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This question asks you to study the so-called Beveridge Curve from the perspective of cointegration analysis. The U.S. monthly data from December 2000 through February 2012 are in BEVERIDGE.RAW.
(i) Test for a unit root in urate using the usual Dickey-Fuller test (with a constant) and the augmented DF with two lags of curate. What do you conclude? Are the lags of curate in the augmented DF test statistically significant? Does it matter to the outcome of the unit root test?
(ii) Repeat part (i) but with the vacancy rate, vrate.
(iii) Assuming that urate and vrate are both I(1), the Beveridge curve,
$$u r a t e_{t}=\alpha+\beta vrate +u_{t}$$
only makes sense if urate and vrate are cointegrated (with cointegrating parameter $\beta<0 )$ . Test for cointegration using the Engle-Granger test with no lags. Are urate and vrate cointegrated at
the 10$\%$ significance level? What about at the 5$\%$ level?
(iv) Obtain the leads and lags estimator with $cvrate_{t}$, $cvrate_{t-1}$ and $cvrate_{t+1}$ as the I(O) explanatory variables added to the equation in part (ii). Obtain the Newey- West standard error for $\hat{\beta}$ using four lags $(\mathrm{so} g=4$ in the notation of Section 12.5$) .$ What is the resulting 95$\%$ confidence interval for $\beta$ How does it compare with the confidence interval that is not robust to serial correlation (or heteroskedasticity)?
(v) Redo the Engle-Granger test but with two lags in the augmented DF regression. What happens? What do you conclude about the robustness of the claim that urate and vrate are cointegrated?

First one. The test for a unit root in series you rate unemployment rate using the usual dickey fuller test with a constant yeah. And the augmented dickey fuller with two legs of change of unemployment rate. I find that seven both times we are unable to reject the now hypothesis that unemployment rate series is a unit fruit. The legs are not significant. However, the significance of the legs matters. So the outcome of the unit root test, we will repeat what we have done in part one two series vacancy rate and report the result in part two. I guess similar result. So the rate is a unit root. Well part one and two. I use package the R. Package A. T. S. A. And the function is a D. F. Dot test. R. Three. We assuming that unemployment rate and vacation re rate are both integrated of level one. We test for co integration using the angle grandeur test with no legs. So the step the steps are as follow. We first regress, you read on the rate then we yet the residual and we run the key fuller has on the residual to see whether the residuals our unit root. I find that you're right and we rate Arco integrated at the 5% level. Yeah Heart Forest. I get the leads and lacks estimator of the change in vacancy rate and I did note that uh CB rates up minus one. This is for the lack and plus one is for the lead. This is a regression result. So the usual centered errors are in green and in round brackets, the robots that Iran's are in blue and in square brackets you can see that the main estimate on vacancy rate is highly significant. This one is not correct. So the centered errol the usual one for the estimate of the first lack of change in vacancy rate is 164 In all cases except for the estimate of the lead of C. V. Right. The robust standard Iran's are larger than the usual standard errors. This is usually the case it happens but rare that the robot standard errors are smaller than the usual standard errors. The r square of this regression is 0.77 So for the rate, because the robot standard error is larger than the usual standard error. So we will get a wider confidence interval if we use a robot standard error and for confidence interval you will run this function in our count in and you impose the name of the regression. It was spits all the 95% confidence intervals for all explanatory variables. The default version is the 95% interval. But because the standard barrel of this estimate is are very close, two versions are very close to each other so the confidence intervals should be roughly equal. Yeah. Last part. What you could say about real business of the claim that you rate and the rate are co integrated. Yeah. When I run the test and good grandeur, the results are not consistent across alternative types of process. In one case I can reject the notion that the residuals are united and for all the cases I cannot reject. So I conclude that the claim that you rate and be rate our co integrated is not robust.

It's a kind of an interesting question if we we would be assuming that the anxiety level if we are going from easy to difficult is equal to the anxiety level. If the questions were difficult to easy and alternately, we just want to know if there's a difference and easy to difficult and not equal to difficult to easy. And so we're going to assume that that difference is actually zero so that the, I'll just write E G minus D. E. That that difference is zero. And when I put this into my calculator, um we have the lists have different sample sizes. We have that first group of the E. T. D. I have, there are 25 numbers versus the D two E. There are 16 numbers and we could use the degrees of freedom of 15, but I'm going to use the degrees of freedom from the formula. And that degrees of freedom ends up being uh 30 almost 39. So 38 point well, I'm just going to call it 39 it's approximately 39 degrees of freedom. And that test statistic that we're going to get, we need to take the mean, which was 27.1152 minus the mean of the other group, which was 31.7 to 8125 And then divided by the square root of. And the first standard deviation all around that a bit is 6.857 one square, divided by the sample size, which was 25. And then the second standard deviation was 4.26 square divided by the sample size of 16. And when I got that test statistic, the test statistic came out to be, I'll just read it up here negative 2.6 566 And so it came out down here. And since we're doing a two tailed test, we also use this one. And so what's the likelihood of getting a test statistic in this distribution that is less than or equal to that negative 2.6566 That gives us this tale and then we want the other tales to double it. And that p value comes out to be a 0.114 So at a 5% significant level, this is smaller than 5%. So we would have sufficient evidence to reject now. Yeah. And say that the mean anxieties are different so that they're different. The means are different. However, at a 1% significance level we would fail to reject the novel. Mhm. And we'd have to say here that they're actually not. They appeared to be the same. So it does depend on our significance level. How did pick you want to be. But it is an interesting idea in all my years of teaching, I never thought about the arrangement of having them all go easy, too difficult or difficult to easy. So what interesting concept.

Part one. This is the result of the simple regression equation. The estimate on education is 0.1 oh one with a standard barrel of 10.7 This variable is highly significant, and you can calculate the 95% confidence interval of the return to education as a range of from 8.7% to 11.5%. Recall that the 95% confidence interval is calculated as the estimate plus or minus the margin of error and the margin of error is the standard error times a critical value. We need to look at the alpha level of 5% because we need the 95% confidence. Yeah, you should have the alpha level because this is a two sided interval. The degree of freedom is the number of observations minus two. We need to degree of freedom or two pieces of information to estimate the slope. And there, uh, the to intercept the intercept and the slope coefficient. So this is what you get. Part two. The simple regression of education on C two. It gives the variable see to it is not significant. The T statistic, which is the ratio of the estimate over the standard error is minus 0.59 And so we cannot use see to it as an instrument for education. Part three. The multiple regression equation estimated by L s is as follows. The coefficient on education is 1.137 experience 0.11 to experience square point oh three. And we have other factors which are not our main focus in this problem. Mhm, even the estimate on education. You can see that the estimated return to education is now higher 13.7%. We convert the estimate on education 2% because our dependent variable, the left hand side variable is in log and education is in level Part four in the multiple regression above. In part two. Mm hmm. No. The one with other explanatory variables. In part three, the coefficient on C two. It is minus point 165 Yeah, and the T statistic is minus 2.8. So an increase of $1000? Yeah. In institution reduces years of education by about Hong 165 Remember that the Tunisian variables are measured in thousands. Part five. Now we estimate the multiple regression model by Ivy using See to it as an I V for education. The I V estimate of beta education is point 25 with a standard error of 250.1 to 2. The point estimate seems large, but the 95% confidence interval is very white from 1.1% to 48.9%. We could reject the value zero for beta education, but this confidence interval is too white to be useful. Lastly, part six very large standard error of the Ivy Estimate in Part five shows that the ivy analysis is not very useful. Yeah, and see to it is not convincing as an I V, as we can see also from part two mhm.

In this part, you are going to show some statistics of the variable net F. A thistle variable indicates the Net total financial assets in thousands of dollars. This variable has a mean of 19 up seven. The standard deviation is 63 Boyne Nice six. The minimum level is minus 502.3 and the maximum level is 1000 536 18 Remember, these numbers are in 1000. In the second part, you are going to test whether the average net total asset differs by 401 k eligibility studies and you can use a two sided alternative. The null hypothesis of this question is the average of net S A for their group that is eligible for for one cave equal that f A of the group that is not Elizabeth. For 41 K, we would do a T test and the T test. We will get a value of 13 11 The P value off the test is very small is way smaller than 0.1 So we are able to reject the non hypotheses that two groups have equal net total financial asset. The difference between the two groups in that total financial asset is well intact. The value of Met Toto A set the average one for the group that is eligible for 401 K, which is 30 points 54 These numbers are generated by their statistical software. Then we subtract from it their average net F A of the group that is not eligible for for a one K, which is 11 going 68 What we get is roughly 18 0.86 So the group that is eligible for for one K has a larger average net total financial asset, and it is larger by the average of the groups ineligible for 41 K by 18.8 $6000. We will estimate a mentally near regression model for net total financial assets. That includes income age eligibility for 401 K, and we also have age and income included in the regression as quadratic form. Okay, we will look at the coefficient of the variable E for a one K. As an indicator of the estimated dollar effect of the eligibility status, the coefficient of E 41 k is nine point 705 It means that if the family is illegible 4401 k, their net total financial asset when increased by $9705 we will add to the model estimated in Part three, the interaction terms of a legibility status and age minus 41. And another term is the interaction of eligibility status and age minus 41 square. This is the regression result. When we look at the interaction terms, we see that only this term it's significant. The term with Asian minus 41 the term with Asia minus 41 square, is not significant. In future models, we can safely drop the last term. Women compare the estimated effect of the allege ability status between model in Part three and model in Part four. The left panel show the regression results from the model. In part, we and the right panel show the results from the model. In Part four, you can tell the difference of the coefficient of E 401 K between two models. In monetary, it is 9.75 A model for it is 9.960 You should note that the meaning of the coefficient of E 401 k, the first between two models in model three. It is the effect for all ages. A model for it is the effect when age equals 40. 1 way will include family size dummies In the regression equation, you can see that in our equation we now have F s two s, three s four and fs five. These are their families side dummies F s two takes a value of one if the family has to people access to equal zero otherwise, after three takes a value of one. If the family has three people and takes a value of zero otherwise ever as far is for family of four. And Verse five is for family with more than five people. Mhm. So you don't cfs one in the equation. That is because Fs one or a family of one is the base group We have to base group So all the coefficient of the dummies should be interpreted as the difference in net total asset compared to the base group. So among the four dummies, you can see that Fs two is not significant. F s 32 Fs five are significant and the level of significance of these dummies increases with their size of the families. So have a sweet It's significant at the 5% level. But s four and s wide are significant at the 1% level. Even the P value you can say that compared to the base group, a family of a single person having three or more people in the family associate with a greater average value of net total financial assets. We can also conduct a F test to see the joint significance of their family size dummies. The null hypothesis is the coefficient of the family size dummy equals each other and equals zero. We will get an F statistic with two degrees of freedom four and NYT housing 265. The statistic is 5.44 and the P value it's very, very small is way smaller than 0.1 So we are able to reject the null hypothesis. In other words, the family size dummies are jointly significant. At the 1% level, we will do a child test forward coefficient equality across five family sized categories For the test statistic. We we need some of square residual from the unrestricted models and re restricted model. The restricted model comes from Part six and we use the phone sample to estimated The sum of square residual for the respective model is 30 million yeah, 213,000 and 115 for the and restricted models. They are five of them. They are estimated separately for each category of family size. The sum of square residual for and restricted models is the sum of the sum of square residual of all five models. Adding them up. You wouldn't get a total of 29 million, 980,000 and 959 now. Given that we have there some of square residual for the restricted model and unrestricted models, we can calculate the F statistic or the child test statistic the child has is based on the F distribution. That's why we have an F statistic. This is their formula for the F statistic. We will take the sum of square residual of the restricted model minus the sum of square of residual of the unrestricted model divided by their sum of square of residual of the unrestricted model altogether multiply with the ratio of the second degree of freedom divided by the first degree of freedom. These degrees of freedom are given from their problem. We have D F one. The first degree of freedom is the number of constraints were going to test, which is 20 and the second degrees of freedom is given as NYT housing and 245. The F statistic would be three point 57 and we can back up the P value using your statistical software. It is a very small number 0.0 03 So we are able to reject the null hypotheses at the 1% level, which means the coefficients are not equal across five family size categories.


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