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Explain what the R-squared in this regression means: Choose the correct answer below:0 A. 88.99 of the variation in the job growth in 2010 is accounted for by the j...

Question

Explain what the R-squared in this regression means: Choose the correct answer below:0 A. 88.99 of the variation in the job growth in 2010 is accounted for by the job growth in 2012. 0 B Overall, the job growth in 2012 is 88.9% of the job growth in 2010. 88.9% of the variation in the job growth in 2012 is accounted for by the job growth in 2010. 0D. Overall, the job growth in 2010 is 88.9% of the job growth in 2012.d) Do these results indicate that; in general, companies with a higher job growth

Explain what the R-squared in this regression means: Choose the correct answer below: 0 A. 88.99 of the variation in the job growth in 2010 is accounted for by the job growth in 2012. 0 B Overall, the job growth in 2012 is 88.9% of the job growth in 2010. 88.9% of the variation in the job growth in 2012 is accounted for by the job growth in 2010. 0D. Overall, the job growth in 2010 is 88.9% of the job growth in 2012. d) Do these results indicate that; in general, companies with a higher job growth in 2010 had a higher job growth in 2012? Explain OA. Yes because there is significant evidence of a linear relationship with negative slope_ 0 B No, because there is insufficient evidence of a linear relationship_ No, because there is insufficient evidence of linear relationship, but the slope is negative. Yes_ because there is significant evidence of a linear relationship with a positive slope:



Answers

For the following exercises, consider the data in Table 2.18, which shows the percent of unemployed in a city of people 25 years or older who are college graduates is given below, by year.
$$\begin{array}{|c|c|c|c|c|c|}\hline \text { Year } & {2000} & {2002} & {2005} & {2007} & {2010} \\ \hline \text { Percent Graduates } & {6.5} & {7.0} & {7.4} & {8.2} & {9.0} \\ \hline\end{array}$$
Determine whether the trend appears to be linear. If so, and assuming the trend continues, find a linear regression model to predict the percent of unemployed in a given year to three decimal places

So I put all of my data in list one and list too. And then hit the linear regression button. And I found that the equation ended up being 3.912 Plus 1.71133 X. And we got a correlation coefficient which it doesn't ask for that. But it's important to look at a .9895. And then I did a stat plot graphing list one versus list too and having wise of one B. This regression equation. And when I look at that graph that graph looks very very linear. And the data points are not all specifically on the line. But when that line is drawn through them it looks very very close and we could tell that by that correlation coefficient that it was quite possible that it was going to look quite linear now for part C. After I've calculated that value for the regression equation, I went to list three and doctor instead of going through and finding the observed well we have the observed value but finding what the expected value was a predicted value by plugging all these list one values back into the equation and then subtracting. We do have that feature under second and list on A. T. I. 84 then the residual at the bottom. And so it's giving me all these residuals. When I look at my uh my chart For my residuals or my staff at it, the first residual comes out to be about .4. The next one is about .24. The next one is negative .58. So that gives you an indication of what the residuals should be. And now I'm going to go back to my staff plot And I'm going to turn my step plot on to go list one vs list three. So list one versus list three. And I'm also going to get rid of my regression equation as wise of one And just the second hair, list one vs list three and hit my wife someone and clear that out. And then again hit zoom number nine zoom number nine. And when I get that residual plot, that residual plot will always have that X axis in the center and my residuals, I have a result. You down here, I have up here, I have done here, down here, there's one up here and down here and I'm really not seeing it's kind of oscillating around this line. So I don't see not a not a particular pattern. There's no curvature in the data. So there's no pattern, which means that the linear model fits quite well, is quite good. Oh, no. All right. So I would use that model as a good predictor.

So I'm using a. T. I. 84. And I put my data in list one and list too. And when I did I find the equation Regression equation to be 21.024 plus .1389 x. And then I grabbed a scatter plot of the data and the scatter plot of the data. The data ends up looking like a pattern kind of like this. And then the regression equation ends up looking like that. So we can see that there is curvature in that data and that regardless of what the correlation coefficient we can see that there's definitely a pattern. And then you are asked to have a residual plot. And so a couple things you can do you can go over here in a to list three and you can um go into your editor and you can find the residuals longhand by putting the regression equation up here so go to variables and statistics and eq. And it will put the equation in there. And the only thing I need to do is change the X. value into list one. And when I do I get my residual excuse me, my predicted value. And that comes out to be 26.3, 25.7 and etc those are the first couple. And then we could take list to minus list three to find my residuals and make sure you're up in this L for icon not to do that and list to minus list three. And when I do I get the residuals of negative 2.3 is my first one, negative 3.7 point 42 and so on and not. That will give you an idea of what the residual should look like. And then I can go back and do a stat plot And my step plot would have list one versus list for. And also I would clear out that Y equals because we don't need that anymore and I don't want that to when I hit zoom and number nine presumed staff, I end up getting a residual plot that looks like clips, we put the axis there we go. And the residual plot will always have that access in the center by the way, and the residual plot has a couple dots and it does this and like so. And anyway, we definitely see a pattern of a curb, a curved pattern on here that tells us that our linear model is really not suited to this, that there's probably a different curved model that would fit this data better.

Part one, Theis estimated equation is as follows. The coefficient on the growth of the minimum wage is 0.151 It implies that a 1% point growth in the minimum wage is estimated to increase the growth in wage of sector 2 32 by about point on 51% points. Part two Yeah, we add the 12 legs of GM wage the growth of minimum wage. We find that the sum of all the coefficient is somewhat higher than what we got from the aesthetic regression. We will use an F test to test for the joint significance of the legs. The P value of the F test is 0.58 showing that these legs are jointly significant at the 10% level. Okay, Yeah, we evaluate the joint significance of the legs, but we also look at the individual significance of each leg. We find that legs number eight to number 12, yeah, have quite large coefficient. And some of the individual T statistic are significant at the 5% level. In part three, we have a new estimated equation. The coefficient on the growth of minimum wage is very small and the teeth statistic is also very small. Okay, Okay. Remember that the T start for the beta coefficient is estimated by taking beta coefficient divided by its standard error. Yeah, another signal is the are square. The art square is almost zero. Meaning neither the growth of minimum wish. And what is the other thing? The other thing, G c p i. So neither of these variables can explain the variation in the growth of employment in sector 2 32 in part four, we will act will add the legs of the growth of minimum wage into the above equation. We find that doing so does not improve the fit of the model. The F test of joint significance of mhm should fix this The joint significance of growth of minimum wage and its lax. The F statistic has a P value of 0.439 Yeah, which means we cannot reject the non hypothesis. Yeah, the coefficients also change sign, and none of the coefficient of the legs is individually statistically significant at the 5% level. Okay. Therefore, there is little evidence that minimum wage growth affect employment growth either in short run or long run

In part one and two women apply the A C F function in our on two variables unemployment rate and vacancy rate. We will find the first order Auto correlation denote row one hat to be 0.986 and point 924 respectively. These correlation coefficients point more toward a unit root process rather than a weekly dependent process for either variable. In part three, we will estimate the beveridge curve. This curve relates the unemployment rate to the vacancy rate with the simplest relationship being linear. This is the old L s result. And looking at the slope coefficient, we find a significant negative relationship. Yeah, heart four. Good. Even though our data confirms that Yuri, we cannot trust this result. The reason is both Siris are highly persistent. They are very close to a unit route, as shown in part one. Mhm. Uh huh. And a unique route process doesn't have weak dependence. Unique route is the opposite of being weakly dependent. Yeah, but weak dependence is important in OLS estimation For time Siri's, we need the data to be weekly dependent so we can apply the standard large sample results, particularly the central limit Terram, and so we can assume the estimates to be as simple optically normally distributed. The confidence interval is built based on normality. Assumption? Yeah, In a regression with unit route, these conditions is violated. Therefore, we cannot trust the confidence interval for beta one in the model in part three. In Part five, we win difference the unemployment rate and vacancy rate. Siri's then run the regression again. This is the result. Okay. As you can see, the estimated slope coefficient compare with Part three is much smaller in magnitude and no longer statistically significant. This example shows that different thing before running and ls. Regression is not always a sensible strategy, but we will learn more about this in Chapter 18.


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