91影视

From the minitab printout, the prediction equation can be written as $\stackrel{铃�}{y}=90.10-1.836x_1+0.285x_2+蔚$.

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.

${\mathrm{H}}_0:{\mathrm{尾}}_1={\mathrm{尾}}_2=0$

H_a: At least one of the parameters ${尾}_1$or ${尾}_2$is non zero

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

Value of F_0.05,15,15 is 2.475

${\mathrm{H}}_0\mathrm{is}\mathrm{rejected}\mathrm{if}\mathrm{F}\mathrm{statistic}>{\mathrm{F}}_{0.05,15,15路}\mathrm{For}\mathrm{伪}=0.05,\mathrm{since}\mathrm{F}>{\mathrm{F}}_{0.05,15,15}$

Sufficient evidence to reject H₀ at 95% confidence interval.

Therefore, ${尾}_1={尾}_2=0$.

$$\begin{gathered} {\mathrm{H}}_0:{\mathrm{尾}}_1=0 \\ {\mathrm{H}}_{\mathrm{a}}:{\mathrm{尾}}_1鈮�0 \end{gathered}$$

Here, t-test statistic =$\frac{\stackrel{铃�}{{尾}_1}}{s\stackrel{铃�}{{尾}_1}}=\frac{-1.836}{0.367}=-5.002$

Value of t_0.025,14 is 2.145

$$\begin{gathered} {\mathrm{H}}_0\mathrm{is}\mathrm{rejected}\mathrm{if}t\mathrm{statistic}>t_{0.05,24,24}.\mathrm{For}\mathrm{伪}=0.05,\mathrm{since}t<t_{0.05,31} \\ \mathrm{Not}\mathrm{sufficient}\mathrm{evidence}\mathrm{to}\mathrm{reject}{\mathrm{H}}_0\mathrm{at}95\%\mathrm{confidence}\mathrm{interval}. \\ \mathrm{Therefore},{\mathrm{尾}}_1=0 \end{gathered}$$

The standard deviation of the regression model can be calculated as$\frac{\mathrm{SSE}}{\mathrm{n}-2}$.

$Here,SSE=1364,s^2=\frac{1364}{13}=104.9230$.