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Use the data in ECONMATH to answer the following questions.(i) Estimate a model explaining colgpa to hsgpa, actmth, and acteng. Report the results in the usual form. Are all explanatory variables statistically significant?(ii) Consider an increase in $h s g p a$ of one standard deviation, about $343 .$ By how much does $\widehat{c o l g p a}$ increase, holding actmth and acteng fixed. About how many standard deviations would the actmth have to increase to change colgpa by the same amount as a one standard deviation inhsgpa? Comment.(iii) Test the null hypothesis that actmth and acteng have the same effect (in the population) against a two-sided alternative. Report the $p$ -value and describe your conclusions.(iv) Suppose the college admissions officer wants you to use the data on the variables in part (i) to create an equation that explains at least 50 percent of the variation in colgpa. What would you tell the officer?

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Chapter 4

Multiple Regression Analysis: Inference

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So this is the last computer exercise of chapter four. So if you already done a bunch of the problems, then great. And if this is your first one, then we'll get through it. So just a quick background of this problem is that we're looking at the function or we're looking at college Deepa as a function of different high school level variables. So, um, things like a C T test scores and high school G p A, um, are the variables that were looking at in predicting or explaining college g p A. So part one literally asks you to just estimate a model of college G, p a and, um, again, the explanatory variables. Here are your high school G p a. Your and your scores. So to get us started, I'll just write out the different the parameter estimates in the standard errors that we're going to get when you estimate this in whatever software program you used. So we have. The intercept is the first part, and we have a positive coefficient on high school G p. A. And it's a large coefficient, so this makes sense that the higher your high school G p a is, the higher your predicted college GPA will be and it pretty small standard air. So this is a definitely going to be a statistically significant coefficient here. So there we go with the high school G p a explanatory variable. We also have our A C T math scores here, and that's also a positive coefficient. So that's good. That makes sense. The higher your seat massacre is, the higher your predicted college GPA will be and looks like that's going to also be statistically significance there. And our final explanatory variable is the other, the other a CT score, which is English. So we would also expect this to have a positive coefficient, and we dio awesome and also looks to be statistically significance. And, of course, I was locked on an air term at the end. Part one is just asking. Are all these explanatory variables statistically significant? And I've kind of already mentioned that, but I'll just circle and go through a high school G P a. Definitely right. That T statistic is going to be very, very large. Not sure exactly what, but maybe approximately what would that be? 12 or so, so very high T statistic, so we could say yes for that one. The T statistic here for a seat math. That's gonna be maybe a little over 2.5. So check that check here on, then the coefficient on a CT. English, that variable. Also going to be live on her 2.5. So we could say that would be statistically significant as well. So pretty quick. Part one there. Moving on a part two part two asks us to consider an increase in high school G p a of about one standard deviation, or about 0.343 And how that would affect predicted college G p A. So before we get into that, I'm just gonna rewrite what are from part one. What are beta hat for high school GPA? A is. And that's 0.659 That's just straight from part one. Right. So what will increase in high school G p A of 0.343? Um, what kind of increase in college? G p A. Will that dio So remember this coefficient here? This beta had this, um, results. I'll just say it this way. Results from a one unit increase in high school GPA. Right? So part two is asking us. Thio, I'm sorry about high school GPA, so party was asking. It's not a one unit increase. Part two is asking us to consider a 0.343 unit increase instead. So the way we're going toe approach How much College GP would increase it? Um, mhm. From from a 0.343 unit increase in high school G p a. Is is the following. We'll just say that the like a little equation. So increase in the predicted college GPA will equal 0.343 times 0.659 Right? So this just coming out of that were just multiplying beta hat of high school GPA by 0.343 instead of by one. Right, So we'll get the increases 0.22 Sorry, there 0.2 to 6, and that will be the first answer again. So we can say that holding the TV sores fixed college G p A. Will increase by about 0.2 to 6, given a 0.343 increase in high school GPA. Second part of the question, though, asks by about how Maney standard deviations would the math. So how much would a seat math have to increase right t to change the productive college GPA by the same amount as we just found? So I can do a similar type of approach, as you did the first little part here for an equation. That's again our increase in predicted college GPA equals, um are Let's see, the standard deviation times, beta hats of a seat. Math. This is the same set up as our as our, um, equation here. Right. And this time are unknown is the increase. Our Sorry that we know the increase in college GPS. What we don't know is the this standard deviation so you can rearrange to g o. Standard deviation equals What's our bait ahead of a C t. Math. So back up here, beta hat was 0.13 Right? So standard deviation equals 0.13 divided by we already know what increase in college people shooting for? No. Sorry. I did that backwards. We're gonna We're gonna resets. So standard deviation actually equals 0.226 That's on the top. So the increase in college g p a over beta hat a CT math, which is 0.13 And once you do that calculation, you will get Let me tape it out really quick. So I have this race about 17. So 17 points three eight is the whole standard deviation. You know, this is the aggregate standard deviation value. But what we have to do at this point is to divide this by the actual standard deviation of a C t. Math, um, to get our final answer. So what we can do is, um, have our last equation. Here's the number of standard deviations that we're looking for equals this aggregate standard deviation value over and in your statistical software program, you can just do some sort of summarize command. That's what state it does. But some some commanding of the descriptive statistics of all your variables. And you can find the standard deviation of a CT math aan dat. Standard deviation turns out to be, um, 3.773 So that will go on the denominator here. 3.773 And so you get 4.6 here. That will be your answer. So 4.6 standard deviations. Yeah, do a quick quick reboot with that and just circle it nicely. There we go. So this is saying that a C T math score would have to increase about 4.6 standard deviations to have the same thio induce the same increase in college deepa as a one standard deviation increase in high school GPA. So why is this? So this is the comment. Ask you to comment Why that might be so. So. One thing you can think about is the the increase in a seat math scores has to be so much larger in comparison, Thio the increase in the high school G p A. Have to be much larger simply because it's not a strong determinant of college. G P a. So to say, because it's not a strong a determinant of college GPA and this, you know, this makes intuitive sense, right? Your GP. Your high school record will likely have greater weight in, um, how you might perform in college. You know, it's it's a instead of, um, your performance in high school overall will probably be a better predictor of your performance in college overall. Instead of just your performance on one standardized test score in one subject. So that's what you can kind of have your comment if you could comment Toe end, part two in. All right. Now that part two is done, we have part three. Part three asks you to test the null hypothesis that the A, c T math and a CT English, um, have the same effect against the to set an alternative. So asking you if a seat math score and a CT English or have the same effect on college g p A. So we can say this as no hypothesis is that beta two equals beta three, where Manitou is associated with your A C T math score, and beta three is associated with your a CT English score. So if you've done previous problems, um, in chapter 40 might have come across this solution or this this approach we're gonna use here, but what we want to do to test this hypothesis is to create a new parameter what's called stated to instead of equal to beta two minus beta three missile out that would allow subs substitute for beta two in a regression. And so you can all, uh, you can alter this. Um this equation here to isolate Beta too, and then plug it in for beta two in this first regression. But I'll just write on how this should look. How this how your regression should look in your, um when you're setting it up, you should have the something like the following. So college GPA equals beta Not plus again. Beta one high school G p A. Not only focused on that one in this in this part, three plus have faded to our new parameter isolated with a CT math and then our last one. We don't have data to anymore. We have beta three at the end, and Beta three is multiplying the some of a CT math and a c t English. And you can check this algebra for yourself for this whole substituting in are our parameter in here. And, of course, an air term here. So once we have these, um, you run that regression in your in your software package and I will write out the parameter estimates you should get. I'll just write out. Uh, I'll just read out theater the fate of two coefficient here because that's what we're interested in. Great. So Once you run, this estimation should get faded. Thio hat equals 0.8 And the standard error on this is 0.85 So this already looks like, you know, this is a very small, small coefficient. The standard air is much larger. And it actually to report the P value for this parameter, and the P value ends up being very statistically insignificant. It equals 0.9 to 5 is what I got to get a P value of 0.9 to 5. Then the conclusion should be for part three. Do not reject h ne h not or the null hypothesis do not reject. And so what does that mean? Just in English. So since we have a high p value on this data to parameter, we should not reject the null hypothesis. So we should say that the math score on a CT English course likely have the same effect on college GPA. That's part three part four. Last part of the problem is asking you to suppose that the college admissions officer wants you to use the data on the variables in part one. So the ones we've been working with to create an equation explains at least 50% of the variation in college Deepa. So just ask you to do, you know, the function function we've been doing, which is just the college g p. As a function of at school G p A. And you're a seat math score and you're a CT English score. I'm sorry, this one up here should be across that out. A CT English. That's my mistake. Sorry about that. Go back up and look at this problem if you need to switch that. Um, Anyway, part four. The college admissions officer wants you to use just these three explanatory variables to explain 50% of the variation in college EPA. So they want you to have an r squared of greater than or equal to 0.5. Right? So, from part one, you could go back and run this regression and look at the r squared that we got from part one. Um, and what I got was the r squared from part one equals 0.256 So that's not really that close 2.5, right? And the most the most you could do other than what we've already done with these three variables, you might think about adding higher powered terms, like squared terms. Um, we don't typically love doing that approach, but to increase the r squared, that could be one thing you could You could use thio address what the college admissions officer wants you to, dio. And so I actually did that? Um, I would say, you know, try it. So add a square, generate a square of, um, high school G p A generator, squared the massacre and generate a square term of a CT English score and added to the regression here at the square terms. But when I when I had in the square terms, um, I'm threatened out. When I added squared terms, the regression the R squared on Lee went up a little bit. So r squared for me only equaled about 0.28 So what you could do is just explain to the college admissions officer it that the following you could just explain with a set of variables. We have the high school GPA and the scores Thio to get to 50% to get thio R squared equals 0.5 or to get to the point where you're explaining 50% of the variation in college GPA, you would need to include mawr relevant explanatory variables so and buy more relevant. I mean, a greater number of relevant explanatory variables that you think would be associated with college GPA. And I'll give a couple of examples of what those might be in a second. But yeah, the only way you're going to get to that high of an R squared will be to include other variables. So you would not be able to use the the variables that we just have from part one. Those won't be sufficient. You will maybe need, um I'll just write down examples of other variables you'd maybe want toe look at would be race. So race of students or gender of students, Or maybe where they lived where they grew up. So region of residents, Um, it was just a few examples of things you would have to include. But you'd have to explain to the college admissions officer that these variables that we have here these three ones, um, high school G p A a. C T scores would not get you to ah explaining 50% of variation and called peer. You need additional variables. So that is the end of the problem.

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