 Hi guys, I want to spend a few minutes talking about a problem that one of your classmates brought up, and it has to do with the finite correction factor that's mentioned in your text as a minor topic, but it is something that you probably want to be aware of. Let's talk about the background for a couple of minutes. Most of the formulas, or many of the formulas that we use to compute the standard errors when we're looking at means and proportions, use the central limit theorem and also assume that there's an infinite population. Now in reality there is no such thing as an infinite population, but if you're taking a sample from a population and you replace that individual sample, then you in essence have an infinite population. But the reality is that when we do surveys and experiments we rarely ever have an infinite population. So we need to be careful to check to see if the sample size is large relative to the population size. When we're doing problems, particularly academic problems for homework and quizzes and exams, you really have to read the problem very carefully. Some problems will be very specific and say check to see if you need the finite correction factor. Others won't. So what you need to do is look at n if you're given that, which is the population size, and if you're given small n which is the sample size, then you can check to see if the sample size is less than 5% of the population size. If n divided by large n is greater than 5% 0.05, then you need to use the finite correction factor as I labeled here, the FCF. And the FCF is really pretty straightforward. It's just the square root of the population n minus the sample n divided by the population n minus 1. Or here I've written that FCF, this is Excel format, square root, parentheses open, parentheses open, big n minus little n, parentheses close divided by parentheses open, big n minus 1, parentheses close, parentheses close. That's your finite correction factor that you use. And you should note, just by looking at it by inspection, that if n is very small relative to big n, then the finite correction factor approaches 1, and therefore the adjustment is insignificant. So that's our equation for FCF. You should know that the finite correction factor applies both to problems involving means and problems involving proportions. And in both cases, we have to adjust the standard error. For the standard error of a mean, we just take the original standard error, which is sigma divided by the square root of n, small n, the sample size, multiply that times the FCF to get you the adjusted standard error of the mean. If we're working with proportions, then the standard error there is just the square root of p, the proportion times 1 minus p divided by small n, the sample size, and multiply that whole thing times the FCF again. So you would get an adjustment, which in general, if n is large compared to capital n, or the population n, then you will reduce the size of the standard error. And we'll see how that works in this problem. I want to give you an example. Now this is a problem that comes from myStatLab, and it is one of the ones that's more obvious because it asks you in the question, determine if the finite correction factor should be used, and if so, use it in your calculations. The problem is about a sample of 800 gas stations. And in that sample of 800 gas stations, the mean price of the gas was $2.82.9 cents per gallon with a standard deviation of 1 penny .010. And then from those 800 stations, they took a random sample, and they saved from this population. And some problems won't be that obvious. But what they're saying is, now we've switched things, this original sample of 800 stations out of all the gas stations in a state, in a country, or whatever. That is now our population, and we are sampling 60 of those stations. And so we need to know, is that sample size of 60 large with respect to the 800 in the now population? So capital N is 800, small n, the sample size is 60, sigma, this is the population standard deviation now, because we redefined that 800 sample gas stations as the population for this second part. And we're given a mean of $2.82.9 cents, and we're asked about the probability of getting a price less than $2.82.6 cents. I always find it extremely helpful, and cuts out a lot of errors, if you can sketch the normal curve, put in the values for your mean, and our mean here 2.829 is right there. I didn't mark it, but it's right in the center. And because we've only got .1 penny for the standard deviations, the $2.82 falls right there. And so it's to the left of the mean, and that means it's on the left tail, and that red area right there is what we're trying to find, which is pretty small. It gives you a hint that it's not going to be .9 or .8 or .7. It's going to be something less than .5, and probably less than .2. So let's check N divided by N. Is that greater than .75? And yes, I just put in a formula, which is N divided by N, and that is .075, which is greater than .05. So yes, we need to use the finite correction factor, and there's the equation again that I showed you on the other slides. And I just took that equation for sigma sub x bar and wrote that in a formula. And I named these values, so it'd be a little bit easier to understand. Sigma sub x bar is equal to sigma divided by the square root of N, that's the first part right there, multiplied times the square root of big N minus little N divided by big N minus 1. And over here, I put the actual cell numbers that we're using, B10 divided by the square root of B9 and so forth. And you don't have to do it, but I went ahead and calculated the z value, which is just x minus mu divided by sigma sub x bar, our adjusted sigma sub x bar, our standard error is adjusted. And then we calculate the probability, and again, we're looking for this from left infinity to the value of x, and to do that we use the norm dot distribution function in Excel. Just type in, I'll just show you there, if you just start typing in equal norm, and it'll start offering up the various types of normal distribution functions we have. You want norm dot dis, and it takes, as its parameters, our value of x, mu, the mean, our adjusted sigma sub x bar, and we use the value of true to give us the cumulative mass function, which is the total probability from left infinity to our x value, make that true. And that gives us a probability of 0.0079, which is pretty small. And I also calculated it using our z value, just uses a different Excel function, if you know the z, use the norm dot s dot dis function. And again, set that to be true because we want the cumulative, everything from the left infinity to the right. So that's how you solve a problem, checking to see if you need to use the finite correction factor, and then actually adjusting sigma, your standard error, for that FCF, finite correction factor, and then solving for a probability the traditional way. I hope this helps.