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Use the second regression analvsis (relatingytoTLand 2)output in answering the questions (d) to (f) below:(d) Give the residual corresponding to the second observat...

Question

Use the second regression analvsis (relatingytoTLand 2)output in answering the questions (d) to (f) below:(d) Give the residual corresponding to the second observation (21 Tz = 23, and y = 18).residual(Round to two decimal places:)(e) Give a 90% upperprediction bound for the next stack loss when 62 and T2 =23.upper prediction bd(Round t0 Lwo decimalplaces)Fitting the complete second-order modely = Bo + 8121 + Bzwz + 83 (21)2 + Ba(e2)? + Bs (2122) + e gave SSR 801.33 and SSE 14.90. Find the value

Use the second regression analvsis (relatingytoTLand 2)output in answering the questions (d) to (f) below: (d) Give the residual corresponding to the second observation (21 Tz = 23, and y = 18). residual (Round to two decimal places:) (e) Give a 90% upperprediction bound for the next stack loss when 62 and T2 =23. upper prediction bd (Round t0 Lwo decimal places) Fitting the complete second-order model y = Bo + 8121 + Bzwz + 83 (21)2 + Ba(e2)? + Bs (2122) + e gave SSR 801.33 and SSE 14.90. Find the value of an F-statistic and degrees of freedom for testing whether all the second order predictors (i.e:. (21)2 , (22)2, and 21 82 can be dropped from the complete secon order model involving all 5 predictors What is your conclusion? observed F value (Round to two decimal places_ Conclusion (Choose one) (The answer to this question is a" or "b") (21)2, (22)2,and (21T2) should be dropped (b) at least one of (21)2, (a2)2 and (21x2) should not be dropped



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. 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.

In question # eight. We have a problem about linear regression and we're trying to figure out what's going to happen when a point is added to this scatter plot. Now the information that were given about the relationship that we're seeing in the scatter plot is that the slope of this least squares regression line would be .377 R squared, which is the coefficient of determination would be 95%, which is very strong. And S which represents the standard deviation of the residuals is .64. Now, when I take a look at this relationship and if I were to just sketch police squares regression line, I would want to imagine where the point that they're giving me would be on this line. The point that they give us is 35 14. And if you actually follow the slope of the data and take a look at the picture, that point would probably be pretty close to the line. So there's a couple things that happen when you add a point that's very close to the least squares regression line in a data set that's already existing. The fact that it's close to the line means that the slope shouldn't change very much because it's following the trend of the data. The rest of the data follows that same pattern. That slope will not change now for our squared R squared is supposed to tell us about the strength of the relationship. And if I take a look at that point, because it's continuing the trend of the relationship, I can confidently say that R squared would actually increase because this point being added so close to the least squares regression line is only going to make our squared more, going to get stronger, it's going to make it higher. So I would predict that this would go up now, s which represents the standard deviation of the residuals essentially. Typically the average distance away from this least squares regression line that all of these points are at this point, since I think it's going to be pretty close to the least squares regression line and that distance between that point at least great regression line, it's residual would be small. That would make me think that the standard deviation of the residuals would decrease a little bit since that distance would be small and therefore we lower that average distance. So, if you take a look at the answer choices, the only answer choice that shows the slope staying the same, Which is 0.377, the r squared value increasing which there's an answer choice that shows r squared becoming 97% and the standard deviation of the residuals s Decreasing to 0.617 is answer choice be so that's the correct answer.

This is Theis. Estimation Result using pulled l s. We have the year 1993 dummy execution on unemployment, we have 102 observations. Our square is point on 02 and adjusted. Our square is 20.7 Yeah. The coefficient on execution is positive, but not significant. There is no evidence of a deterrent effect. Hard to we difference away the observed state effects which can include historical factors that lead to higher murder rates. An aggressive used of capital punishment. We get a different story. The first difference estimates are here. As you can see, the number of observations we can use has dropped by half The R square and adjusted our square are roughly the same. Now we find a deterrent effect even that the change in execution is negative and significant. This coefficient can be integrated as follows. One more execution in the prior three years is estimated to decrease the murder rate by about 0.1 or about one murder per million People remember that murder rate is measured as murders per 100,000 people. The T statistic on the change in execution is about minus 2.4 and so the effect is statistically significant. In Part three, we will run test for hetero scad elasticity. The BP and the white tests are similar. You win used there fitted residuals of the regression and you regress that on its past value or the set of independent variables. Right? And you in calculate an F test to see whether the coefficients are jointly significant. Yeah, the BP test gives an F statistic of 0.6, and the white test gives an F statistic of 0.58 very close to each other. Both have P values above 0.5. Yeah, so we can conclude that there is no evidence of hetero ski elasticity in the first difference equation. Part four. We win. Calculate the hetero Scholastic City Robbers T statistic on the change in execution, and that value is minus 6.11. This is a huge increase in magnitude. Remember that the usual T statistic is minus 2.4. Yeah, so, in other words, the T statistic here has almost triple, So this result is kind of strange because from Part three we see that there is no evidence for hetero Scholastic city, and it is rare to find a hetero scholastic city robots standard error that is much smaller than the usual L s standard error. Part five. Given what we know so far, we should go with the usual l S T statistic. It is important to remember that the robot standard error our Ali justified in large sample. And here our simple is not very large. We have Ali 51 observations. So with this small sample, it is not easy to produce reliable, hetero, sick elasticity robots statistics.

This is the result for Part one. We use the full sample, which has 177 observations. From this estimation, we obtain the student ized residuals and we call them as t r. Supply. The number of student ized residuals, which are above 1.96 in absolute value, is nine. If the student ized residuals were independently drawn from a standard normal distribution, we would expect about by percent of our sample. Sir, 177 times 5%. You will get a number between eight and nine. It is 8.56 something we would expect between eight and nine cases of student ties residuals to be above two. It is so because in a standard normal distribution, about 95.5% of the observations are within two standard errors off standard deviation are within two standard deviation. And in a standard normal distribution, the standard deviation is one. So 95.5% of the observation are within two equivalently. It means 5% up to 5% of the observations are either above two or less than minus two. That's why we have the 5% number here and as just say, um, right here to be above two in absolute value, you can check. You can fact check this statement and you will find that there are eight observation with student ties. Residuals above two in absolute value for three. The student ties residuals are used to detect are liars. We will drop. There are liars, which are defined as observation with student ties. Residuals above 1.9 16 absolute value so we wouldn't drop there NYT, cases we find out in Part two, we will re estimate the model in part one again using 169 observations. This is the result. Compare with the regression. In Part one, we find that the main coefficient become significant at the 1% level. Let me come back to part one, so the first one lakh of sales is significant at the 1% level I windy note with three stars. Lakoff M. Kate Evil is significant at the 5% level, so two Stars CEO 10. Thank you significant at the 1% level, and CEO 10 square is significant at the 5% level. Back Thio, Part three, lock of sales is significant at 1% level. Still lock of em Katie Value before it was significant at the 5% level. Now it is significant. At the 1% level, nothing changed in terms of significance level for C E. 0. 10 and for CEO 10 square. Okay, nothing changed. It is still significant at the 5% level. Yeah, so we have beta head of lock of sales Mhm and CEO 10. They have the same level of significance. The exactly value is different, but not too substantial to give them a new level of significance. The estimates on em, Katie Value increase in magnitude and significance level. Okay, You may also notice that the magnitude of the estimates on sales and CEO 10 decrease, but not too much. And the coefficient of CEO 10 square does not change in terms of magnitude. Now we will use least absolute deviation to estimate the regression in part One again, this is part four. We re news all the data Here is the result. The l. A. D method is, um, estimated with with a different, um methodology. So it doesn't have the are square. It is estimated by maximum likelihood So to measure the fit of the model, you will need to look at their results generated by their statistical software. And you will look at the lock likelihood value. I would not report that here because we don't care about the fit of the model in this problem, we care about the estimates of their explanatory variables. Compare this regression with previous regression where we use l s. We see that beta one. The coefficient of lack of cells is closer to that of the restricted sample and the restricted. Simple is the regression. In part three, where we drop the outlier observations. We don't have the same observation for beta three beta three hat. The coefficient of CO 10 is actually closer to the estimate in the full sample. Even these results part five, we will be able to evaluate this statement dropping our lawyers based on extreme values of student ties. Residuals makes the resulting L s estimates closer to the L A G estimates on the phone sample. This statement is not always true. It is not true to every estimate


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