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I was working on a statistics homework and this is one of the questions:

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One of the parts asks to calculate a 95% confidence interval for the slope of each predictor. We are given that the sample variance of the number of games won is $12.11$, and the residual sum of squares is $111.30$. I understand that the formula for the confidence interval is $\hat{\beta_i}$ +/- $t_{\alpha/2, n-p-1}SE($$\hat{\beta_i}$). However, I'm not sure how I can find the standard error. I am told that the formula for the standard error is $S\sqrt{(\mathbf{X'X})_{ii}^{-1}}$, and I will need to use the variance-covariance matrix to find the variance. Can someone explain to me what I need to do? I am very confused right now.

Thanks!

peco
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  • You don't have enough information to determine the standard error, s.e. of $\beta_j$, since the s.e. of $\beta$ is requires statistics based on the predicted y-values, $\hat{y}_i$. Everything above is based on the "observed" y-values, $y_i$. Your $S$ is based on the average of the residuals, $e_i=y_i-\hat{y}_i$. –  Mar 15 '20 at 03:30
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    @NXGLogic whoops, we are also given that the sample variance of games won is 12.11 and the residual sum of squares is 111.3. I will add that in the question! – peco Mar 15 '20 at 15:45
  • The $\mathbf{X}^\top\mathbf{X}$ matrix is a 2x2 matrix, which has to be inverted to get $(\mathbf{X}^\top\mathbf{X})^{-1}$. Then, the square root of the diagonal entries of $(\mathbf{X}^\top\mathbf{X})^{-1}$ will be the $SE(\beta_j)$. Do you have the data? –  Mar 15 '20 at 17:23
  • See the bottom of [this](https://stats.stackexchange.com/a/44841/116195) answer. You need to expand $(X'X)^{-1}$ into its cell values, which you can then substitute using the values you were given. [This page](http://faculty.cas.usf.edu/mbrannick/regression/Part3/Reg2.html) does something similar with two predictors. – Noah Mar 17 '20 at 02:27

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Recall, there will be three regression coefficients for your model, $\beta_0$, called the y-intercept or constant term and $\beta_1$ and $\beta_2$. For a two coefficient model with only $\beta_0$ and $\beta_1$, the standard errors of coefficients estimators are:

\begin{equation} SE(\hat{\beta_0}) = \sqrt{\frac{\sum x_i ^2 \sigma^2}{n \sum x_i^2 - (\sum x_i) (\sum x_i)} } \end{equation}

and

\begin{equation} SE(\hat{\beta_1}) = \sqrt{\frac{n \sigma^2}{n \sum x_i^2 - (\sum x_i) (\sum x_i)}} \end{equation}

The residual sum of squares, $\sigma^2$, is 111.30. For 3 total variates, the answer will be a variation on a theme.

Mentossinho
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