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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 $\frac{1}{\sigma^2}(SSE_R-SSE_F) = \frac{SSE_R-SSE_F}{df_R - df_F}$?

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  • $\begingroup$ 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? $\endgroup$ – Omar123456789 Oct 8 '18 at 16:17

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