4

Points/GameYards/Game25.8345.614.4303.618.6306.917.7286.716.5286.120.1290.324354.225.3346.929.6355.818.6346.122.4317.725.9374.222334.827.6352.323.8357.115.9270.614....

Question

Points/GameYards/Game25.8345.614.4303.618.6306.917.7286.716.5286.120.1290.324354.225.3346.929.6355.818.6346.122.4317.725.9374.222334.827.6352.323.8357.115.9270.614.9289.623.5341.937.2419.724.8368.923.4321.918.7299.419.3290.722.3350.322.7330.525.3322.314.9231.622.93451730321333.519.430919.8330.91- Develop a model for predicting points/game as afunction of yards/game.points/game = ( ? ) + ( ? ) ×yards/game(Round to three decimal places as needed.)2- Determine the coefficient of determination,

Points/Game Yards/Game 25.8 345.6 14.4 303.6 18.6 306.9 17.7 286.7 16.5 286.1 20.1 290.3 24 354.2 25.3 346.9 29.6 355.8 18.6 346.1 22.4 317.7 25.9 374.2 22 334.8 27.6 352.3 23.8 357.1 15.9 270.6 14.9 289.6 23.5 341.9 37.2 419.7 24.8 368.9 23.4 321.9 18.7 299.4 19.3 290.7 22.3 350.3 22.7 330.5 25.3 322.3 14.9 231.6 22.9 345 17 303 21 333.5 19.4 309 19.8 330.9 1- Develop a model for predicting points/game as a function of yards/game. points/game = ( ? ) + ( ? ) × yards/game (Round to three decimal places as needed.) 2- Determine the coefficient of determination, R2, and interpret its meaning. Select the correct choice below and fill in the answer box to complete your choice. (Round to three decimal places as needed.) A.The coefficient of determination is R2= ( ? ). This value is the proportion of variation in yards/game that is explained by points/game. B.The coefficient of determination is R2= ( ? ). This value is the proportion of variation in points/game that is explained by the variation in yards/game. C.The coefficient of determination is R2= ( ? ). This value is the probability that the slope of the regression line is statistically significant. D.The coefficient of determination is R2= ( ? ). This value is the probability that the correlation between the variables is statistically significant.



Answers

Consider the following scatterplot and regression analysis of 15 data points: Slope $b=0.377$ Coefficient of determination $r^{2}=95.0 \%$ Standard deviation of the residuals $s=0.640$ With the addition of a data point at $(35,14),$ which one of the following choices gives the most likely new regression statistics? (A) Slope $b=0.377$ Coefficient of determination $r^{2}=97.0 \%$ Standard deviation of the residuals $s=0.663$ (B) Slope $b=0.377$ Coefficient of determination $r^{2}=97.0 \%$ Standard deviation of the residuals $s=0.617$ (C) Slope $b=0.377$ Coefficient of determination $r^{2}=93.0 \%$ Standard deviation of the residuals $s=0.640$ (D) Slope $b=0.377$ Coefficient of determination $r^{2}=93.0 \%$ Standard deviation of the residuals $s=0.663$ (E) Slope $b=0.397$ Coefficient of determination $r^{2}=95.0 \%$ Standard deviation of the residuals $s=0.640$

What we are given the following data points X. Y. Listed at the top of this white board. And we want to use that answer information to answer the following six questions A through F. As follows. First in part A on the left, we want to produce a scatter plot of this data. We've already done so with the scattered provided right below and the data points marked with X's or crosses next. We want to compute the sums and the correlation coefficient are on the right. I've already listed the sums out their computers simply by following the formula sum of all X values, some of our Y values and so on. The correlation coefficient. R. Is given by this formula which makes use of the sample size and and the sounds we just computed, plugging in these values, we get articles 0.9 98. Next part C. We want to find the X. Meanwhile, I mean and the constant related to the equation line of best fit. So exciting. And what I mean are simply given as follows. Remember that the being a value they're given by the following formulas. He takes his input and the sums. It's very similar to the correlation coefficient are plugging in. We get the equals 4.509 and plugging in Wiebe R. E. And explode at a gives 33.696 Guest we have our equation for the line of best fit. White hot equals 33.6964 point 509 X. Next we want to plot this. Why had onto our scatter plot, Making sure to include our expire and are Y. Bar we do sell it as is observable here. Next let's calculate R squared and interpret so R squared to simply 0.9954 That means that approximately 99.54% of our data can be rather 99.54% of variations in our data can be explained to the data itself, and roughly half of a percent cannot be explained. Finally, for F we want to project Why, for X equals 12. Using Ry had equation. Doing so, we obtain 87.804.

What we are given the following data points X. Y. Listed at the top of this white board. And we want to use that answer information to answer the following six questions A through F. As follows. First in part A on the left, we want to produce a scatter plot of this data. We've already done so with the scattered provided right below and the data points marked with X's or crosses next. We want to compute the sums and the correlation coefficient are on the right. I've already listed the sums out their computers simply by following the formula sum of all X values, some of our Y values and so on. The correlation coefficient. R. Is given by this formula which makes use of the sample size and and the sounds we just computed, plugging in these values, we get articles 0.9 98. Next part C. We want to find the X. Meanwhile, I mean and the constant related to the equation line of best fit. So exciting. And what I mean are simply given as follows. Remember that the being a value they're given by the following formulas. He takes his input and the sums. It's very similar to the correlation coefficient are plugging in. We get the equals 4.509 and plugging in Wiebe R. E. And explode at a gives 33.696 Guest we have our equation for the line of best fit. White hot equals 33.6964 point 509 X. Next we want to plot this. Why had onto our scatter plot, Making sure to include our expire and are Y. Bar we do sell it as is observable here. Next let's calculate R squared and interpret so R squared to simply 0.9954 That means that approximately 99.54% of our data can be rather 99.54% of variations in our data can be explained to the data itself, and roughly half of a percent cannot be explained. Finally, for F we want to project Why, for X equals 12. Using Ry had equation. Doing so, we obtain 87.804.

But one this is the estimated equation. The usual S standard errors are in green and in round brackets. The hetero skid elasticity, robust standard Iran's are in blue and in square bracket. As you can see the robot standard Iran's are somehow larger in all cases on variable lock of E. X. B pp. Yeah the robust T statistic is 1.5 not too high. So we are not very convinced if performance its link to spending hard to. The f statistic has a value of 132.7 and a p value almost zero. Therefore there is the strong evidence of heterocyclic elasticity for three and four. This is the same equation estimated by weighted least square usual. Standard herons are in round and green brackets and weighted least square standard Iran's are in blue square brackets. We don't care about it. Their standard barrel over there. Constant term the intercept. So I don't report the centered errol oh, weighted least square for it. Yeah. Oh sorry the round one hour the usual wait at least square and the square one is robust. Type all part three. Mhm. We have to compare the estimates from weighted least square approach to L. S. Approach. Yeah. And we see that the old L S n W L. S coefficients on lunch or the same. But for other variables they differ. The waiting list square estimate on lock of spending is much larger than the old L. S coefficient. Based on this estimate a 10% increase in spending. Let me write that down. That is equivalent to an increase of 0.1 in lot of spending. Let's do you a pulling 65 percentage point increase in the math past rate the weighted least square T statistic is much larger too. Before we get 1.5 now we get 3.8. More than double past four. We compare the robot standard Iran's with the usual W. L. S. Standard Iran's and we find that for the key variable lack of spending, the robust standard error is somewhat larger than the usual one robust scented errol, our lock of enrollment also somewhat higher and for lunch, the robust version is slightly lower than the usual one, part five, the weighted least square estimated is more precise and by precise, I mean, robust standard errol is smaller than the robust version of the old LS estimate. For the first one we got 1.82 and for the second one two point 35


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