Why do we use n-1 for sample denominator?
Posted 22 August 2012 - 01:38 PM
So I understand the concept of degrees of freedom, as in we are not free to choose x many variables. Like if there are three numbers and the mean is 20, the degree of freedom is 2, because if we choose 2 numbers, the third one is defined.
I don't get, however, why sample's of a population use n-1 in the denominator, when populations only use 1.
Lets say this is a population: 1,3,5,2,7,8,6,9,2,7. The mean would be 50/10=5. The variance is 7.2.
Let's take a sample of this population. 5+7+6+8+7.
Let's take a sample of this population. 5+7+6+8+7. The mean would be 33/5=6.6. The variance, using N IN THE DENOMINATOR (which I know is incorrect), would be 1.04. With N-1 as the denominator, the answer is 1.3. I fail to see how 1.3 is any more accurate of the population than 1.04, since neither is anywhere close to 7.2. I guess because it is ever so slightly closer to 7.2?
I don't really get it conceptually.
Thanks in advance.
Also, if the sample was 1,3,7,8,9. The mean would be 5.6. The variance using N would be 9.44. The variance using N-1 would be 11.8. So in this case, N is actually closer to the real deal of 7.2, than n-1 would be
Posted 22 August 2012 - 02:30 PM
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Posted 22 August 2012 - 03:09 PM
since the residuals, , must sum to zero, the entire vector is determined fully by the first N-1 entries.
[14:02] <Sato> want
[14:02] <Sato> Schroedinger
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