# How to find $s_{xx}, s_{xy}$ from linear estimate

If I have the least squares estimate, can I derive either $$s_{xx}$$ or $$s_{xy}$$ from it? For example, given the following linear model: $$-2.367=\frac{s_{xy}}{s_{xx}}$$ would give me the value for the estimate, however without either $$s_{xx}$$ or $$s_{xy}$$ known, I cannot see how I can derive this.

I am testing for the hypothesis that \begin{align}H_0&:\beta_1 = -4 \\ H_1&: \beta_1 \ne -4 \end{align}

and using the following test statitic $$T = \frac{\hat{\beta_i}-\beta_{i0}}{S\sqrt{c_{ii}}}$$

where $$c_{11}= \frac{1}{s_{xx}}$$, but I cannot get this from the model shown.

$$T = \frac{\hat{\beta_i}-\beta_{i0}}{S\sqrt{c_{ii}}} = \frac{-2.367-(-4)}{1.168\sqrt{c_{11}}}$$

When not including $$\sqrt{c_{11}}$$ I get $$t = 1.398$$, with $$t_{8, .025}=2.306$$, there is weak evidence to reject the null hypothesis that $$\beta_1 = -4$$. However, I am unsure whether that I can remove $$\sqrt{c_{11}}$$ from the calculation.

$$s_{xx} = \sum (x - \bar{x})^2 = \sum x^2 - \frac{(\sum x)^2}{n}$$ $$s_{yy} = \sum (y - \bar{x})^2 = \sum y^2 - \frac{(\sum y)^2}{n}$$ $$s_{xy} = \sum (x - \bar{x})(y - \bar{x}) = \sum xy - \frac{(\sum x)(\sum y)}{n}$$
Then the linear least-squares estimate is $$y - \bar{y} = \dfrac{s_{xy}}{s_{xx}} (x - \bar{x})$$ (or in slope-intercept form $$y = mx + b$$, then $$m = \dfrac{s_{xy}}{s_{xx}}$$ and $$b = \bar{y} - m \bar{x}$$), and Pearson's correlation coefficient is $$r = \dfrac{s_{xy}}{\sqrt{s_{xx} s_{yy}}}$$.
So, if you can find $$s_{xx}$$, you could calculate $$s_{xy} = ms_{xx}$$, and then $$s_{yy} = \frac{m^2s_{xx}}{r^2}$$. And you're given $$m = -2.367$$ and $$r^2 = 0.3392$$.
I'm having trouble figuring out a way to get $$s_{xx}$$, though.
• There likely is not a way to find these just from the linear model in R. However, it turns out that including $\sqrt{c_{ii}}$ is not necessary, at least the solutions to my past exam papers suggest so as they do the calculations without it. May 31 at 18:14