 Hi, this is Dr. Don. I have a problem out of Chapter 5, Section 4, on sampling distributions. This one is about a sampling proportion. And the problem says, according to a recent poll, 68% of adults have used the internet to download music, and a random sample of N equals 750 let p hat represent proportion who pay to download music. The first question is, fine, the mean of the sampling distribution p hat, well, in these, the mean of that sampling distribution is just equal to the sampling proportion, which in this case is 0.68. To find the standard deviation of the sampling distribution, sigma sub p hat, and to answer these other questions, we're going to do that using Excel. Okay, I have an Excel spreadsheet that I've already worked on here, and I'll walk you through how it's constructed. In my Excel spreadsheets, or my, I call them sometimes my Excel calculators, things that are in blue are data inputs. Things that are in white are yellow, pickly yellow, are outputs which are calculated from the inputs using formulas. So here you can see I entered the N, which for this case is 750, and our p is 0.68. And I'm looking down here, we're going to check out two limits on p hat 0.75 and 0.57. So I'm going to enter those. Now I've added some additional values there, I call them x's, x1 through x4, just to show you something additional that is not in this particular problem, but you may see in a problem down the line, so I wanted you to have a tool for that. Okay, we have to use some formulas here. The expected value, which is the mean of the sampling distribution of a proportion, p hat, is just equal to the proportion that we were given, which in this case is 0.68. And down here in my yellow area, I've got that the expected value of p hat is equal to 0.68, and you can see there the formula is just equal b2 up there. I like to figure or calculate q, which is 1 minus p, and I've done that in that cell there, 1 minus p is equal to 1 minus b2. So our sigma sub p hat, using this formula, and again, 1 minus p is just q. So it would be equal to the square root of b2, which is our p times q, 0.32, divided by n, which is b1, and take it the square root of that whole thing, gives me a sigma sub p hat of 0.17. Okay, let's go back over here, and they asked a question about the central limit theorem, and that's the key to this. When you're dealing with proportions, and on most of the problems you'll run into, when you're given a proportion, you use the binomial distribution to model it. But in this case, they want us to model it using what is known as a normal approximation to the binomial. I don't know if your book says that, but that's what we're doing here. And the reason we can do that is because the central limit theorem says, as n goes large, the shape of the sampling distribution will approach normal, or will be approximately normal. And generally we say if n is greater than 30, then the distribution is approximately normal. Here we've got 750, so the central limit theorem says as the sample size grows large, the sampling distribution of the proportion p hat will become approximately normal. So that allows us to use the normal distribution to solve the two final problems. Calculate the probability that p hat is less than 0.75, and the probability that p hat is greater than 0.57. Over here in my Excel spreadsheet, I put the first cell there, the probability of p hat less than or equal to x1, x1 is 0.75. And we get that answer using the norm.dis function. And the norm.dis function uses these inputs, the x value, the mean mu, and the sigma for whatever the means are mused. In this case, our mean is our expected value and our sigma is our sigma sub x hat. And that's what I've got here. The norm distribution of b3, b9, b10, b3 is x1. B9 is our expected value, which is the mean, and b10 is sigma sub x hat. The final parameter is true, which means we want the cumulative probability from the left infinity all the way up to that point. And so we just put that formula in. In this case, it tells us that the probability is approximately one, that more than likely the sampling proportion is going to be less than 0.75. The other question is the probability of greater than x2, x2 being 0.57. In this case, we want to the right end of the probability distribution, the normal distribution, so we use a one minus. We subtract that left tail from the overall distribution to get the right tail. But again, this is so extreme that it's approximately one. The last thing I did there, which this question doesn't ask for, is if you were asked what is the probability of p hat being in a range between two values of x. In this case, I just picked out 0.59 to 0.62. And there you just subtract the two normal distribution values. Normal distribution of the 0.62 minus the normal distribution of the 0.59 gives you the area between 0.002. So that's how you do this using Excel.