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3. In your answer to Question 2, you should have found that alinear trendline fits the data well. Compare the slope ofthe graph from the trendline equation with the...

Question

3. In your answer to Question 2, you should have found that alinear trendline fits the data well. Compare the slope ofthe graph from the trendline equation with the appropriateslope from Newton’s Second Law. Remember, you areplotting F versus a. Hint: Write Newton’sSecond Law in the form y =mx where 𝑦 represents F and 𝑥represents a and m represents the slope.Extract the slope from the equation and plug in the values you usedin your simulation. Show all your work in thespace provided

3. In your answer to Question 2, you should have found that a linear trendline fits the data well. Compare the slope of the graph from the trendline equation with the appropriate slope from Newton’s Second Law. Remember, you are plotting F versus a. Hint: Write Newton’s Second Law in the form y = mx where 𝑦 represents F and 𝑥 represents a and m represents the slope. Extract the slope from the equation and plug in the values you used in your simulation. Show all your work in the space provided below. You may either type your calculation into this space or insert a picture of a calculation you did on paper here. Do not forget to convert all values into SI units! (3 points) For Reference: Trendline Equation is: y = 25x +0.5 (the regression value is r^2=1) Data from the Table is: F (N) a (m/s2) 5.5 0.2 15.5 0.6 25.5 1.0 35.5 1.4 45.5 1.8 4. State what your slope represents. (3 points)



Answers

For each of the following data sets, a. Create a scatterplot. b. Use $\operatorname{LinReg}(\text { ax }+\text { b) to determine the best fit line and } r .$ Does the line seem to accurately describe the pattern in the data? c. For each of the different choices listed in the above chart, find the equation of the best fit curve and its associated $r^{2}$ value. Of all of the curves, which seems to provide the best fit? Note: The $r^{2}$ -value reported in each case is NOT the linear correlation coefficient reported when running $\operatorname{LinReg}(\mathrm{ax}+\mathrm{B})$ Rather, the value will typically change depending on the curve. The reason why is that each time, the $r^{2}$ -value is measuring how accurate the fit is between the data and that type of curve. A value of $r^{2}$ close to 1 still corresponds to a good fit with whichever curve you are fitting to the data. $$\begin{array}{|c|c|} \hline x & y \\ \hline 1 & 16.2 \\ \hline 2 & 21 \\ \hline 3 & 23.7 \\ \hline 4 & 24.8 \\ \hline 5 & 23.9 \\ \hline 6 & 20.7 \\ \hline 7 & 15.8 \\ \hline 8 & 9.1 \\ \hline 9 & 0.3 \\ \hline \end{array}$$

Working with a data table and were asked to determine the line of best fit and also the our value. So this scatter plot WAAS created with Excel and Excel has the nice future that when you do what's called adding a trend line, it'll do. It'll add this little blue dash line, which is your linear regression, and it will allow you to display the R squared value right on the charge and the equation. And it turns out that our is equal to the square root of r squared, and it has either a positive or a negative, depending on Dez Y decrease. Is it downward sloping or is it upward sloping? And in this case, why decreases as X increases? Therefore, this will be negative. So if we take negative the square root of 0.85 to 4, we will get that the the are the coefficient is negative. 0.9 23 What's a pretty good our value. And when you look at the data and take a fairly decent fit, none of the point is too far away from the line. But you probably also notice when you look at this, that actually there appears to be a bit of a curve here, so that might be some other functions of polynomial. So we're going to try other functions with Excel. As with your tea I calculator, you can choose different shapes for the best fit. So let's take a look at a quadratic. So this is a polynomial of degree to here. You can see that's already a terrific fit. R squared is 0.9936 And except for this data point right here, the rest of, um, lie right on the line. So that's a wonderful fit. Let's see what happens if we bring the polynomial up to degree three. So this is a Cuba occurred polynomial degree three, and it looks like our improved just a little bit. And it looks like this point right here is a little closer to the polynomial. So we're improving. What happens if we make it a degree for polynomial so X to the fourth? Uh, polynomial? Well, then we get a perfect fit. R squared equals one, and we see every single point lies exactly on the trend line, so it's pretty hard to get better than that. In fact, you can't have better than R squared equals one, so that's probably going to be our winner. But let's see how the other shapes dio this is. If we did a natural log, it still has a pretty good fits. You can see right here R squared equals 0.9776 But we do have a few of the points, not quite on the line. It's still a decent fit in a pinch. We could certainly use it for some predictive value, at least within a short domain of access. Finally, what is the power look like? Power function. This particular one, um R squared is 0.94 Also a good fit, and again the points are all close to it on several of the points or even on the line. But that said our best fit came from our a polynomial degree for, and this one would be extremely confident if we wanted to use the equation for predictive value to find the wise that go with exes. We didn't have because our spurred actually equals one

So given a data table and asked to create a scatter plot and determine the linear regression and then the our value that goes with that. So this this scatter plot was created. Using Excel on Excel has a nice feature that with a few clicks of the mouse, you can add a trend line, and it will do. The regression for you will even allow you to choose which shape function. It doesn't have to be a line. Furthermore, if you want the R squared value displayed on the graph, it's another click. Same with the equation. So for us to determine our we're going to take advantage of that R squared R is equal to the square root of R squared, and it's either positive or negative. In this case, it will be positive because there's a positive correlation as X increases. Why increases another way to look at? It is the trend is sloping upwards. So if we take 0.942 and take the square root of it, then we're going to get seat 0.9 for to take the square root, we get 0.9706 so the R values 0.9706 That looks like a pretty good fit. There are other places where we would be pretty happy with that. We look at the data, I want her. I'm sure you noticed that. Really? It's got some curvature to it. I wants to be some function that looks more like that, which is not a line. So sometimes with data, you want to try a different shape, a question we are going to start off trying. What does a quadratic dio of X squared function parabola and what we see here is, all of a sudden, all of the data points are quite near to the curve, and our our our value shoots up 2.9805 That's much better. Um, see what happens if we go degree three. If we make it a cubic, we could raise it one more degree. Looks like it gets even better a lot of times with these higher degree polynomial ills. For a small region of the data, you can get a fantastic fit like this, Um, and it does look like every single dot lies on the curb. So it's tempting to choose that one. Let's see what happens if we raise it one more degree? So now we're looking at an extra. The Four Excel believes that the R squared value is one for this particular, um, fit. And if you verify that on another graphing utility, you'll see it's something like 0.9998 which is essentially one. And again. Every single dot seems to be right on the line. It was a slightly complicated shape, but it does seem to fit very well. Let's look at natural log. This one has also a very, very high correlation. 0.9988 That's extremely close to one, and furthermore, this shape is a little simpler, and it very much seems to represent the data. So even though degree for appears to have a better R or R squared, the natural log just looks like a better fit. All right, so finally, let's take a look at a power function. This one Excel didn't do such a great job. The the exponents is pretty near one, which makes this look very much like a lot, and we can clearly see that this is a curved kind of function. So the power best fit really wasn't our option. And in fact, we could see the arse words the lowest of the six things we tried. So when we're coming down to pick, which is the best model when could either choose one of the higher degree polynomial is because the R squared is better. Or we could choose the natural log because not only is the r squared very close to one, but also the shape just appears to represent the data better, and I'm inclined to choose natural Log. One could probably debate that for a while, but this just appears toe match the trend very well. And in either case, we have a fairly high degree of certainty. If we were using the equation to predict a Y value for one of the next that we didn't already have, we would we would have some high degree of confidence that we would get an accurate result. The, uh, further away you get from the range of exes that you had actual data for this, you know, between one and 16 the last good your model is. My gut tells me the natural logs going to do a better job with it. So that's what we're gonna choose

Part one. We use the ADF regression with one leg and a time trend. We get the coefficient on the locker wage in area 2 32. A Tam T -1 to be -1056 and the T statistic is minus one point 39 The estimated route is Very close to one .9944. The coefficient on the lack of employment in the same area and same period in the A. D. Every question is slightly positive .0008. The route is estimated to be one and the team study is also very small. We don't care about that. We have a little evidence against a unit root. So we should treat these time series as having unitards are too without a time trend. The regression of delta U. T. Hat on U t minus one, had Delta UT -1 Hat and delta U. T monastery hat. Mhm yield A coefficient on UT -1 hat Of .00008. And so the estimated route is one beauty. Here are the residuals from the co integrating regression. When time is added to the equation, they need show regression. The coefficient on U. T minus one hat Is 0- two. And again there is no evidence for co integration or three when we use the real wish and a time trin the coefficient on new t minus one hat is minus point 044 with a T. Statistic of -3.09. The 10 critical value for the test with the time trend is -3.5. And so we cannot recheck the no hypothesis that the two series lock of wage and lack of employment are not into co integrated lot of wish and lack of employment and lack of real witch. Sure. A factor that is submitted from the co integrating relationship could be labor productivity. Mhm. We could also include other sources of income, say non wage income. This factor can affect the supply of labour.

All right. So we know that one over why is equal to one over eight times the coastline of the X or we can be right. This as the coastline of the X is equal to the sign off the X plus pi over to substituting the this into this we end up with one over why is equal to one over eight times the sign off b x plus pi over too.


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