Question

A population has standard deviation o=17.5.

Part 1 of 2
(a) How large a sample must be drawn so that a 99.9% confidence interval for u will have a margin of error equal to 4.1? Round the answer up to the nearest integer. (Round the critical value to no less than three decimal places.) A sample size of is needed to be drawn in order to obtain a 99.9% confidence interval with a margin of error equal to 4.1.
Part 2 of 2
(b) If the required confidence level were 95%, would the necessary sample size be larger or smaller? (Choose one) V , because the confidence level is (Choose one) V .

Answer 1

a) A sample of 198 must be drawn.

b) Smaller, because the confidence level is smaller.

Explanation:

Question a:

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

`\alpha = (1 - 0.999)/(2) = 0.0005`

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 - 0.0005 = 0.9995`, so Z = 3.291.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

How large a sample must be drawn so that a 99.9% confidence interval for u will have a margin of error equal to 4.1?

This is n for which M = 4.1. So

`M = z(\sigma)/(√(n))`

`4.1 = 3.291(17.5)/(√(n))`

`4.1√(n) = 3.291*17.5`

`√(n) = (3.291*17.5)/(4.1)`

`(√(n))^2 = ((3.291*17.5)/(4.1))^2`

`n = 197.3`

Rounding up:

A sample of 198 must be drawn.

(b) If the required confidence level were 95%, would the necessary sample size be larger or smaller?

We have that:

`M = z(\sigma)/(√(n))`

Solving for n

`√(n) = (z\sigma)/(M)`

That is, n and z are directly proportion, meaning that a higher value of z(higher confidence level) leads to a higher sample size needed.

95% < 99.9%, so a smaller confidence interval.

Smaller, because the confidence level is smaller.