4
$\begingroup$

I'm reading a textbook and I see this question:

enter image description here

So there are 200 women, and the DF is 196, implying that the equation for DF is $n - k - 1$. There are 3 variables: bp, age, and type so $k == 3$. What's the intuition behind this?

Also, why is the degrees of freedom for linear regression n - 2?

enter image description here

$\endgroup$
  • 3
    $\begingroup$ if n=200 and k=3 df =196 which is n-k-1.as you mention in your question and not 4n-k-1 as you mention inside your question. Get this straightened out and then we can consider the explanation. Use the self study tag. $\endgroup$ – Michael R. Chernick May 1 '17 at 21:03
  • 1
    $\begingroup$ Based on your edit I would say that the example leading to n-2 has only 2 parameters but the text book question has 4 parameters. $\endgroup$ – Michael R. Chernick May 1 '17 at 21:06
  • $\begingroup$ My mistake! I changed it to $n-k-1$ $\endgroup$ – Jwan622 May 1 '17 at 23:24
  • $\begingroup$ Would you please provide the name/link of the book? $\endgroup$ – yaojp Oct 4 '19 at 14:02
13
$\begingroup$

In linear regression, the degrees of freedom of the residuals is:

$$ \mathit{df} = n - k^*$$

Where $k^*$ is the numbers of parameters you're estimating INCLUDING an intercept. (The residual vector will exist in an $n - k^*$ dimensional linear space.)

If you include an intercept term in a regression and $k$ refers to the number of regressors not including the intercept then $k^* = k + 1$.

Notes:

  • It varies across statistics texts etc... how $k$ is defined, whether it includes the intercept term or not.)
  • My notation of $k^*$ isn't standard.

Examples:

Simple linear regression:

In the simplest model of linear regression you are estimating two parameters:

$$ y_i = b_0 + b_1 x_i + \epsilon_i$$

People often refer to this as $k=1$. Hence we're estimating $k^* = k + 1 = 2$ parameters. The residual degrees of freedom is $n-2$.

Your textbook example:

You have 3 regressors (bp, type, age) and an intercept term. You're estimating 4 parameters and the residual degrees of freedom is $n - 4$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Why including an intercept? Where does that even come from? $\endgroup$ – Jwan622 May 1 '17 at 23:23
  • $\begingroup$ @Jwan622 The intercept is another term you have to estimate. $\endgroup$ – Matthew Gunn May 2 '17 at 15:45

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.