Question

Use 6.84 days as a planning value for the population standard deviation. a. Assuming 95% confidence, what sample size would be required to obtain a margin of error of days

Answer 1

The sample size required is
`n = ((1.96*6.84)/(M))^2`, in which M is the desired margin of error. If n is a decimal number, it is rounded up to the next integer.

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 Z-table as such z has a p-value 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.

Use 6.84 days as a planning value for the population standard deviation.

This means that
`\sigma = 6.84`.

What sample size would be required to obtain a margin of error of M days?

This is n for the given value of M. So

`M = z(\sigma)/(√(n))`

`M = 1.96(6.84)/(√(n))`

`M√(n) = 1.96*6.84`

`√(n) = (1.96*6.84)/(M)`

`(√(n))^2 = ((1.96*6.84)/(M))^2`

`n = ((1.96*6.84)/(M))^2`

The sample size required is
`n = ((1.96*6.84)/(M))^2`, in which M is the desired margin of error. If n is a decimal number, it is rounded up to the next integer.