Question

A research firm wants to compute an interval estimate with 90% confidence for the mean time to complete an employment test. Assuming a population standard deviation of three hours, what is the required sample size if the error should be less than a half hour?

Answer 1

n=97

Explanation:

1) Notation and definitions

`\sigma=3` population standard deviation known

Confidence=90% or 0.9

n sample size required (variable of interest)

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The sample mean have the following distribution

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

2) Calculation for the sample size required

In order to find the critical value we need to take in count that we are finding the interval for the mean with the population deviation known, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by
`\alpha=1-0.90=0.1` and
`\alpha/2 =0.05`. And the critical value would be given by:

`z_(\alpha/2)=-1.64, t_(1-\alpha/2)=1.64`

The margin of error for the sample mean interval is given by this formula:

`ME=z_(\alpha/2)(\sigma)/(√(n))` (a)

And on this case we have that
`ME =\pm 0.5` and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=((z_(\alpha/2) \sigma)/(ME))^2` (b)

And replacing into equation (b) the values from part a we got:

`n=((1.64(3))/(0.5))^2 =96.83`

And rounded up we have that n=97