Question

It is desired to estimate the average time it takes Statistics students to finish a computer project to within two hours at 90% reliability. It is estimated that the standard deviation of the times is 14 hours. How large a sample should be taken to get the desired interval? n = 231 n = 189 n = 325 n = 133

Answer 1

n = 133.

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 - 0.9)/(2) = 0.05`

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 - 0.05 = 0.95`, so Z = 1.645.

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.

Standard deviation of the times is 14 hours.

This means that
`\sigma = 14`

How large a sample should be taken to get the desired interval?

Within 2 hours, which means that we want n for which M = 2. So

`M = z(\sigma)/(√(n))`

`2 = 1.645(14)/(√(n))`

`2√(n) = 1.645*14`

Dividing both sides by 2

`√(n) = 1.645*7`

`(√(n))^2 = (1.645*7)^2`

`n = 132.6`

Rounding up, n = 133.