Question

Determine the minimum sample size required when you want to be 75% confident that the sample mean is within 30 units of the population mean assume the population standard deviation of 327.8 in a normally distributed population

Answer 1

The minimum sample size is 158.

Explanation:

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

`\alpha = (1-0.75)/(2) = 0.125`

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
`1-0.125 = 0.875`, so
`z = 1.15`

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.

In this question:

The minimum sample size is n.

n is found when M = 30.

We have that
`\sigma = 327.8`

So

`M = z*(\sigma)/(√(n))`

`30 = 1.15*(327.8)/(√(n))`

`30√(n) = 1.15*327.8`

`√(n) = (1.15*327.8)/(30)`

`(√(n))^(2) = ((1.15*327.8)/(30))^(2)`

`n = 157.9`

Rounding up

The minimum sample size is 158.