why do we divide by N-1 and not by N to get Standard Deviation?

Answer 1

if the N numbers represented the entire population, then you would divide by N, since x-bar in the formula for standard deviation would actually be the population mean mu.

However, if the N numbers are a sample of size N taken from a larger population, then the "common sense" standard deviation (dividing by N) is a biased estimator for the true population standard deviation. This can be proved algebraically, but an intuitive explanation that works for me is the following:

Since x-bar was computed FROM the N sample points, then the deviations from x-bar will tend to understimate the deviations from the true population mean. (For example, if the data are all atypically large, x-bar will "follow the data" and be large as well, so the deviations will be moderate.) Thus the sum of the squares of the deviations from the sample mean will tend to underestimate the total squared deviations in the population. Dividing by N will then give an underestimate for the sample variance. As it turns out, dividing by N-1 is the right correction to make the variance just a little larger, so that it is an unbiased estimator for the population variance.

This is the meaning of "degrees of freedom" in the previous post. If you know x-bar and you know N values, then you really only have N-1 independent data values, since knowledge of x-bar allows you to find the remaining value.

Answer 2

Explaining this is not easy. Be kind to your teacher. Most books skip over this, or they call it unbiasing the estimator. I'll try to explain--- A deviation (from the mean) is defined as xi - Xbar, where xi are the individual measurements and Xbar is the sample average. There are n - 1 INDEPENDENT deviations. If you add all these deviations, the answer is always zero. Therefore, the value of any particular xi is always equal to the negative of the sum of the other n - 1 deviations. In other words, if you know n -1 deviations, then the remaining deviation is FIXED and the degrees of freedom is n - 1.

Answer 3

If you have 5 numbers, you take the 5 numbers, add them up and divide the sum by 5 to get the average number which is an artificial number you made up. Every of the 5 numbers contributes to the average so you divide the total by 5. The larger than average reduces its value and the smaller than average increases its value. .

In the standard deviation, you take the 5 numbers. Then you find the difference of each number from the "average number".

So out of the 5 numbers you reduce it to 4 numbers instead of the orginal 5 numbers.


You think you have 5 numbers contributing to the calculation but actually only 4 contribute to the standard deviation.

This is explained by "the sum of the rest of any of the four is equal to the negative of the fifth." The fifth number is not contributing. Once you subtract each number from the average number you automatically loose the contribution of one of them without you knowing it until you think about it. A similar view is that you have five fingers. There are only four gaps between them not five.

Please remember that the key to explain and understand this is that "the sum of the rest of the four is equal to the negative of the fifth." Only four of the five numbers not five contribute to the standard deviation.

Hence if you have n number, only n-1 number is contributing.

The reason is based on the understanding of how the standard deviation is defined whether sample or population.

Answer 4

because you are dividing by degrees of freedom, and you must have one less degree of freedom.

Answer 5

Can you say, "fudge factor"?