Finding the sample size needed to estimate a mean confidence interval of student completion rates
Howdy! I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help. Today we're going to learn how to find the sample size needed to estimate a mean student completion rate confidence interval. Here's our problem statement: In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full time college students who earn a bachelor's degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.05 margin of error, and use a confidence level of 95%. Complete Parts A through C below.
Part A says, "Assume that nothing is known about the percentage to be estimated." Well, to solve this problem there is a sample size estimator inside StatCrunch. So let's pull up StatCrunch. I don't think it'll be of much use to us because --- I'll show you in a moment. It actually requires us to know the standard deviation of our sample size, and there's nothing about standard deviation there in our problem statement. So that option there in StatCrunch isn't all that useful to us. But I'll show you here. We can move this around, resize that so we can see a little bit better what's going on. And there we go.
So if I go to T Stats --> One Sample --> Width/Sample Size, you can see here. Standard deviation is required to actually make the estimate for the sample size. Well, we don't know anything in here about standard deviation. There's nothing in here that indicates that, and there's nothing that we can use to estimate standard deviation. So this option is not going to be very useful for us. This means we have to go old school. So here's our equation for estimating sample size. We need a z score, which we then square, multiply by p-hat, which is the percentage that we want to be successful, and q-hat is the compliment of p hat. And then we divide all that by the margin of error squared.
First up, let's go get our z score, which means I need to go back into StatCrunch and pull up my Normal calculator. Remember that z scores come from the standard Normal distribution, which has a mean value of zero and a standard deviation of one. So here I've got the default values, and that's exactly what I need for my standard Normal distribution. I'm going to select the Between option here because we want a z score coming from two tails here. And the reason why I know that is because if I go back and look at my equation, I've got z alpha over two. That means only half of the alpha is in the right tail of my distribution. So there must be another half in the left tail of the distribution. I've got alpha split between two tails. So I want to use that Between option there.
In StatCrunch, this option gives us the percentage of the total area that is between the boundaries here, between the tails. So that in this case, that's going to be equal to the confidence level that we want. So I'm going to come down here and replace this with my 95% confidence level. And now when I hit Compute!, the z scores that come out, which is the boundary on these tails. This is the z score that I need. So I'm going to pull out my calculator. I'm going to put 1.96 squared (because the 1.96 is coming from that z score here; it's on the right tail of my distribution). So I've got the 1.96 squared. Now the rest of the formula says I need p-hat, q-hat. Well, we don't know anything about the percentage of what's successful. And so in that case we're going to estimate p-hat q-hat with 0.25. So I multiply this by 0.25, and then I'm going to divide by the square of my margin of error, which here in the problem statement is 0.05, so divide it by 0.05 squared. And now I get three 384.16. I'm instructed here to round up to the nearest integer. So that's going to give me 385. Nice work!
Now Part B says, "Assume prior studies have shown that about 55% of full time students earned bachelor degrees in four years or less." Now we have a value for p-hat q-hat. So if you want, go ahead and run through the same calculation that we just ran through for Part A using this same formula. You can start from the very beginning if you want, or you could follow the shortcut that I'm going to show you here. I know that the new p-hat is 55%. So I'm gonna take this value that's still in my calculator, and I'm going to multiply it by 0.55 (55%), and then I'm going to multiply that by q-hat. And to do that, I'm going to take the compliment of p-hat. So I can either do that in my head and get the 45% that I know it is, or I can let the calculator do that for me and say one minus 0.55. It's going to give you the 0.45. And now I'm going to divide this by the old p-hat q-hat, which was the 0.25, and there's my new value. Round up to the nearest integer. E voila! Well done!
Now Part C says, "Does the added knowledge in Part B have much of an effect on the sample size?" Well, we went from 385 down to 381. So our sample size got reduced by, like, four people. That's not a very large reduction; it's only slightly reduced. So I would say it doesn't have that much of an effect. And I'm going to select the answer option here that basically says that --- "No, it only slightly reduces the sample size." Well done!
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6/27/2019 09:55:09 pm
I am an adult student and I struggle with math. I am grateful for your videos and want to thank you for taking the time to make them. I would love to have you walk through problem 7.1.30 on both Statcrunch and the TI84 if possible?
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Frustrated with a particular MyStatLab/MyMathLab homework problem? No worries! I'm Professor Curtis, and I'm here to help.