Problem

In exercise $1,$ the following estimated regressi…

05:28

Problem 18 Easy Difficulty

Refer to exercise 10 , where data were reported on a variety of statistics for the 29 teams
in the National Basketball Association for a portion of the 2004 season $(\mathrm{NBA}$ website.
January $3,2004$
$$\begin{array}{l}{\text { a. In part (c) of exercise } 10, \text { an estimated regression equation was developed relating the }} \\ {\text { proportion of games won given the percentage of field goals made by the team, the }} \\ {\text { proportion of three-point shots made by the team's opponent, and the number of turn- }} \\ {\text { overs committed by the team's opponent. What are the values of } R^{2} \text { and } R_{\mathrm{a}}^{2} \text { ? }}\end{array}$$
b. Does the estimated regression equation provide a good fit to the data? Explain.

Answer

a) $R_{a}^{2}=0.511 \%$
b) good fit

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Chapter 13

Multiple Regression

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Video Transcript

we know why is equal to negative 1.2 three plus four point 82 Text one minus 2.59 x two plus your 0.0 344 x So we know that I m s r is equal to SSR for P, which is your friend zoo. June 30 for six over three is equal to zero ones. You're one A to else Another confined ski just s e over n minus p one, which is your point, too. 3635 Those air 29 three minus one. It was approximately. Is your prints here is you nine for five. So the hypothesis to be tested is that age one is eucalyptus created to which is you called debated three. The terms two purposes is just that one or more for others is not equal to show you. So we can now calculate which is just a mess. Are those are e sure you calculate here, Andi, if you do that, we'll get 10.7 seven. And for the f distribution with P equal to three degrees of freedom in the numerator and en minus P minus one degrees of freedom in the denominator. The area in the upper tail corresponds to the test statistic. F is equal to 10.7 as less than 0.1 from for the appendix Be so the P value is less than 0.1 on exactly value is zero points. Yours, your zero. So we can reject H not at Alfa Physical. Tuju opens your five level of significance and conclude that at least one of the parameters not equal to zero. So therefore we can conclude that there is a significant relationship between X. Why on the tube? Independent variables X one takes two next to okay. No, we know that s one physical. 1 ft one to be has to as you did you 7 June for one B three sequiturs Your print you won 253 on again the buses to be tested. Our alternative that 10 China wants is a one is not so. Now we know that T is equal to be won over speed just for plain 817 over 113 which is equal to four. It's from table two of Appendix B T distribution with n minus one Mons and minus P minus one is equal to 25 degrees of freedom in the upper tail value. Corresponding statistic team is equal to 4.7, which is that stands Europeans. Here's your five so we can reject the null hypothesis at this year's brings your five level of significance. I conclude that parameter b one No. One is significant.

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