Tollowin naspa ŷ = 17.6 + 3.8x, - 2.3x, + 7.6x3 + 2.7x4 The values of SST and SSR are 1,803 and 1,756, respectively. (a) Compute R2. (Round your answer to three decimal places.) R2 = (b) Compute R,. (Round your answer to three decimal places.) R2 (c) Comment on the goodness of fit. (For purposes of this exercise, consider a proportion large if it is at least 0.55.) O The estimated regression equation provided a good fit as a small proportion of the variability in y has been explained by the estimated regression equation. O The estimated regression equation did not provide a good fit as a small proportion of the variability in y has been explained by the estimated regression equation. O The estimated regression equation did not provide a good fit as a large proportion of the variability in y has been explained by the estimated regression equation. O The estimated regression equation provided a good fit as a large proportion of the variability in v has been explained by the estimated regression equation.

Tollowin naspa ŷ = 17.6 + 3.8x, - 2.3x, + 7.6x3 + 2.7x4 The values of SST and SSR are 1,803 and 1,756, respectively. (a) Compute R2. (Round your answer to three decimal places.) R2 = (b) Compute R,. (Round your answer to three decimal places.) R2 (c) Comment on the goodness of fit. (For purposes of this exercise, consider a proportion large if it is at least 0.55.) O The estimated regression equation provided a good fit as a small proportion of the variability in y has been explained by the estimated regression equation. O The estimated regression equation did not provide a good fit as a small proportion of the variability in y has been explained by the estimated regression equation. O The estimated regression equation did not provide a good fit as a large proportion of the variability in y has been explained by the estimated regression equation. O The estimated regression equation provided a good fit as a large proportion of the variability in v has been explained by the estimated regression equation.

College Algebra

7th Edition

ISBN:9781305115545

Author:James Stewart, Lothar Redlin, Saleem Watson

Publisher:James Stewart, Lothar Redlin, Saleem Watson

Chapter1: Equations And Graphs

Section: Chapter Questions

Problem 10T: Olympic Pole Vault The graph in Figure 7 indicates that in recent years the winning Olympic men’s...

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Olympic Pole Vault The graph in Figure 7 indicates that in recent years the winning Olympic men’s pole vault height has fallen below the value predicted by the regression line in Example 2. This might have occurred because when the pole vault was a new event there was much room for improvement in vaulters’ performances, whereas now even the best training can produce only incremental advances. Let’s see whether concentrating on more recent results gives a better predictor of future records. (a) Use the data in Table 2 (page 176) to complete the table of winning pole vault heights shown in the margin. (Note that we are using x=0 to correspond to the year 1972, where this restricted data set begins.) (b) Find the regression line for the data in part ‚(a). (c) Plot the data and the regression line on the same axes. Does the regression line seem to provide a good model for the data? (d) What does the regression line predict as the winning pole vault height for the 2012 Olympics? Compare this predicted value to the actual 2012 winning height of 5.97 m, as described on page 177. Has this new regression line provided a better prediction than the line in Example 2?
The admissions officer for a certain college developed the following estimated regression equation relating the final college GPA to the student's SAT mathematics score and high school GPA. ŷ = −1.39 + 0.0234x1 + 0.00482x2 where x1 = high-school grade point average x2 = SAT mathematics score y = final college grade point average. #1)A high-school average 84 corresponds to x1 = 84 and a score of 535 on the SAT mathematics test corresponds to x2 = 535. Substitute these values into the estimated regression equation to find the final college GPA, rounding the result to two decimal places. GPA = −1.39 + 0.0234x1 + 0.00482x2 = -1.39 +0.0234 (_____________) + 0.00482 (535) = __________________
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The admissions officer for a certain college developed the following estimated regression equation relating the final college GPA to the student's SAT mathematics score and high school GPA. ŷ = −1.39 + 0.0234x1 + 0.00482x2 where x1 = high-school grade point average x2 = SAT mathematics score y = final college grade point average. #1) The given regression equation follows where x1 is the high-school grade point average, x2 is the SAT mathematics score, and y is the final college grade point average. ŷ = −1.39 + 0.0234x1 + 0.00482x2 We are to predict the final college GPA for a student who has a high-school average of 84 and scored 535 on the SAT mathematics test. Thus, the corresponding values to substitute into the regression equation are x1 = and x2 = .