To estimate the mean age for a population of 4000 employees, a simple random sample of 40 employees is selected.
a. Would you use the finite population correction factor in calculating the standard error of the mean? Explain.
b. If the population standard deviation is $\sigma=8.2$ years, compute the standard error both with and without the finite population correction factor. What is the rationale for ignoring the finite population correction factor whenever $n / N \leq .05 ?$
c. What is the probability that the sample mean age of the employees will be within $\quad \pm 2$ years of the population mean age?
All right. So this question asks us about a sample of 40 employees from pool of 4000. In part a asks us if we would use the finite population correction factor. And in this case, yes, we would, because we're sampling 40 from a pool for 1000. And if we do that to find the proportion, that would be sampling 40 divided by 4000 is 40000.1 So it's still 1%. So by our 5% rule, we could ignore it. But mostly since we have access since we have access to the total number of employees, it will make our result more accurate. All right, so now it wants us to calculate the standard air with and without the correction factor. So with the correction factor, we'll call this Sigma. Acts corrected is equal to this correction factor, which is the total number minus our sample size. Divided by our degrees of freedom of our population, times our formula for standard air and then putting all the numbers in their crack spot. We get 4000 minus 40 over 4000 minus one all times our population sigma of 8.2, divided by the square root of our sample size, which with the correction factor, this is one point two zero at one point 2901 And that's the corrected. And then our normal standard air is just population Sigma over the square event, which is just the previous formula without that square factor that we gonna multiply by. So in this case, it's just 8.2 over Route 40 and that turns out to be 1.2 965 So, as you can see, based on our two numbers here, there is a bit of a discrepancy. Not too much. They're both still 1.29 but there's a little bit of a discrepancy, and that's because we're only sampling from 1% of the population, so they're pretty close, but it still doesn't hurt to use the correct. So now it wants the probability that the sample man is within two years, which is the same thing, is asking probability that mu minus two Sigma standard air is there a lower bound and then you plus two sigma standard air is our upper bound and they don't actually give us value for this mean here, which could be a bit tricky, but based on how we know normal curves work, we could take whatever value we want from you. As long as we use the right value for our standard air, we're going to get the same answer. It doesn't matter if we have army and be 30 years, 40 years, 50 years or 1000 years. So in this case, just to make things easy, let's just pick let Mu equal of 40 cause that's an easy number to work with. And it's logical. But so we want the probability that were between 38 in 42 which this is, of course, a probability question with a normal curve. So we use normal CDF. We have our lower bound, our upper bound are mean, and now we have to make a choice. What do we want to use for our standard deviation? So we have two choices. We have the standard air, and we have the corrected standard air. So since we already computer the corrected standard error, we'll just use that. So at one point thio 901 And what does this work out to be? Well, it works out to be 0.878 eight