Question

Find the sample size necessary to estimate the mean arrival delay time for all American Airlines flights from Dallas to Sacramento to within 6 minutes with 95% confidence. Based on a previous study, arrival delay times have a standard deviation of 39.6 minutes.

Answer 1

The sample size necessary is of 168.

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.

Based on a previous study, arrival delay times have a standard deviation of 39.6 minutes.

This means that
`\sigma = 39.6`

Find the sample size necessary to estimate the mean arrival delay time for all American Airlines flights from Dallas to Sacramento to within 6 minutes with 95% confidence.

This is n for which M = 6. So

`M = z(\sigma)/(√(n))`

`6 = 1.96(39.6)/(√(n))`

`6√(n) = 1.96*39.6`

`√(n) = (1.96*39.6)/(6)`

`(√(n))^2 = ((1.96*39.6)/(6))^2`

`n = 167.34`

Rounding up:

The sample size necessary is of 168.