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Consider the following regression modelYi = β0 + β1X1i + β2X2i + β3X3i + β4X4i...

Question

Consider the following regression modelYi = β0 + β1X1i + β2X2i + β3X3i + β4X4i + β5X5i + uiThis model has been estimated by OLS. The Gretl output isbelow.Model 1: OLS, using observations 1-56coefficientstd. errort-ratiop-valueconst-0.38590.5198-0.74240.4613X10.30650.20781.47540.1464X20.79820.15895.02410.0000X30.10950.55030.19900.8431X42.71510.219212.39000.0000X50.97900.35952.72340.0089Mean dependent var1.9809S.D. dependent var2.4792Sum squared resid68.239S.E. of

Consider the following regression model Yi = β0 + β1X1i + β2X2i + β3X3i + β4X4i + β5X5i + ui This model has been estimated by OLS. The Gretl output is below. Model 1: OLS, using observations 1-56 coefficient std. error t-ratio p-value const -0.3859 0.5198 -0.7424 0.4613 X1 0.3065 0.2078 1.4754 0.1464 X2 0.7982 0.1589 5.0241 0.0000 X3 0.1095 0.5503 0.1990 0.8431 X4 2.7151 0.2192 12.3900 0.0000 X5 0.9790 0.3595 2.7234 0.0089 Mean dependent var 1.9809 S.D. dependent var 2.4792 Sum squared resid 68.239 S.E. of regression 1.1682 R-squared 0.79814 Adjusted R-squared 0.77795 F(5, 50) 39.539 P-value(F) 0 Log-likelihood -84.995 Akaike criterion 181.99 Schwarz criterion 194.14 Hannan-Quinn 186.7 (5 marks) Construct a 99% confidence interval for β4β4. (5 marks) Using the critical value method and a significance level of 5%, test H0:β2=1H0:β2=1 against H1:β2<1H1:β2<1.



Answers

Use the data in BARIUM for this exercise.
(i) Estimate the linear trend model chnimp_{t} $=\alpha+\beta t+u_{r}$ using the first 119 observations (this excludes the last 12 months of observations for 1988 ). What is the standard error of the regression?
(ii) Now, estimate an $\mathrm{AR}(1)$ model for chnimp, again using all data but the last 12 $\mathrm{months.}$ . Compare the standard error of the regression with that from part (i). Which model provides a better in-sample fit?
(iii) Use the models from parts (i) and (ii) to compute the one-step-ahead forecasr for the 12 months in 1988 . You should obtain 12 forecast errors for each method.) Compute and compare the RMSEs and the MAEs for the two methods. Which forecasting method works better out-of-sample for one-step-ahead forecasts?
(iv) Add monthly dummy variables to the regression from part (i). Are these jointly significant?
(Do not worry about the slight serial correlation in the errors from this regression when doing
the joint test.)

Part one. The regression of U. T. Had on U T minus one had and the change in unemployment rate gives a coefficient on U T minus one hat of 0.73 with a T statistic of 0.42 Okay, therefore, there is little evidence, a first order serial correlation. Like part two. We have a simple regression. We regret UT head square on the change in unemployment rate and the regression gives a slow coefficient. The beta head of the change in unemployment rate yeah of 2.452 and the teeth statistic of 2.7. So at the 5% significance level, we find the evidence for hetero Scott elasticity and because the beta head of the change in unemployment is positive. So the variance of the error appears to be larger when the change in unemployment is larger. Part three, the hetero scholastic city robbers standard error is 30.2 to 3 and the usual old L s standard error is 0.182 So the hetero scholastic city robots standard error is more than 20% Okay, larger then the usual l as one. And of course, a larger standard. Errol leads to you a wider confidence interval for beta one

Part one. This is the L S estimate We've been used. The regression results in part one for a white test. The white test is for hetero scad elasticity. For the white test, we would run another regression. The dependent variable in this regression is the square of the residual from the previous regression. Let's say it's you square and we will regress You square on the fitted value and square a fitted values from the previous equation. So we have college G p A at and College GP a head square. This is a regression with an intercept and the white test has the non hypotheses of no hetero scholastic city in mathematic form. This now hypotheses imposes doubt. Taiwan equals Delta two equals zero and data one and data to are the coefficients. I was fitted value terms. In this equation, we have to restrictions, so we will calculate the F statistic. The related degrees of freedom are two and 158 respectively. You may find that the F statistic is 3.58 and it's P value is 0.31 Given this p value Bush, we are able to reject the null hypotheses at the 5% level. Okay? Meaning there is significance evidence. What happened of hetero Scholastic city in the errors of the college G p A equation. Yeah. Okay. To fix this problem, we will run weighted least square. First, we need to examine the fitted values from the regression. In the first part, the minimum value of the fitted value is polling 027 and the maximum fitted values. The maximum fitted value is 0.165 So we can confirm that fitted values are within zero and one. Yeah, we are able to use thes fitted values as the weights in a weighted least square regression. And this is the result for after we run weighted least square regarding PC variable, there is a very small difference in its estimated coefficient. The LST statistic and W L s T statistic are very close. The R square in this regression is somehow larger than that from the regression in part one. But the old ls result and WLS result are not quite comparable. In part four. We win estimate the equation again with all l s, but with Robert standard error. Okay, you should get exactly the same estimate s in part three now. It should be the same result as in part three. Uh, not part one. So if you get a different any different estimated coefficient, you must do something wrong. The only difference is is the standard error. You wouldn't get a larger standard, Errol for almost all estimated coefficients.

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.


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