## Question 506:

1

We need a sample size which will generate a margin of error of no more than +/- 5 units (which I interpret to mean vacation days). The margin of error is calculated by multiplying the standard error of the mean by the critical value of the t or z-distribution. As the sample size increase the values of t and z converge. If this is the population standard deviation then we'd use z. For a two-sided 99% confidence interval, the normal deviate or z-score is 2.33.

We can then just setup and equation and solve for the unknown:

1. The margin of error is the standard error of the mean SEM * the critical z value of 2.33, so working backwards from a margin of 5, the SEM = 5/2.33= 2.15.
2. So the SEM is 2.15 which is made up of the SD/ SQRT(n). Using the given standard deviation of 6 units and solving for the unknown sample size n we have the equation 6/sqrt(n) = 2.15.
3. Squaring both sides = 36/n  = 4.62
4. Isolating n gets us 4.62n = 36 or n = 36/4.62 = 7.79
5. Since we cannot ask part of a person, we rounding up to the nearest person to get 8, which will also slightly lower the margin of error from 5 to 4.9

So a sample size of 8, given our estimate of the standard deviation of 6 will generate a +/- 4.9 unit of vacation margin of error, which would be no more than the +/-5 required.

Caveat:
Keep in mind the assumption that the standard deviation given is the known population standard deviation, which is not always the case (in fact rarely the case in sample size planning). If we treated the given standard deviation as an estimate of the population standard deviation, then we'd use the t-distribution to build the margin of error. This would increase the width of the confidence interval. Using the same method as above and substituting the critical value which would satisfy the margin of error of +/-5 we'd insert a t of 3.01 instead of the z of 2.33. This would increase our sample size to 14.