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(13 points) Suppose you have a simple linear regression model such that Y; = Bo + B18: +€4 with and N(0,0%) Call: 1m (formula
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(a)\\ $regression equation is given by$\\ \hat y_i=8.00967-0.62009x_i\\ (b)\\ TSS=SSR+SSE\\n-1=k+(n-k-1)\\ y_i=\beta_0+\beta_1x_i+\epsilon_i\;;$here k=1$\\ $degrees of freedom for error is n-k-1=197$\\ n-2=197\Rightarrow n=199\\ (c)\\R^2=0.52\;$means this regression model explaines 52 percent of variance present in the original data. $\\ (d)\\$we know that $\\ r=\sqrt{R^2}=\sqrt{0.52}=0.72\\ (e)\\R^2=\frac{SSR}{TSS}=\frac{SSR}{SSR+SSE}\Rightarrow \frac{1}{R^2}=1+\frac{SSE}{SSR}\\ F=\frac{MSR}{MSE}=\frac{SSR/k}{SSE/(n-k-1)}\;;k=1 \\ F=(n-2)\frac{SSR}{SSE}=(n-2)(\frac{1}{\frac{1}{R^2}-1})=(n-2)\frac{R^2}{1-R^2}\\ (f)\\ $hypothesis under testing are$\\ H_0:\beta_1=0\\ H_1:\beta_1\neq 0\\ $test statistic is$\\ F=(n-2)\frac{R^2}{1-R^2}\\ n=199\\ $df for numerator is 1 , denominator is 197.$\\ R^2=0.52\\ F=(199-2)\frac{0.52}{1-0.52}=213.417\\ P- value=P(F_{1,197}>213.417)=0\\ (g)\\ $p value is less than 0.05 and 0.01 hence we reject the null hypothesis at 5, 1 percent level of significance$

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