Question

The travel-to-work time for residents in Ventura County is unknown. Assume the population variance is 39. How large should a sample be if the margin of error is 1 minute for a 93% confidence interval

Answer 1

A sample size of 128 is needed.

Explanation:

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

`\alpha = (1-0.93)/(2) = 0.035`

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
`1-0.035 = 0.965`, so
`z = 1.81`

Now, find the margin of error M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population(square root of the variance) and n is the size of the sample.

How large should a sample be if the margin of error is 1 minute for a 93% confidence interval

We need a sample size of n, which is found when
`M = 1`. We have that
`\sigma = √(39)`. So

`M = z*(\sigma)/(√(n))`

`1 = 1.81*(√(39))/(√(n))`

`√(n) = 1.81√(39)`

`(√(n))^2 = (1.81√(39))^2`

`n = 127.8`

Rounding up

A sample size of 128 is needed.