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Refer to the data presented in exercise $2 .$ The estimated regression equation for the these data is$$\hat{y}=-18.4+2.01 x_{1}+4.74 x_{2}$$Here $S S T=15,182.9,$ SSR $=14,052.2, s_{b}=.2471,$ and $s_{b n}=.9484$$\begin{array}{l}{\text { a. Test for a significant relationship among } x_{1}, x_{2}, \text { and y. Use } \alpha=.05 \text { . }} \\ {\text { b. Is } \beta_{1} \text { significant? Use } \alpha=.05 \text { . }} \\ {\text { c. Is } \beta_{2} \text { significant? Use } \alpha=.05}\end{array}$

a) The test statistic value is greater than the critical value. Reject the null hypothesis. Therefore, itcan be concluded that a significant relationship is present between $y$ and two independentvariables $x_{1}$ and $x_{2}$b) The test statistic value is greater than the critical value. Reject the null hypothesis. Therefore, itcan be concluded that the parameter $\beta_{1}$ is statistically significant.c) The test statistic value is greater than the critical value. Reject the null hypothesis. Therefore, itcan be concluded that the parameter $\beta_{2}$ is statistically significant.

Intro Stats / AP Statistics

Chapter 13

Multiple Regression

Descriptive Statistics

Linear Regression and Correlation

University of North Carolina at Chapel Hill

Piedmont College

Cairn University

Lectures

0:00

05:28

In exercise $1,$ the follo…

So if you refer to data presented next size too, we know that the estimated regression equation for this data is can be modelled by this expression. So negative 18.37 plus 2.0 x one plus 4.74 Thanks to here, we know that s a T is equal to 15. 1 82.9 Assist our as equal to 14 points. Your five to a speedy one is equal to jerk. Went two for 71 Andi SB two is equal to 0.9 for 84 Does foot 10,052 going to? Okay, so now this some and the air of squares is just as e which is equal to SST minus s. Are these values in get as s E is equal to 1130. The corresponding standard airs are as b Do you gee cooled to 471 and 0.9484 Again, This is just que From this we can calculate. Yes, our which is just as our okay. Each is wanted 452 0.2 over to which will give us 7026. No, I mean of standard the mean squares due to air E is equal to S S E and my P minus form 1137 over 10 to 1. Then my simmers 17 So 1 1030.7 divided by seven. Give us 100 61.5286 No, we know that the no and alternate hypothesis are h if you won and you beat it too is equal to zero And alternative had once his age is a one or more of the perimeters is not equal to zero. Cheers. There's not equal, Did you that for the test statistic can be calculated Just s are over, which is 7026.1 over 1 60 1.5 three but you'll give us approximately 43.5. So now we can find the degrees of freedom due to their aggressions. Degrees of freedom and numerator is equal to p two that the use of freedom in the denominator is equal to end minus P to sit. Let's let the critical value of significant level of significance be Alfa. That is your prints your five. So now we confined and the corresponding critical, which is just going to be for seven. For So this test statistic value is greater than the critical value, so we can reject the No Both says, if we could be concluded that a significant relationship is present between why and to independent veils X one and x. Okay, no for port. See, we know that h not. It is just that B two is equal to zero on each one. You know, baited two is not people because you, the G test statistic, is just be too over. Speak to that's 4.74 over June 9484 which will give us approximately 4.9 98 degrees of freedom seven on If he let the level of significance be your front five again, find a T critical value for them to tell critical values. Charges. Placer Miners G off to equal to plus or minus t Sure, sure to five, which is just plus or minus 2.365 Since the statistic buys get in critical value in rejecting a lot but says therefore, it can be concluded that the parameter B two is statistically significant

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