Question

How large a sample should be selected to provide a 95% confidence interval with a margin of error of 10? Assume that the population standard deviation is 40. (Round your answer up to the nearest whole number.)

Answer 1

We need a sample of at least 62.

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`.

So it 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.

How large a sample should be selected to provide a 95% confidence interval with a margin of error of 10?

We need a sample of at least n.

n is found when
`M = 10, \sigma = 40`

So

`M = z*(\sigma)/(√(n))`

`10 = 1.96*(40)/(√(n))`

`10√(n) = 40*1.96`

Simplifying by 10

`√(n) = 4*1.96`

`(√(n))^(2) = (4*1.96)^(2)`

`n = 61.47`

Rounding up

We need a sample of at least 62.