Howdy! I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help. Today we're going to learn how to find a confidence interval given summary sample data. Here's our problem statement: Refer to the accompanying data display of results from a sample of airport data speeds in megabytes per second. Complete Parts A through C below.
OK, Part A says, "Express the confidence interval in the format that uses the less than symbol given that the original list of data use one decimal place around the confidence interval limits accordingly." OK. Typically when students encounter a problem like this, by this part of the term you're looking at Chapter 7 here, you've gotten into the habit of running to StatCrunch or running to Excel to solve a problem. And here you just need to take a step back and realize the answer is sitting right in front of you.
They're giving you a data display, and look — "t interval" OK, an interval using the, the uh, I guess the Student t distribution, which makes sense because we're looking at means here, and here's our lower and upper limit for a confidence interval. So it's just staring you right in the face. All you gotta do is follow the instructions here, round to two decimal places as needed. So that's what we're going to do. Excellent!
Now Part B says, "Identify the best point estimate of mu and the margin of error to get the best point estimate of μ." We need to recognize μ is the population mean, and we learned back in Chapter 6 that the mean is an unbiased estimator. That means the sample mean is going to target the population mean; it makes the best point estimate for your population parameter. So the sample mean, which we represent with x-bar, is listed here in my problem statement. So I'm just going to round that to two decimal places. Fantastic!
And now the margin of error. Don't be taking this number here and running with it. That's not your margin of error. Your margin of error is not represented with S(x). This is a standard deviation, so don't be using that. That's not your margin of error. We're going to have to actually calculate margin of error, which we can get pretty easy enough. We've got the point estimate, so we've got the central point for our confidence interval. And we've got the upper and lower bounds here. So if I just take this point estimate and subtract out the lower limit, that distance is going to give me my margin of error.
Now alternatively, I could be subtracting the point estimate from the upper limit, but it gives you the same number. The other way to go about it is to just average the two out. But then that just gives you this point estimate here. Then you've got to actually, you know, do one or the other with it. So I just find it easier to just take the point estimate and then subtract out the lower limit. Well done!
And now the last part, Part C, asks, "In constructing a confidence interval estimate of μ, why is it not necessary to confirm that the sample data appear to be from a population with a normal distribution?" OK, let's look at each one of these answer options and see which one represents our best option here.
Answer option A says, "Because the sample is a random sample, the distribution of sample means can be treated as a normal distribution." Well, OK, we would hope that the sample is obtained through some random sampling method, but that's just saying how we get the data. The distribution comes from what the data represents, not how we get the data. And so there's no real connection between the two here. So Answer option A isn't going to work for us.
Answer option B says, "Because the sample standard deviation is known, the normal distribution can be used to construct the confidence interval." OK, so here we've got a ... sample standard deviation is known. It's listed up here in our summary stats table. But again, in all, this is talking about how the data is spread out. And the distribution comes from what the data represents. So again, there's a faulty logical connection here. Answer option B isn't going to work for us.
Answer option C says, "Because the population standard deviation is known, the normal distribution can be used to construct the confidence interval." Well, same thing here. We don't really know what the population standard deviation is, but if we did, that doesn't necessarily tell us what distribution we're supposed to be using to represent our data. Again, the distribution comes from what the data represents, not from how it's sampled and not from what the numbers are --- although, you know, it kind of plays in part to what the numbers are, but it's more representative of what the numbers are represented. That's where the distribution type comes from.
So now we're just left with Answer option D: "Because the sample size of 50 is greater than 30, the distribution of sample means can be treated as a normal distribution." Oh, bingo! That's what we're looking for right here. Remember, the Central limit Theorem says that, when your sample size is greater than 30, you can use the normal distribution as an approximation for whatever distribution that data actually represents. So here we've got n = 50, so our sample size is greater than 30, and that means the Central Limit Theorem applies. So we're going to just select Answer option D. Good job!
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Frustrated with a particular MyStatLab/MyMathLab homework problem? No worries! I'm Professor Curtis, and I'm here to help.