91Ó°ÊÓ

From the minitab printout, the prediction equation can be written as$\hat{y}=90.10–1.836x_1+0.285x_2+\mathrm{Î\mu }$

Value of R² is 0.916 meaning that approximately 92% of the variation in the regression is explained by the model. Higher the value of R², better fit the model is for the data. Since 91.6% is a very high number, it can be concluded that the model is a good fit for the data.

$H_0:{\mathrm{β}}_1={\mathrm{β}}_2=0$

H_a: At least one of the parameters${\mathrm{β}}_{1}or{\mathrm{β}}_2$is non zero

Here, F test statistic = $\frac{SSE}{n-(k+1)}=\frac{1364}{15-3}=113.667$

Value of F_0.05,15,15 is 2.475

H₀ is rejected if F statistic > F_0.05,15,15. For α=0.05, since F > F_0.05,15,15 Sufficient evidence to reject Ho at 95% confidence interval.

Therefore, ${\mathrm{β}}_1≠{\mathrm{β}}_2≠0$

H₀:${\mathrm{β}}_1=0$

H_a: ${\mathrm{β}}_1$≠ 0

Here, t-test statistic =$\frac{{\widehat{\mathrm{β}}}_1}{s{\widehat{\mathrm{β}}}_1}=\frac{-1.836}{0.367}=-5.002$

Value oft_0.025,14is 2.145

H₀ is rejected if t statistic > t_0.05,24,24. For α=0.05, since t < t_0.05,31 Not sufficient evidence to reject H_o at 95% confidence interval.

Therefore, ${\mathrm{β}}_1$=0 .

The standard deviation of the regression model can be calculated as $\frac{SSE}{n-2}$Here, SSE = 1364, s²= $\frac{1364}{13}=104.9230$