# Calculating the deviance when performing a hypothesis test of a parameter in a linear regression model

I want to make sure I am using and calculating the deviance correctly when performing a hypothesis test of the significance of a parameter in a linear regression model.

Suppose that I have two models at hand. $$\mu_F = \beta_0 + \beta_1x_1 + \beta_2x_2$$ and $$\mu_R = \beta_0 + \beta_1x_1$$. I want to test the hypothesis that $$H_0: \beta_2 = 0$$ v $$H_1: \beta_2 \ne 0$$. I can use the deviance of the two models as my test statistic to test this hypothesis, correct?

If so is this how I can calculate my deviance?

$$D = \frac{Likelihood_R}{Likelihood_F}$$ => $$-2D = -2*log(\frac{Likelihood_R}{Likelihood_F}) \sim \chi^2$$

$$L_F = \prod\frac{1}{\sqrt(2\pi\sigma^2)}exp(-\frac{1}{2\sigma^2}(y_i - \mu_i)^2$$ = $$\prod\frac{1}{\sqrt(2\pi\sigma^2)}exp(-\frac{1}{2\sigma^2}(y_i - (\beta_0 + \beta_1x_1 + \beta_2x_2)^2)$$

$$L_R = \prod\frac{1}{\sqrt(2\pi\sigma^2)}exp(-\frac{1}{2\sigma^2}(y_i - \mu_i)^2$$ = $$\prod\frac{1}{\sqrt(2\pi\sigma^2)}exp(-\frac{1}{2\sigma^2}(y_i - (\beta_0 + \beta_1x_1))^2)$$

$$D = 2[log(L_F) -log(L_R)] = 2[\sum-log(\frac{1}{\sqrt(2\pi\sigma^2)})-\frac{1}{2\sigma^2}(y_i - (\beta_0 + \beta_1x_1 + \beta_2x_2))^2 +log(\frac{1}{\sqrt(2\pi\sigma^2)})+\frac{1}{2\sigma^2}(y_i - (\beta_0 + \beta_1x_1))^2] = \frac{1}{\sigma^2}(SSE_R-SSE_F) \sim \chi^2((N-2)-(N-3)=1)$$

Is this how the deviance is calculated to test the hypothesis $$H_0: \beta_2 = 0$$? Isn't this test usually calculated as $$\frac{SSE_R-SSE_F}{df_R - df_F}$$? Does $$\frac{1}{\sigma^2}(SSE_R-SSE_F) = \frac{SSE_R-SSE_F}{df_R - df_F}$$?

• I've noticed up to 10 views on my question and 0 input. Just want to be sure and ask is everything clear in this post? – Omar123456789 Oct 8 '18 at 16:17