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Use the user-defined function SysZODEsRK4 (Program 10-6 in Example 10-8) to compute the distance J(t) , the velocity j(t) from to [ 8s ,using step size O.Is and plo...

Question

Use the user-defined function SysZODEsRK4 (Program 10-6 in Example 10-8) to compute the distance J(t) , the velocity j(t) from to [ 8s ,using step size O.Is and plot "(t ) and J()on lwo separale ligures (only for this one sep Size ) What is J(3)How does it compare lo "(3) given in Part (2) for the step size 0.ls discussed in the context of relative percentage error: Ne code submission needed for this part:This should beUsing the resuks_fro Put (Labove: perform leasLsquare regression f

Use the user-defined function SysZODEsRK4 (Program 10-6 in Example 10-8) to compute the distance J(t) , the velocity j(t) from to [ 8s ,using step size O.Is and plot "(t ) and J()on lwo separale ligures (only for this one sep Size ) What is J(3) How does it compare lo "(3) given in Part (2) for the step size 0.ls discussed in the context of relative percentage error: Ne code submission needed for this part: This should be Using the resuks_fro Put (Labove: perform leasLsquare regression for the discrete duta (results ) thatyou have obtained for the distunce { ,J6;) usIng quadrutic polynomial. f(t). Provide details on the procedure and the final fon of the quadratic polynomial (i.e after finding all coeflicients). Plot this polynomial ugainst the original data. Wilis f()2 How does compare t0 J())? Using_the results_from Park(3Labovc: for the diserete data for the Kelocity {o,jo) _ integrate from ( = 040 '=35 using composite Simpson $ MArule Then compure the result to "(3) given in Purt (3) This should be diseussed in the context of a relative percentage error;



Answers

Use the data set GPAl for this exercise.
(i) Use OLS to estimate a model relating $\operatorname{col} G P A, A C T,$ skipped, and $P C$ . Obtain the OLS residuals.
(ii) Compute the special case of the White test for heteroskedasticity. In the regression of $\hat{u}_{i}^{2}$ on $\widehat{\operatorname{col} G P A}_{i}, \overline{c o l G P A_{i}^{2}},$ obtain the fitted values, say $\hat{h}_{i}$
(iii) Verify that the fitted values from part (ii) are all strictly positive. Then, obtain the weighted
least squares estimates using weights 1$/ \hat{h}_{1}$ . Compare the weighted least squares estimates for the effect of skipping lectures and the effect of PC ownership with the corresponding OLS estimates. What about their statical significance?
(iv) In the WLS estimation from part (iii), obtain heteroskedasticity-robust standard errors. In othe
words, allow for the fact that the variance function estimated in part (ii) might be misspecified.
(See Question $8.4 . )$ Do the standard errors change much from part (iii)?

Part one. The test with strict exhaust Janet E gifts is row hat equal to minus 0.97 and the T value of minus 2.41 The regression that includes growth of minimum wage and growth of C P. I gives row hat equal to minus point 098 and it is statistic of minus 2.42 Roughly the same T values and roll had value. Therefore, we find evidence of some negative serial correlation and it does not matter which form of the tests we use. Yeah, this is the regression equation for part two and three with three types of standard. Erin's Yeah, note that the estimates are not changing. Only the standard errors change. The first line of standard error is the old LS the usual and probably incorrect type. The second line is the new E West standard payroll type, and the last line is hetero skate elasticity robots standard errors we have over 600 observations and the are square are the same for three types of standard herons, which is 30.293 So comparing the new e West standard error and the usual L s standard error for the variable growth of minimum wage. We find that the new E West than it Errol is much larger than the old l s one roughly four or five times larger. But for the variable growth of C P I, the new E West standard error is actually smaller than the old l s one that is the answer for part two. And for part three, we consider the hetero Scholastics city robots standard error with the newly West standard error, we don't find much difference. Yeah, So the difference between the two type of standard error is that, um, the last one only controls for different variants of the Errol terms. That's why it's called hetero Scholastic City. Robust, but new E was standard. Errol does more than that. The new E West Standard Erin's our robots to both hetero Scholastic City and serial Correlation, as we find little difference between the two, is probably because the negative serial correlation adjusting the standard error on, um, CP, I actually reduces it. Hetero Scholastic city does not have a major effect on the growth of CP I standard error. That is part three and part four. We run a BP test, We get F statistic equal to to 33.8. Yeah, which means the P value is almost zero. There is a very strong evidence of hetero ski elasticity. Part five, The usual F test is 4.53 with a P value of pointing 058 So, in the static model using the hetero Scholastic City Roberts T Statistic lead Teoh a less significant minimum wage effect. Okay, Actually, this one is there. This is for the usual F test and this is through the hetero stick elasticity. Robust test. All right, so the hetero see elasticity robots test for the legs show a very strong significance of the wage effect. Part six, the new we West version of theme F statistic is about 7.79 which show even mawr Statistic. Significance, then just the hetero Scholastic City. Robust statistic. So at just in the F start for hetero scholastic city or hetero ski elasticity and 12 order serial correlation leads to the conclusion that the LAX are very statistically significant. March 7 with 12 legs, the estimated long run propensity is about 0.198 and without the legs. The estimated L R P is just the coefficient on the growth of minimum wage. Yeah, which is 0.151 So when we include the LAX, L R P is about 30% larger using the new We West standard error the 95% confidence interval for the l r P. ISS from yeah, hauling 111 to you Point Thio 84 which easily contains the estimate from the aesthetic model.

Hello, everyone, Uh, this is C seven off computer excites and check the tree. So the person we wanted to use to estimate the model that max 10 TV cool to beta zero plus beta one. A lot of expenditure plus beta to lunch program. Plus you. Here. Matt Kenney is the percentage of students passing the media Matt can like Spanish is a lot of expenditure and explained that trees in it's for per student in terms of dollars and lunch program is the percentage of students in the school lunch program. So first, we estimate the model that last one is equal to constant plus bait along like expanded. A possibility, too. This is the code progress Matt Salmon Longest finishing lunch program. So the costume tomatoes you had is minus 20.36 based on one hat, which is the car coefficient on. Well, I expect she's 6.23 and made a two hand, which is the coefficient on lunch program is negative point Trio five and our squad from dis regression ease points 18 are the signs of stop coefficients what they expected. So let's see, um, Beta one had, which is a coefficient of long extended. She's positive, which means that spending more per student will increase the percentage of students passing the test. This is something would expect. Beta two hat has a negative, Uh, is the negative side. So this is the coefficient of lunch program betweens. As a percentage of students in school lunch program increases the percentage of students passing that test decreases. That's, um Well, that could be because, uh it could mean that in some schools where the percentage of students intervention program, it might be that they're spending that there could be a lot of factors, but this is a little bit weird. Maybe unexpected. Uh, OK. Second question is, what do you make with the intercept you estimated in part one in particular. Does it make sense to set it to explain it? Three variables to zero. So because your hat in the first part we find is minus 20.36 which means one large expenditure at lunch program is he called zero. Then the percentage of students passing that test is gonna be negative. 20 which, of course, doesn't make sense, because it should be in between 0 100 but ah, So let's see if that if it makes any sense explanatory variables to equal to zero. And when we actually look at the date that we see that the minimum that the large expense you get this it points something. So it doesn't make sense to set the vehicle to zero. And, uh, also in the daytime, a minimum. Uh, Thea Lunch expenditure lunch program percentages 1.4. So he could make sense to set the lunch from that vehicle to zero. That's like nobody is in the school lunch program, but from later we see that doesn't make sense to set of a stag nature to be equal to zero. Okay, okay. The question is, now run the SIM progression of maths final, just large expenditure, and we'll make lunch program and we see that now the constant is minus 69 Betas. You Matilda point beautiful on basil until days 11 point of 16. And within a compared the slope coefficient to estimate we find in the part one and we are asked to, uh is the estimated spending effecting a larger or smaller than the part one. So in part one, the crawfish most 6.23 now in part to impart tree. It is 11.16. So the effect of expenditures on students passing the math test it's bigger. 11 is bigger than 6.23. And, uh, okay, and part four is asking us to find the correlation between large expenditure and lunch program. And instead I just fruits coronation might be spending much program. It gives me coalition here, give me the coalition table. Some people expenditure and lunch programs is gonna be minus 0.19 is the coronation and the park five is asking this to use this correlation to explain the difference between these two qualifications from two different Rick rations. So in ah, regression in turn part, we have only one extra natural available large expenditure, and in part one, we have to log expended on lunch program. So when we don't include lunch program in this regression, this knocked expenditure is also gonna capture the effect of one from him on that test. There's gonna be only too available by us to lunch. For them has a negative effect. All mad at 10 and two expended Chandler for the negative correlation. Overall, we're gonna have negative negative. We're gonna have a positive by us. So which means that log, the effect of flogged expenditure on math 10 is overestimated compared to the 1st 1 So in the first part we have 6.20 to know you have 11 point of 16 which is higher, just overestimated compared to the first part. And that is because lunch program even negatively affecting that and and not explain the children's program has a negative correlation. So overall, these two negatives makes it positive, and we have a positive bias. All right. I hope that help. Thank you for watching.

Part one. The estimated equation is we have math for as the dependent variable. This is the number of students that passed fourth grade math test. This variable is a function of the lock of expenditure per student, lock of school in Roman and lunch, which denotes the percent of students eligible for free or reduced price lunch. Lastly, we have the intercept or the constant term. I went right there, estimated coefficients in blue, and it's their standard, Errol's in Green. So we have 3.52 minus 5.40 yeah, minus 0.449 and 91.93. The standard errors are as follows. 2.1 0.94 0.15 and 19.96 This regression has an R square of 0.372 The number of observations is 1000 206 196 92. So you may see from your statistical package when it runs this regression. The two variables lock of enrollment and lunch. They are individually significant at the 5% level. Regardless of whether we use a one sided or two sided tests, they are significant they have a very small P value, but the other explanatory variable lock of expenditure per student. It has a T statistic of 1.68 It is not significant against a two sided alternative, so the two sided alternative here is beta equals zero. However, it's one sided. P value is about 10.47 so it is statistically significant at the 5% level against the positive one sided alternative that is Part one and we will move on to Part two. The range of fitted values is from 42.41 to 92.67 which is much narrower than the range of actual math pass rate in the sample, which is from 0 to 100. In Part three, we will do a residual examination. The largest residual is about 51.42 and it's belonged to building code 11 for one. This residual is the difference between the actual pass rate and our best prediction of the past rate. If we think that per student spending, enrollment and the poverty rate are sufficient controls, the residual can be interpreted as value added for the school that ISS for school. 1141 It's past rate is over 51 points higher than we would expect based on its spending size and student poverty. Heart. Four. We would work with quadratic six and we will test a null hypotheses were the beta of the quadratic quadratic variable, our own zeros. You know, the joint F statistic with two degrees of freedom three and 1685 is 0.52 and it's P value is 0.67 Okay, so we are unable to reject the novel hypotheses. It means that the quadratic six are jointly insignificant, and so we can drop Drop them from the model Let's move on to part five. So in this part we regressed the lock of wage on points and we've been regressed. We regret the same model as in Part one. We will divide the dependent variable and each exponentially variable by its sample standard deviation. So this is a technique usually used when your explanatory variables are not are not comparable, so you can rerun their regression in part one, and you will include an intercept unless you also first subtract the mean from each variable in terms of standard deviation units. Come on. You may find that lunch has by far the largest effect. Okay, Right. So beta had the coefficient of locks. Spending lock expenditure per student is 0.35 Estimated coefficient of lock of school enrollment is minus 0.115 Yeah, and the coefficient of lunch is minus 0.613 So in terms off standard deviation units, lunch has by far the largest effect. The spending variable has the smallest effect, this one, and it is smallest by absolute terms.

This is the regression result for Part one. We have the dependent. Variable is the number of the percent of student who passed the fourth grade math test, and we have the lock of expenditure per student, lock of enrollment and whether the student can receive free or reduced lunch. Reduced price lunch. So I have the estimates in blue and the standard arrows in green and in bracket. The R square of the regression is 0.373 and we have almost 1700 observations in the regression. As you may see from your statistical package, we have thes two variable lock of enrollment and lunch are individually significant at the 5% level. Okay, but for lock of expenditure per student, these variable has a T statistic of 1.68 And so it is not significant against a two sided alternative that is part one. And for part two, we will get a fitted value of math. Four. From the regression, we see that the range of fitted value runs from 42.41 to 92 point 67. This is this range is much narrower than the range of actual math pass rates in the sample, which is from 0 to 100. In Part three, we will do a residual examination. We find that the largest residual is about 51.42 but this residual belongs to you, building code 11 41. But the residual is the difference between the actual past rate and our best prediction of the past rate. If we think that per student spending enrollment and the lunch indicator as sufficient controls, then the residual can be interpreted as a value added for the school. So for school 11 41 it's past rate is over 51 points higher than we would expect based on its spending size and student poverty. Part four. We will do another regression. We went at the quadratic six. Let's see the quarter tricks off all explanatory variables, and we will test for joint significance of these quadratic six. So we will do an F test and are not. Hypothesis is all quadratic. Six, um are not meaningful. We will get an F statistic with two degrees of freedom. The first one is three, and the second one is 1600 85 the F statistic is about 0.52 and the P value is 0.67 So we are unable to reject the null hypothesis, which means the Quad atrix are jointly very insignificant. And so we should drop them from the model for the last part. You win, do another regression. What? You in return to the model in part one and divide the dependent variables and each explanatory variables by ease by its central division. Then you re run the regression. This is a very common technique when you have, um, incomparable variables. Okay, so we don't care about the estimation of the intercept. We only care about the slopes we have there. Oh, beta coefficients. For lot of expenditure per student of 0.53 four. Lock of enrollment as minus 0.115 and four lunch. The beta coefficient is minus 0.613 So instead, activation units lunch has by far the largest effect. The spending variable has the smallest effect


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