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5. (15 Points). Let gradbe a dummyvariable for whether a student-athlete at a large universitygraduates in five years. Let hsGPA and SAT behigh school grade ...

Question

5. (15 Points). Let gradbe a dummyvariable for whether a student-athlete at a large universitygraduates in five years. Let hsGPA and SAT behigh school grade point average and SAT score, respectively. Letstudy be the number of hours spent per week in anorganized study hall.(i) (5 Points) Let Φ(﹒) be the standard normal cumulativedistribution function and ϕ(﹒) be the standard normaldensity function. Please write down the Probit model to describethe probability of graduating in five years

5. (15 Points). Let gradbe a dummy variable for whether a student-athlete at a large university graduates in five years. Let hsGPA and SAT be high school grade point average and SAT score, respectively. Let study be the number of hours spent per week in an organized study hall. (i) (5 Points) Let Φ(﹒) be the standard normal cumulative distribution function and ϕ(﹒) be the standard normal density function. Please write down the Probit model to describe the probability of graduating in five years in terms of hsGPA, SAT and study. Please write down the partial effect of study on P(grad = 1|hsGPA, SAT, study). (ii) (5 Points) If we use a Linear Probability Model instead, write down the model that you’ll estimate. What limitations does this model have? (iii) (5 Points) Suppose that, using data on 420 student-athletes, the following logit model is obtained: P(grad = 1|hsGPA, SAT, study) = Λ(-1.17 + .24hsGPA + .00058SAT + .073study) where Λ(z) = exp(z)/[1 + exp(z)]. Please give the expressions for P(grad = 1|hsGPA, SAT, study)/P(grad = 0|hsGPA, SAT, study) and log ( P(grad = 1|hsGPA, SAT, study)/1-P(grad = 1|hsGPA, SAT, study)). Please show your work as detailed as possible.



Answers

Let grad be a dummy variable for whether a student-athlete at a large university graduates in five years. Let $h s G P A$ and $S A T$ be high school grade point average and SAT score, respectively. Let study be the number of hours spent per week in an organized study hall. Suppose that, using data on 420 student-athletes, the following logit model is obtained: $$\widehat{\mathbf{P}}(g r a d=1 | h s G P A, S A T, s t u d y)=\Lambda(-1.17+.24 h s G P A+.00058 S A T+.073 \text { study })$$, where $\Lambda(z)=\exp (z) /[1+\exp (z)]$ is the logit function. Holding $h s G P A$ fixed at 3.0 and $S A T$ fixed at $1,200,$ compute the estimated difference in the graduation probability for someone who spent 10 hours per week in study hall and someone who spent 5 hours per week.

Okay, So in this problem, instructor has found a way to calculate Ah students probability of passing a class based on variables of s and a where s is the students score on ah, departmental placement exam and is a number of semesters of mathematics passed in high school and a is the students mathematics s a T score. So I have that form your function working down here 0.5 times a plus six times s times end to 1/2 power. Now this problem has three parts A B and C part A wants you to find the probability that a student passes a class with we'll score of six on the placement exam four semesters of high school math and s a T score of 4 16 So they want you to find f of six for for 60. And all you do is simply plug in these about values for the variables into our function. So if we do that, we get 0.5 times a which is for 16 or 60 plus six times the square root of size and which is four. And if we were to multiply its all out, we get an answer of 52.39%. So this student in particular would have a 52.39% chance of passing a math class. Um, eso not party for TB. Ask you to find probability for another student. Another student this time with math. Score off. Uh, with the last score of four. Please don't score. Four with five years of 55 years, five semesters of high school math and a score of 300 it s a T. So all we gotta do again is play this into the formula. Pretty simple. Your points. You're a fine times 300 plus six times the square root of 20. All right, so if we were to calculate all this, this person has a 41.83 41.83 percent chance of feeling, So this person has more likely to fail the class and pass it. Now. Part C Part C asks you to find an interpret the partial prove it is with respect to end and a at the 0.45 480. So, if we were to take the partial derivative with respect and and find the value. At 45 for 80 we're gonna first need to take the partial derivative with respect to end. Let's go ahead and do that. If we took the derivative with respect to end, all were differentiating. Is, uh this terms? This is the only turn with on invariable in it, and everything else is constant. So if we were to differentiate that we get, we had to bring down the two. So we've got three times s times on to the native. 1/2 power times times s because off general. Right. So this is our part of the related with respect to when and if you plug in the 0.45 480 to arm for me long, we get two point 683683 And this is our partial derivative. Well, just to pay. No, no. They also want you to interpret what this value means. That we're gonna go ahead and do that. Type that here in this text box. Yes, text box. And it says, and this would be a great change. Read a change in the probability of passing her unit change in the lumber. Oh, it's a mess. Race a school while keeping well. Keeping reveals constant everything else. I mean, the other two variables. All right, so this is our This is our interpretation for the function, the partial derivative of the function with respect to end. Now, the other thing they want us to do is take the partial derivative with respect to A with respect to a at the point, the same point for five for 80. All right, so first, we're gonna need to take the partial derivative respect A and thankfully, we just have to differentiate this. So are partial derivative is simply yes. 0.5 So there are no variables. So our partial derivative at the 0.4 580 is just 0.5 If we were to interpret this, we're contemplative. Since we can't interpret this as well, um, this is just going to be a rate of change. Greater shame off. Read a change in a probability. Passing her unit change in the students in the students, Uh, s a T score. Well, keeping There you go. This is our interpretation for the partial derivative with just like to a and I believe that is it for this problem

Cool. Okay. Sixty. A developmental mathematics instructor at a large universe that has determined on a student's probability off success. You know, you're Estes Pass or fail, Matthew. Our general course is a function of s and an early where s is a student school auntie departmental placement exam and is a lumber up its master Sof masked man. It's past in high school and a is a student's mathematics and Santy score estimates. She asked him, Mr P, the probability off passing the course in person will be their function. And we have a range of a twenty, two hundred and hundreds as his front there to turn a frontier eight and way. Want Teo finds Feli A for students Score six hundred placement exam has taken force master of high school mass and high school and he's along grossness through. Okay, so any for no. Okay, we're just plugging numbers for A and equals four. Score six on the replacement exam, which is ass because six and s A T scores for sixty Shit. That's a if for sixty now a plaque in p equals zero point zero five. What flower for sixteen applause. That's a lie. Six times. What hundred six with one half. Now I just need to calculate that he's in our It was a dream of flying here. Yeah, close. Thanks. Twenty four a half. One for the zero ah, years What I ask you, but oh P's in person. That's right. Yeah. So he's going to be fifty two. Two point through knife, fifty two percent approximately. So relax for eight. Not be. It is similar, but different. A lumber. So this time five masters off high school mass, which is on he was five s equals four A equals three hundred. Ah, Now blocking P equals syrup on zero five. Twenty. Plus we're grew close. Six months. Fly forward tongues fine. Toothy one half. Now, if we can correct that. Ah, ze aeroplane. Fried flies three hundred a quarter Rosie hair appointment here does it for me in there. What a one point takes three percent. Not this car's nowhere probability to pass that eight three. Now. See, we want to find f sub end off this lumber and f sub bay. So we need to first find its up and And if something so f sub urn you cause differentiate this function with respect and eso, This party's become constant this part of we need to use Okay, we can write off how frequent their plans. Their five a day pull us six s water and one off into this part thereof thiss part constant. One of us is far channel for end one half on next to one House three. This is the one half. And to collect you one half we want to will have four five, seven for eighty Now, this is a This is can This is class. So we're plucking. We have three multiply forces. One half butterfly End was in Nephew five ninety one have now every Congress. That's just too right. So it caused six divided by a squared five. Um cock anus. These are spooks. Bye, squirt. Fine. Oh, which is Troop on fifth? Take Not That means we want to interpret it, right. That means one. A CE stays out of four estates. Before it was for eighty. When and increased by one. You got one more semester of mass, Angus, your person did go up by two point six. Yeah, so that's for FC on. Have sub a Yeah way. First find out someday equals it's just their appointed. There are five, right? This part is this constant way back in. That's all right. I hear this guy course they're part of their body. Doesn't even matter. So what? This means one your ass. And you was flying a five. Whenever you're a increased by along, that's your s a T score increased by one. Your probability off passing this art protest increased by three point zero five. Yeah. Yes, that's it for us.

Referring to an example in the section. Example 442 Yeah, for the by variant normal model for the random variable X, which was the S a T critical reading score and the random variable wide It was the s a T mathematics score. Define the random variable bull W to be the S a T writing score. The third component of the S A. T, we're told us is a mean of 488 and standard deviation of 114 were asked to assume that X and W So the critical reading and the writing portions heavy by various normal distribution where the correlation coefficient of the X and W is 0.5 in part a were asked to find the distribution of X plus w Well, we know that X and W are by vary. It's normal. This is what we're given as an assumption. It follows that you some X plus w has a UNA vary it normal distribution, in particular with a mean of the expected value of X plus w, which recalled the expected value of a sum is the same as thes some of expected values. So this is this expected value of X plus The expected value of w were given that the expected value of X previously Excuse me. Yeah, was 496 were given that the expected value of W is 488. So we have 496 plus 488 and this sums up 2 984. This is the mean and it has a standard deviation well determined by the variance. So we'll actually find the variance first, The variance of a some of variables. This is going to be the some of the variances so variance of X plus the variance of W plus to times the coefficients Next w, which is one and one times the co variance of X and w. This is the same as the variance of X plus the variance of W plus two times the correlation coefficient rho of X and W times the standard deviation of x times the standard deviation of w By the definition of correlation, coefficient and substituting, we have that the variance of X with standard deviation of x 114 This is would be 114 squared staring deviation was of W. Was also 114. And so this is 114 squared, plus two times the Correlation coefficient Rho, which we're told was 0.5 times the standard deviation of X, which again is 14 or 1 14. I mean times the standard deviation of W, which is also 1 14 and this adds up to 38,988. Therefore, it follows that the standard deviation of X plus w is the square root of this number. Yeah, this gives us 197.45 approximately so we have that. X Plus W is normally distributed with a mean, which we calculated to be 984 and a standard deviation of approximately 197.45 next in part B. Whereas to calculate the probability that thes some of the reading and writing scores is greater than 1200. Well, because the sum of X and W, we determined is normally distributed probability that X plus w is greater than 1200. This is the same as one, minus the probability that it's less than or equal to 1200 which is one minus five of 1200 minus the mean, which is 984 over the standard deviation, which is approximately 197.45 This reduces to approximately one minus Phi of 19 and using a table or a computer to calculate this value, we find that this is approximately 0.1379 Next, in Part C were asked to find the combined reading and writing score that separates the top 10% of the students from the rest. In other words, we're looking for the 90th percentile of this distribution. So the 90th percentile of the normal distribution with mean of 984 standard deviation of 197.45 So to find this, we'll set 0.9 for 90 percentile, equal to five of and then are variable X minus 984 over staring deviation 197.45 This is approximately five of Well, we could calculate it that way, or we could again using computer. You want to find the value of X minus 984 for 1 97.45 such that why of this number will be approximately equal 2.9. Seizing a computer, a table we have that X minus 984 over 197.45 is approximately equal to 1.28 and therefore, using some algebra, we find that X is equal to 984 plus 1.28 times 197.45 And this is approximately 1237. So this is what the 90th percentile looks like. Score of 12 37 in the combined reading and writing.

Okay. What do we have with us? We have X as a random variable. That represents the average daily temperature in Para Night in January. For the town off Hannah, Now X has a mean off approximately 68 F. Okay, so we can see that the mean is 68 all right? And the standard deviation sigma is approximately 4 F. All right, Now, there is a 20 year study that is one of 620 January days, and we have their observations with us in the table. All right, so it is Just look at the table. Okay, So the first column is reason under. Of course. Okay, after that, we have the temperature. Then we have expected person from the normal. Then we have the expected person from normal curve. Okay. After that, we have the observed values. After that, we have the observed values. We have the observed values. All right. Now, the region under the curve is the first category Is the reason between three standard deviation and to standard deviation towards the left. Okay, so this is going to be new, minus three sigma. And let me just write this anti. Call him. It'll just take a minute. Then we have new minus sigma. Then we have new. We just have milk. Then we have new plus sigma. And then we have mute lis. All right, X, then we have new minus two sigma. Mean minus two Sigma new minus Sigma men, New plus sigma. New plus two Sigma New plus three sigma. All right. Now, what are the exact boundaries this is given in this column? We have 56. We have 60 than 64. 68 72 76. All right, these are all less than equal to science. Okay, 80 76 70 to 68 64 60. All right. Now, if the normal distribution actually fix the, you know, actually fix the observations that we have, then what should be the expected percentage from the normal Come? Like if I draw this normal distribution? Let's say that this is the mean right? This is my one standard deviation of it. I'm to standard deviation of I'm 300 deviation of. Okay, so this one is mu minus three sigma. And this one is mu minus two. Signal what is actually the area what percent off values lie in this region. If this actually is a normal distribution, the expected present attribute 2.35%. Right? So these are the values over here in this column. So this is 2.35%. And since it is symmetrical, the last one will also be 2.35%. Then it is 13.5% 13.5%. And since this is symmetrical, this is also 13.5%. This is 64% and this is also 64%. Now, what is exactly the observed values that we have with us? Okay, The observed values are 14 86 207 Then we have 215 Then we have 83 then we have 15. All right, so now let's just look at the questions. What exactly are they asking us? They're saying impact one that mu and sigma this and this. Now what? Our columns 12 and three. In the context of this problem, well, call them one. What has actually happened happened over here is they have divided the entire distribution into various categories. So column one is going to tell us that this particular reason the first region is the reason between three Sigma and two Sigma towards the left like and this value is the difference between two sigma and one Sigma towards the left and so on. Okay, so this is column one column six Give Sorry. This column to gives us the exact boundaries and column three is nothing but the expected value. If this really were to follow normal distribution, what is the percentage of values that we expect to find between new minus three Sigma Nu minus two Sigma that is in this region, right? What is expected amount? It is 2.35 What is expected amount in this region, that is between 23 sig mind to sigma it is 13.5%. Right. And then over here we have the observed values, the values that have actually been observed. All right, now, what is this question to question two says we have to use a 1% level of significance, which means what is our Alfa R Alfa for this question is 0.1 all right to test the claim that the average daily January temperature follows a normal distribution with Mu 68 sigma for All right. So in orderto do this, in order to perform this analysis, we're going to use the chi Square statistic. And the first time of the Chi Square statistic is to find the expected values to find the expected values to find the expected values. And the formula for that is the total sample sizes. The total sample size multiplied by the probability multiplied by the probability multiplied by the probability off each category off each category. Okay, off each category. All right, so what exactly is going to be my sample size? If I add all of these up, this should be 6. 20 as it is given to us, right? 6, 20 days. All right, so let us just use that formula way. This is the expected values for all the categories expected values. All right, so this comes over here, so let's put the formula into action. Okay? So I'm going to use my calculator for this. I want to 0.35% off 6. 20. So this is 0.235 multiplied by 6. 20. This is 14.57 expected values. 14.57 Then I have 13.5% off. 6. 20 0.135 in 26 20. This is 83.7, 83.7. Then we have 64% or 0164 into 6. 20. This is 3 96 3 96 point eight. All right. Just a moment. This is 0.6. For how can this be 0.64 This has to be 34. My mistake. I apologize for this. This should be 34 right? So this is 34. 34%. All right, so this is 0.34 into 6. 20. This is 210.8. So if I just erase this, this thing is 210 pointed Now, since this is symmetrical, this is a normal distribution. So this will also be 210.8. This will be 83.7 and this will be 14.57 These will be the expected values. Now what now? In order to calculate the Chi Square statistic for all the categories we're going to apply the formula observed value minus the expected value. Whole square divided by the expected value. And in the end, I'm going to sum them Allah, and it will give me the overall chi square statistic for my question. All right, so let's just look at this inaction. These will be the individual chi square value, the individual chi square values for all the categories. We're going to send them all up in the end. So this is the difference between 14 point 57 and 14 and we square this and this. I'm getting us 0.32 and I divide this by the expected value, that is 14.57 So this is 0.2 0.2 Similarly, this is 86 minus 83.7. I square this and I divide this by 83.7, 83 0.7. This is 0.6 0.6 Then we have the difference between 210.8 on 207 I square this and I divide this by 210.8. So this is 0.685 Recognized this as zero point 0.7 All right, then we have difference between two and five and 210.8. We square this and divide this by 210.80 point 80.8 Okay, then The difference between these two is 20.7, and we square this divide this by 83.7. So this is 0.0 0.55 Texas. Right. This is six. And then the difference between 15 and 14.57 We square this and divided by 14.57 So this is 0.1 0101 Now we, some all of these up. So this is 0.22 plus 0.6 plus 0.7 plus 0.8 plus 0.6 plus 0.1 And this is 0.248 So I can say that my guys question to stick for this problem is 0.248 All right, now that I have Mike ice Question District, what else do I need in order to get my p values? I need the degrees of freedom DF I need d f. And how do I find d f? This is given by the formula Number of categories. Number off categories minus one. What is the number of categories that we have? We have six different categories, Right? 123456 So this is going to be six minus one six minus one. Or I can write this as five. So my degrees airfield I was five. Now I have my kaist question the steak. I have my degrees of freedom. So you can either use a chi square table and it will give you an approximate P value. Or what you can do is you can use an online software or statistical tool so that you get your exact P value. My guys question is thickest 0.248 This is 0.248 and my degrees of freedom is five, as we just saw are significant level is 0.1 and we can see that a P value is 0.9985 My P value is 0.9985 What was my Alfa? My Alfa was 0.1 hence I can say that as my P value is greater than Alfa I failed to reject. I fail to reject my null hypothesis. H not Okay now I think we also forgot to write the hypothesis initially. Okay, so let's just look at it briefly. What is going to be my null hypothesis for this question? What was my null hypothesis? My hypothesis waas that normal distribution normal distribution fits the distribution off days Just a moment for the distribution of the average daily January temperature fits the distribution fits the distribution off the average daily January temperature, January temperature and what would be my alternative hypothesis? My alternative hypothesis would be that the are not similar, right? That the normal distribution does not fit the average daily January temperature. So they are not similar. Let me just write it like this. They are not similar. All right? So I cannot reject my null hypothesis. So what is going to be my answer? This line is very important. I will say that I don't have I don't have enough statistical evidence. Enough statistical evidence to suggest I do not have enough statistical evidence to suggest that the normal distribution dozen doesn't fate The January temperatures. The distribution doesn't fit the distribution off average daily January temperatures, and this is how we go about doing this question.


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