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

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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?

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    $\begingroup$ 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
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    $\begingroup$ 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

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 $k^* = k + 1$.


  • 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.


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 $n - 4$.

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  • $\begingroup$ 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

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