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 R²and 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.

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

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.

$H_0:{\beta }_1={\beta }_2=0$

H_a: At least one of the parameters${\beta }_{1}or{\beta }_2$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, ${\beta }_1\ne {\beta }_2\ne 0$

H₀:${\beta }_1=0$

H_a: ${\beta }_1$≠ 0

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

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, ${\beta }_1$=0 .

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