Econometric Methods Essay

Part A. quadruple choice interrogative sentencesAnswer each(prenominal) question by circling match slight and only one answer. Each question is worth 3 marks (total 30 marks).1. When estimating a linear opportunity work utilise OLSa. The estimators be molded beca character breaks atomic number 18 necessarily heteroskedasticb. The slope coefficient estimates tail assemblynot billhook changes in the presageed prospect of Y=1c. The estimators underside be asymptotically normally distributed d. All of the to a high place2. When internal validity is forayda. OLS coefficients no long-dated measure the partial correlational statistics between the informative variable star and the capable variableb. The population break depots cannot be normally distributed c. The dependent variable necessarily becomes skewedd. None of the preceding(prenominal)3. Which of the sideline dependent variables is least like a limit dependent variable? a. Wagesb. Net assets of a househ s r. (total assets minus debts)c. Number of take cargons to the dentist in a yeard. An index of cheer where happiness is rated 1 to 104. A variable Y is a Bernoulli variablea. Its distribution has the usual 2 independent parameters representing the mean and the varianceb. Its judge tax jibes the dimension of the fortune of Y=0 to the hazard of Y=1c. Its variance equals the product of the prob skill of Y=0 and the prospect of Y=1d. All of the above5. In the probit amaze catch outn in classa. The variance of the error precondition depends on the vector of instructive variables b. The variance of the error term is assumed to be 1c. The variance of the error term does not need to be stipulate because of the normality assumptiond. The variance of the error term can be estimated from the variance of the estimated residual6. In panel data, the problem of attrition refers toa. The presence of great measurement error in key variablesb. The correlation of measurement errors with e xplanatory variablesc. The misclassification of key dummy explanatory variables collectable to measurement error d. None of the above7. In the probit modela. The partial effect of a single continuous explanatory variable X on the predicted chance has the uniform sign as the estimated coefficient on Xb. The test statistic constructed by the ratio of the estimated coefficient to its meter error is normally distributed because we ar employ the normal distribution to model the expected measure out of the dependent variablec. The partial individualal effects of an explanatory variable are quantitatively close to energy when the standard error of the coefficient on this variable is real large.d. All of the above8. You fuck off data on a experiment of 95 managers working in large firms in Australia. You estimate a logit model of Y= 1 if earning $500,000 per annum using as explanatory variables F=1 if the manager is female (0 otherwisewise) PHD=1 if the manager has a PHD (0 oth erwise) an interaction variable FPHD=F*PHD TEN= kick upstairs with the firm measured in years (a continuous variable). You find the fall outing estimatesIndexi = 0.053 0.095 Fi + 0.020 PHDi + 0.007 FPHDi + 0.0015 TENi (0.002) (0.011) (0.009) (0.003) (0.0005)where the standard errors are denoted in parenthesis. You want to test H0 land tenure has no effect on the probability of earning $500,000 per annum versus H1 tenure has a positive effect on the probability of earning $500,000 per annum. You pass on use a 5% take aim of significance to conduct this test. You get an asymptotic t-stat equal to 3.0. Using the tables provided at the end of the exam, choose one of the following as an appropriate life-sustaining esteem to conduct this testa. 1.662b. 1.645c. 1.987d. 1.960e. 5.02399. Refer to the model and estimates in the anterior question. Ceteris paribus, according to these estimates (and ignoring statistical significance)a. Women without PHDs buzz off a higher probability of earning $500,000 than men without PHDs.b. Men with PHDs have a lower probability of earning $500,000 than men without PHDs. c. Women with PHDs have higher probability of earning $500,000 than women without PHDs. d. Women with PHDs have higher probability of earning $500,000 than men with PHDs.10. Refer to the model and estimates in the previous question. You want to test that ceteris paribus, men and women have the very(prenominal) probability of earning $500,000. Under the null, the Wald test statistic is asymptotically chi-squared distributed with a. 1 degree of emancipationb. 90 degrees of libertyc. 93 degrees of freedomd. 2 degrees of freedome. 3 degrees of freedomPART A. Multiple Choice1. C2. D3. B4. C5. B6. D7. A8. B9. C10. DSOLUTIONSPart B. business (Total 30 marks)Equity of admittance is a primary goal of many health systems. Determining whether Australias system (Medicare) meets this goal is an important research question. check the case of main course to general p ractitioners (GPs). The probit results presented below in Table 4 are part of an outline aimed at answering whether there is equitable access to GP services where access is defined on the basis of health needs rather than ability to pay. The data consists of a savor of 3207 single females who were surveyed end-to-end Australia in 1995. The dependent variable for the study was VISIT, an indication variable that was equal to one if the women had chitchated a GP in the last deuce weeks and postcode otherwise. The sample has been divided into cardinal subsets depending on whether the women are less(prenominal) than 40 years mature (the unripened sub-sample) or whether they are greater than 40 years gray-headed (the one- clip(a) subsample). Table 4 presents estimation results (variable definitions follow the table).YoungOldTable 4 Probitestimates for call downto GP* VariableIntercept-0.7910 (0.1602)-1.1570 (0.2495)AGE-0.0060 (0.0064)0.0055 (0.0033) health0.3930 (0.0687)0.6131 (0.0746)KIDS0.1651 (0.0881)-0.1479 (0.1159)INCOME0.0003 (0.0032)-0.0052 (0.0037)TERTDUM0.0120 (0.1042)0.0844 (0.1509)TRADEDUM0.1842 (0.0884)0.2399 (0.1013)DIPDUM0.0077 (0.1281)0.0478 (0.1422)PHI0.0258 (0.0783)0.1781 (0.0768)Observations17171490Log-likelihood-935.52-892.24PART B.i. (8 marks) establish the effects of PHI on the probability of confabing a GP and equalize these effects for the two subsamples of offspring and old women. Repeat the representative for the KIDS variable. Do you think that these variables are apt(predicate) to violate the zero conditional mean assumption? DiscussPHIIn both subsamples, the estimated coefficient on PHI is positive ceteris paribus the probability of visiting a GP is higher for those with PHI than without. The sizing of coefficients whitethorn be discussed using the rule of leaf but these must not be crushed with partial effects. The effect is statistically significant among the old while the opposite is true for the novel. In the ups tart subsample, the coefficient is insignificant at any conventional train (t statistic for testing irrelevance of PHI against the 2-sided alternative is 0.3295 1.645) whereas in the old subsample it is significantly different from zero at the 5% significance level (t statistic = 2.319 1.96).The sign is as expected since PHI makes it cheaper to use GP services and women who expect to visit GPs to a greater extent often are more credibly to purchase PHI. The latter implies that ZCM whitethorn be violated due to a survival effect.KIDSIn the young subsample, the coefficient on KIDS is positive and statistically different from zero at the 10% level (t statistic = 1.874 1.645) the probability of GP visit is higher for those with dependent chelaren.In the old subsample, the sign of the coefficient indicates that the effect is negative but the coefficient is statistically insignificant from zero at conventional levels (t stat = 1.276 1.645). A priori, the expected sign is ambiguous women may visit GPs for childrens medical care as headspring as their own (positive) but at the same time they may become busier due to child rearing (negative). For the old sample, KIDS may be previous(a) and hence mothers no longer visit GPs for the childrens health. Other reasonable explanations are acceptable. You can grapple both ways on the ZCM assumption for example, you can argue that fertility decisions are exogenous to GP visits. You could also argue that there is an omitted variable stroke (KIDS is picking up some unascertained serving e.g. better health measurement than what is being captured by the existing explanatory variables). Also if the true implicit in(p) relationship depends on the number of resident dependent children, KIDS is top-coded at 1, causing the ZCM assumption to fail due to a measurement error correlated with this variable. supererogatory satisfyingYou could also earn marks (lost elsewhere in the question) by discussing the coat of the eff ects. For example, the effect among the young seems non-trivial in the sense that the coefficients magnitude is approximately over 40% of that of the coefficient on the poor health indicator (HEALTH) while for the old, the variable seems far less economically relevant relative to HEALTH. ii. (5 marks) If there is candour of access so variables related to income, education and unavowed health insurance should not move visits to GPs. When the models are re-estimated without these variables (i.e. with only AGE, HEALTH and KIDS included) the log-likelihood values are 937.92 for the young sub-sample and 898.63 for the old. Using these results evaluate the null hypothesis of impartiality of access.Statement of the hypothesesCalculated statisticsLR test statisticsLLRYOUNG = 2(-935.52+937.92) = 4.8LLROLD = 2(-892.24+898.63) = 12.78.Distributions of the test statistics and particular valuesThey are asymptotically chi-squared distributed with 5 degrees of freedom under the null.The ap propriate 10% and 5% critical values are 9.2364 and 11.0705 respectively. Decision rules and conclusionsSince LLRYOUNG 9.2364, we fail to lower the null at 10% level in the young subsample there is not enough state to leave off that income, education and PHI variables affect young womens GP visits.Since LLROLD 11.0705, we reject the null at the 5% significance level in the old subsample and conclude that there is some evidence against equity of access among the old women. iii. (4 marks) Consider two types of women type 1 where AGE = 20, HEALTH = 1, INCOME = 20 and all other variables = 0 type 2 is identical moreover that AGE = 60. Write down the equation(s) you would use to compare the probability of visiting a GP for these two types of women. Using the probit results can you determine which of these two types of women are more likely to have visited a GP in the last two weeks? If your answer is yes then make the comparison, if your answer is no then explicate what informatio n you would need to make the comparison.One workable answer is to use the index and argue that the rank by the probabilities will be the same as that provided by the indexIndex for type 1 = -.791 + -.006*20 +.3930 + 0.0003*20 = -.5120 -.51 Index for type 2 = -1.1570 + .0055*60 +.6131 0.0052*20 = -.3179 -0.32 Since the standard normal CDF increases in the probit index, type 2 woman is more likely to visit GPs than type 1 woman.Another attainable answer is to write down the normal CDF for the two types and argue that the equation for type 2 will be greater than type1.Additional materialYou could also earn marks (lost elsewhere in the question) by calculating the dispute in the probabilities using the table on p.10 of the exam paper i.e. the difference in the predicted probabilities can be evaluated as (.5-.1255)-(0.5-.1950) = 0.0695 .07 higher for type 2. iv. (6 marks) In determining the sample to be used for estimation, any individual who did not story their income or reported zero income was deleted from the analysis. Do you see any real or potential problems with this imitate decision? Can you provide an alternative regularity to deal with this problem?Likely problems (one of the following or other sensible problem) -The potential selection bias which arises when the decision to report zero income or spurn reporting any is correlated with the decision to use GP services. For instant, top incomegroups may be more jealous of their income information and at the same time more likely to be health advised and visit GPs in consequence excluding the said individuals would affect all coefficient estimates as the model would have to predict a lower probability of GP visit on average.-The decrease in the sample size and the resulting increase in standard errors. The incomplete cases may still provide useful information on the effects of other variables on GP visits and the police detective has discarded this information.Alternative solutions (one of the fol lowing or another sensible solution) Use other information to pass judgment the abstracted information Use dummy variables for get bying income. More sophisticated imputation methods Estimate a selection model (this is covered in more level later in the class but you may know about it from reading or elsewhere)v. (7 marks) excuse how you would construct and use a hit and miss table to compare the performance of the models for the two subsamples of women (young and old). (You do not have to actually construct a table.)Step 1. Calculate a predicted probability for each person in the relevant subsample. Step 2. fetch a predicted binary outcome for each person using a classification rule if person is predicted probability exceeds c, the predicted outcome is 1 and otherwise 0. It is ok if you use 0.5 or the sample mean.Step 3. For each subsample, tabulate frequencies of predicted and actual binary outcomes in the following formPredicted01Observed 0 A B1 B AwhereA (A) = the total number of women whose predicted and discovered outcomes are 0 (1) B (B) = the total number of women whose observed outcome is 0 (1) but the predicted outcome is 1 (0).Step 4. Now, compare the relative frequencies of correct predictions for each subsamples i.e., compare (A+A) / (A+B+A+B) across the subsamples. This tells us how well one model performs relative to another in terms of predicting the observed outcomes. It is ok to describe the comparisons of the predicted 0s severally from the predicted 1s (ie the comparisons of A / (A+B) and A / (A+B) across subsamples) but this is not needed for in full marks.