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Based on 107 sludenn` eCic Was catimated by OLS:the first exam In = COuIYEcconomctrcu_ the following modelscore 2.178 469e[3core 3.369sudy 3.05+gpa (0.090) (0,456] (1.45710-107.R- 0.686In this model scare is the aniul SCOfe On 4n exini 4pcor thc expected score; Jtud is number of ttudy boWrS Pcr Wicck_ and &pa [9 thc atudcnt" gradc Point a1criec, Thc crandard cnor: Ier cuch Cengnet parenthes? . bclov tle estimate . Assume the regreesdou model crrors 4rc hrrelculc(4pous) Mikc j studetlt

Based on 107 sludenn` eCic Was catimated by OLS: the first exam In = COuIYE cconomctrcu_ the following model score 2.178 469e[3core 3.369sudy 3.05+gpa (0.090) (0,456] (1.4571 0-107.R- 0.686 In this model scare is the aniul SCOfe On 4n exini 4pcor thc expected score; Jtud is number of ttudy boWrS Pcr Wicck_ and &pa [9 thc atudcnt" gradc Point a1criec, Thc crandard cnor: Ier cuch Cengnet parenthes? . bclov tle estimate . Assume the regreesdou model crrors 4rc hrrelculc (4pous) Mikc j studetlt with an cxpected scote of 50. uho studics 10 hous > Pct Wcck: Kd_ EPe of 3 $ Stcve has art expected score of SU. irudicee Ioua Per Wcck_ 4nd ha: Epa = Construct = 9590 confiderce intcyal for thc diffcrence thcir ectual Etm eotes poinin Lstn? = an F-tcst; Icat thc null hypothcsis thet 8j against the ultemuatita that i i not aks Iiel pointz) Joitly Ical Ine null hypothesie that p, Ind & Mainet [ the ulteruttc that orke both of tose {E trictions - not true Atslizne the t-statistiee for festing cuch {catriction todit icunlly bate eottclation 0f 0.5.



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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.

Yeah. All right guys, the first thing we need to do here is we do identify on are null and alternative hypothesis are no hypothesis. Excuse me being that the variations are going to be the same. And the turn of prosthesis is that they're going to have different variations as you can see right here. Okay, so the next thing I did to solve the problem was I opened a google she and I am putting a lot of the data. You'll notice all the data is and put it right there. Okay. And then, so what I'm gonna do from there, what I'm gonna do from there is this I'm gonna go ahead and calculate the variances for each. The first variance I'm gonna calculate using google sheets. So I'm gonna type equals var open parentheses and then just highlight everything I want. I know have an extra box in there. That's not a big deal. And there is my first variant. Second variants will be found the exact same way it looks like it already knows what I want, which is great. And there is our second variance. Alright, so from there I need to find my f statistic. Okay, so my F statistic is just going to be the greater variance divided by the smaller variance and there we go, there's my f statistic. Once I found my F statistic, I'm ready to find my P value. Actually. Sorry, scratch that. Once I find my F statistic, I'm almost ready to find my P value. But there's one more step in order to find a P value. Using google sheets with an F statistic. First need to calculate degrees of freedom. Okay, very easy to do in google sheets. So degrees of freedom is just all the data, the amount of data you have minus one. So how do you find that? Google sheets is equals count. Just highlight everything. I have noticed how I do have an extra box there. That's not a big deal to subtract one. And that I have 20 in this case 24 degrees of freedom. And in our second case again equal count Open parentheses, highlight everything and I do need to subtract one. And the other case have 15 15 degrees of freedom. Okay, so now I'm ready to find my P. Value. So my P value uh through here is going to be something called an F. Distribution or F. Dist. And then open parentheses. I need to put three values. Number one, my F statistic. The second value is going to be the number of entries from S one squared. I'm sorry the degrees of freedom from s one square in this case is 24. And my last and final value is the degrees of freedom from my S two squared. So that's right here. I close all that up and there is my P value. However, I'm not quite finished yet. Okay, that is a P value for a one tailed distribution. What do I mean by that? That's a good P value. If that was a greater than or less than sign, this is a not equal to sign. Okay. That's not equal to sign. Which means um uh that which means that I have actually had the wrong P value right now. Since this is for a one tailed distribution and I need a two tailed distribution because they're not equal sign, I need to take this value multiplied by two. So equals I click on the value multiply symbol in google sheets. Asterisk hit A two and I'm done. That is my P value. Notice that P value is just a touch above 0.05, meaning I'm going to fail to reject the no hypothesis. I'm gonna fail to reject. And all hypothesis

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.

Following is the solution in number 14 at one way Innova test. Uh and this is about the mean sales prices for three cities. And the null hypothesis here is that the mean sale prices are the same for these certain houses. And then the alternative is that at least one of them is different. The second step is to find the critical value and you need three pieces of information to find the critical value. One is your alpha, Your significance level, that significance level. In this case, that's usually given to you is 10. They also needed the degrees of freedom for the numerator, which is the number of categories in this case, the number of cities minus one. So there were three cities that we looked at minus one is two, so degrees of freedom for the numerator is two degrees of freedom for the denominator is the total number of data values minus the number of categories. So in this case, if you counted up those data values, there were 31 -3 cities that we looked at. So 31 -3 is 28. So that's what we need. So from there you can use a table or you can use software. I'm gonna use software. So I wrote a program and I called it inverse. F. I'm not going to show you how to write this program. You can youtube it if you wish. But um it makes Makes it easier for me. So the area is the alpha value. So we'll put in .10 for that. Degrees of freedom from the numerator was too. And then degrees of freedom for the denominator was 28. And that's going to spit out my f. star my alpha value and my uh critical value which is about 2.503. Let's call it 2.503 is my F. stars 2.503. So anything greater than 2.503. We're going to reject h not anything less than 2.503. And we're gonna fail to reject the null. The next step is to find the f statistic and you can do that manually but especially with bigger data sets that can be really time consuming. So I went ahead and punch this into stat. Edit. And these are my data values. So these are the I think these are in thousands of dollars but these are the mean sale prices and if you go back to stat and then tests and the very last one in nova And you put in your columns just make sure you separate them. This is on the T. 84 by the way but make sure you separate them by commas otherwise it's not gonna read it right so nova for those three columns and that's gonna give us everything we need to. The f statistic is about 0.966 Let's go and write that down. So 0.9 66 which is somewhere over here. So that lands in the non rejection region. So that's actually gonna tell us why our fourth step which is the decision and we're going to fail to reject H not since the F statistic is less than the critical value. Now you can also use the P value method that's what this second piece of information is good for it. Now this other stuff doesn't really matter. Um You can just kind of ignore it because this is really what we need. We need the F statistic and we need the p value and the p values pretty large. It's about 0.39 And what you do is you explicitly compare the P value with your alpha value. So the p value in this case is greater than your alpha value. 0.39 is bigger than 0.10. And any time you're P values greater than alpha, you failed to reject H nine. If it's less than alpha then you you reject. And then the final step is to conclude this, you know, with actual words and bring it back to the question at hand. And so what we're going to say is that there is not enough evidence or there is not sufficient statistical evidence. So there's not sufficient evidence to suggest that the mean sales price prices of houses In the three cities are different. Okay that's the five step in Nova process.


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