Testing the Model

To say the model is useful is to confirm whether $x$ can really be used to explain $y$. Basically, changing the values of $x_i$ should relate to change in $y_i$, if $x$ is really an explanatory for $y$.

If ${\beta }_1\approx 0$, then $x$ doesn’t affect $y$, which means the model is not useful.

This means, to assess the usefulness of the model, we need to test whether ${\beta }_1\ne 0$.

Hypothesis

• $H_0:{\beta }_1=0$
• $H_1:{\beta }_1\ne 0$

Test statistic

$$t=\frac{{\hat{\beta }}_1}{s_{{\hat{\beta }}_1}}\sim t_{\text{df}=n-2}$$

Where:

$$\begin{array}{rl}{s}_{{\hat{\beta }}_1}& =\frac{s}{\sqrt{S{S}_{xx}}}\\ S{S}_{xx}& =\sum _{i=1}^n(x_i-\bar{x}{)}^2\\ s^2& =\frac{{\sum }_{i=1}^n{ϵ}_i^2}{n-k}\end{array}$$

Decision rule

Reject $H_0$ if:

• $t<-t_{\alpha /2}$ or $t>t_{\alpha /2}$ for 2-sided test
• $t<-t_{\alpha }$ or $t>t_{\alpha }$ for 1-sided test

Assessing the model using confidence interval

We can construct the $(1-\alpha )\times 100\%$ confidence interval for ${\beta }_1$, which is

$$\begin{array}{r}({\hat{\beta }}_1-t_{\alpha /2}\times s_{{\hat{\beta }}_1},{\hat{\beta }}_1+t_{\alpha /2}\times s_{{\hat{\beta }}_1})\end{array}$$

If the value of ${\hat{\beta }}_1$ is outside the confidence interval, then the model is not useful.