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I just wanted to know how to explain "degrees of freedom" to a non-statistics person without mentioning any statistical terms. How to explain this?

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  • $\begingroup$ In some cases, referring them to a text on mechanics would be sufficient. I doubt this is quite what you are seeking. $\endgroup$ – Nick Cox Mar 21 '15 at 9:31
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    $\begingroup$ I think the emphasis on avoiding statistics terms makes this a bit different. The question seems to me self-contradictory in that there is little point to, or little scope for, explaining degrees of freedom without linking it to the statistics a person already knows or needs to know, but that in turn is a different issue. $\endgroup$ – Nick Cox Mar 21 '15 at 10:56
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You get the next 7 days off work, but you use the first day planning the rest of the days, so you have 6 days free.

"the number of observations minus the number of necessary relations among these observations." -Walker

Degrees of freedom can be very complex though and contrary to popular belief, see this answer for more.

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Definition for Layman

Degrees of freedom is the number of values that are free to vary when the  value of some statistic, like $\bar{X}$ or $\hat{\sigma}^2$, is known. In other words, it is the number of values that need to be known in order to know all of the values.

I will provide two examples. The first example is appropriate for a layman with a mathematical mind and the second example is probably not good for a layman but I give it anyway.

Example with $\bar{X}$

As a very simple example, consider $\bar{X}$ with a sample size of $n$. If you know the values $X_1, X_2,..., X_{n-1}$ then you also know the value of $X_n$. So, we say that $\bar{X}$ has $n-1$ degrees of freedom.

$$X_n = n(\bar{X}) - \sum\limits_{i=1}^{n-1} X_i$$

Example with Simple Linear Regression

As a slightly more difficult example, consider the simple linear model $Y_i =\alpha+ \beta x_i + \epsilon_i$ for $i=1,...,n$. Recall the following two identities in linear regression

$$e_1 + e_2 + ... + e_n = 0$$

$$x_1e_1 + x_2e_2 + ... + x_ne_n = 0$$

where the residuals $e_i = \hat{Y}_i - Y_i$. Notice, if we know $e_1,..., e_{n-2}$ but $e_{n-1}$ and $e_n$ are unknown then the above identities give us two equations with two unknowns. Since the number of equations and the number of unknowns is the same, we can solve for $e_{n-1}$ and $e_n$. So, knowing the $n-2$ values  $e_1,..., e_{n-2}$ allows us to know all of the values $e_1,...,e_n$ and so the residuals have $n-2$ degrees of freedom.

This can be easily generalized to a multiple linear regression situation with $p$ variables if one recalls that $X^Te = 0$.

 

 

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  • $\begingroup$ Reasons for downvotes are always welcome so the OP can improve their answer. $\endgroup$ – Momo Mar 21 '15 at 14:27
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    $\begingroup$ I guess someone (not me) voted this down for not answering an impossible question. I reversed the downvote, as I think this could be useful to many people. $\endgroup$ – Nick Cox Mar 21 '15 at 15:15
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The Wikipedia article Degrees of freedom (statistics) is pretty good at explaining it (see the first few paragraphs).

Imagine you have some system or a black box which behavior is defined by a number of values (or parameters). The concrete values of the parameters set one particular mode of operation. Another set of parameters will define another mode of operation.

The catch here is that even for a simple system (say with one parameter only) you can "create" many versions of that parameter (e.g. multiple, added constant etc). Here the degrees of freedom is minimal number of parameters that are independent and are free to vary (under some constraints if any).

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