Question

An investigator wishes to estimate the mean number of flight hours of pilots who fly for regional airlines. Assuming that the standard deviation of the number of hours is 120 hours, the minimum sample size needed in order to construct a 99% confidence interval for the mean number of hours of flight time of all such pilots, to within 50 hours, is about

Answer 1

To construct a 99% confidence interval for the mean number of flight hours of pilots who fly for regional airlines, the minimum sample size needed to estimate the mean within 50 hours is approximately 43.

Step-by-step explanation:

To construct a 99% confidence interval for the mean number of flight hours of pilots who fly for regional airlines, the minimum sample size needed to estimate the mean within 50 hours can be found using the formula:

n = (Z * σ / E)^(2)

where:

• n is the minimum sample size
• Z is the Z-value corresponding to the desired level of confidence (in this case, 99%)
• σ is the standard deviation of the number of flight hours (120 hours)
• E is the desired margin of error (50 hours)

Substituting the values, we have:

n = (2.576 * 120 / 50)^(2) ≈ 42.03

Therefore, the minimum sample size needed to construct the 99% confidence interval is approximately 43.

Answer 2

The minimum sample size needed is 39.

Step-by-step 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 - 0.99)/(2) = 0.005`

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 - 0.005 = 0.995`, so Z = 2.575.

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.

Assuming that the standard deviation of the number of hours is 120 hour

This means that
`\sigma = 120`

The minimum sample size needed in order to construct a 99% confidence interval for the mean number of hours of flight time of all such pilots, to within 50 hours, is about?

The minimum sample size needed is n.

n is found when
`M = 50`

So

`M = z(\sigma)/(√(n))`

`50 = 2.575(120)/(√(n))`

`50√(n) = 2.575*120`

`√(n) = (2.575*120)/(50)`

`(√(n))^2 = ((2.575*120)/(50))^2`

`n = 38.19`

Rounding up

The minimum sample size needed is 39.