1
$\begingroup$

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:

enter image description here

$$-2.367=\frac{s_{xy}}{s_{xx}}$$ 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.

Additional:

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

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

Recall that if you define:

$$s_{xx} = \sum (x - \bar{x})^2 = \sum x^2 - \frac{(\sum x)^2}{n}$$ $$s_{yy} = \sum (y - \bar{x})^2 = \sum y^2 - \frac{(\sum y)^2}{n}$$ $$s_{xy} = \sum (x - \bar{x})(y - \bar{x}) = \sum xy - \frac{(\sum x)(\sum y)}{n}$$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$
    – tesla john
    May 31 at 18:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.