# Standard error and confidence interval at a point for the fitted value

I have a simple linear regression model $$G: Y = \beta_0 + X\beta_1 + \epsilon$$. I have found least square estimates for the coefficients, i.e. $$\hat\beta_0 = 32.1359$$ and $$\hat\beta_1=-14.5388$$. I have also found that the standard error for the estimate is $$\hat \sigma$$=2.72.

I have to find the standard error at a point $$x=1.63$$ and construct a 95% confidence interval for the expected value at that same point.

What I know is that $$\hat Y \in N_n(X\beta, \sigma^2I_n)$$, so $$Y_{1.63}$$ should also be normally distributed with expected value $$\mu=(1, 1.63)\beta$$ and variance $$\sigma^2$$.

How do I find such standard error and such confidence interval?

Assuming the errors are i.i.d $$N(0,\sigma^2)$$ with $$\sigma$$ unknown.

The mean response at $$x=x_0$$ is $$E(y\mid x_0)=\beta_0+\beta_1x_0$$

We estimate $$E(y\mid x_0)$$ from the fitted model by $$\hat\mu_{y\mid x_0}=\hat\beta_0+\hat\beta_1x_0$$

Show that

\begin{align} \operatorname{Var}(\hat\mu_{y\mid x_0})&=\operatorname{Var}(\hat\beta_0)+x_0^2\operatorname{Var}(\hat\beta_1)+2x_0\operatorname{Cov}(\hat\beta_0,\hat\beta_1) \\&=\sigma^2\left[\frac{1}{n}+\frac{(x_0-\bar x)^2}{S_{xx}}\right] \end{align}

, where $$S_{xx}=\sum\limits_{i=1}^n (x_i-\bar x)^2$$ as usual.

Since $$\sigma$$ is not specified, the estimated standard error of $$\hat\mu_{y\mid x_0}$$ is $$\text{S.E.}=\hat\sigma\sqrt{\frac{1}{n}+\frac{(x_0-\bar x)^2}{S_{xx}}}$$, where $$\hat\sigma^2=\frac{1}{n-2}\sum\limits_{i=1}^n(y_i-\hat\beta_0-\hat\beta_1 x_i)^2$$ is the residual variance.

Since $$(\hat\beta_0,\hat\beta_1)$$ is jointly normal, we have $$\frac{\hat\mu_{y\mid x_0}-E(y\mid x_0)}{\sigma\sqrt{\frac{1}{n}+\frac{(x_0-\bar x)^2}{S_{xx}}}}\sim N\left(0,1\right)$$

The above again is independent of $$\frac{(n-2)\hat\sigma^2}{\sigma^2}\sim \chi^2_{n-2}$$

So you have the pivot $$\frac{\hat\mu_{y\mid x_0}-E(y\mid x_0)}{\hat\sigma\sqrt{\frac{1}{n}+\frac{(x_0-\bar x)^2}{S_{xx}}}}\sim t_{n-2}$$

Hence a $$100(1-\alpha)\%$$ confidence interval for $$E(y\mid x_0)$$ has confidence limits $$\hat\mu_{y\mid x_0}\mp (t_{\alpha/2,n-2})\hat\sigma\sqrt{\frac{1}{n}+\frac{(x_0-\bar x)^2}{S_{xx}}}$$

Relevant threads (also check out the linked posts):