of pollution produced by a mining operation to 1.7 mg/L with 95 % confidence.

(Round up to the nearest specimen.)

Answer 1

Given that, standard deviation $(\sigma)$ = 7 mg/L

margin of error ( E ) = 1.7 mg/L

A 95% confidence level has significance level $(\alpha)$ = 0.05 and critical value is, $Z_{\alpha/2} = 1.96$

We want to find, the sample size ( n ),

$n = (Z_{\alpha/2} *\frac {\sigma}{E})^2 = (1.96 * \frac {7 }{1.7})^2 = 65.1344$

$n \approx 65$

Therefore, The company must sample 65 water specimens in order to estimate the mean daily rate of pollution produced by a mining operation to 1.7 mg/L with 95 % confidence.