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 4.5 mg/L. What is the minimum number of samples required to estimate today's level to within 0.4
mg/L with 95% confidence? (Don't forget to round Zc to two decimal places!)

Answer 1

The minimum number of samples required is 487.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1 - 0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.025 = 0.975`, so Z = 1.96.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

Standard deviation 4.5 mg/L.

This means that
`\sigma = 4.5`

What is the minimum number of samples required to estimate today's level to within 0.4 mg/L with 95% confidence?

This is n for which M = 0.4. So

`M = z(\sigma)/(√(n))`

`0.4 = 1.96(4.5)/(√(n))`

`0.4√(n) = 1.96*4.5`

`√(n) = (1.96*4.5)/(0.4)`

`(√(n))^2 = ((1.96*4.5)/(0.4))^2`

`n = 486.2`

Rounding up

The minimum number of samples required is 487.