Question

Suppose you used Minitab to fit the model $y={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2+\epsilon$

to n = 15 data points and obtained the printout shown below.

1. What is the least squares prediction equation?
2. Find R² and interpret its value.
3. Is there sufficient evidence to indicate that the model is useful for predicting y? Conduct an F-test using α = .05.
4. Test the null hypothesis H₀: β₁ = 0 against the alternative hypothesis H_a: β₁ ≠ 0. Test using α = .05. Draw the appropriate conclusions.
5. Find the standard deviation of the regression model and interpret it.

Short answer

1. From the minitab printout, the prediction equation can be written as $\overbrace{y}=90.10-1.836x_1+0.285x_2+\epsilon$.
2. 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.
3. At 95% confidence interval, it can be concluded that${\beta }_1\ne {\beta }_2\ne 0$.
4. At 95% confidence interval, it can be concluded that${\beta }_1=0$
5. $s^2=104.9230$.

Step-by-step solution

Least square prediction equation

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

R2 interpretation

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.

F-test

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

H_a: At least one of the parameters ${\beta }_1$or ${\beta }_2$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

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

Sufficient evidence to reject H₀ at 95% confidence interval.

Therefore, ${\beta }_1={\beta }_2=0$.

Significance of β1

$$\begin{gathered} {\mathrm{H}}_0:{\mathrm{\beta }}_1=0 \\ {\mathrm{H}}_{\mathrm{a}}:{\mathrm{\beta }}_1\ne 0 \end{gathered}$$

Here, t-test statistic =$\frac{\overbrace{{\beta }_1}}{s\overbrace{{\beta }_1}}=\frac{-1.836}{0.367}=-5.002$

Value of t_0.025,14 is 2.145

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

Standard deviation

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

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