1. ## Population and Sample

Dear All,

I am very new to statistcs and just started learning it.
The concept of SD is defined in two ways - separately for the whole i.e. population and the sample thereof. In first case the square sum of variance from mean is divided by "n" i.e. total number of variables, but the second case is slighty different since it propose such division by "n-1".

What is the actual significance of this difference? Yes the concept of degree
of freedom is there to explain it,i as a newcomer i feel it little bit unintelligible.
Is there any simpler approach or analogy to grasp the pith of the concept?

My simple query is what if the dividor is "n" in both the case? What, otherwise, adverse bearing will be on the result?

Kind Regards

Gulam Dastgir

2. It's all about the bias of the estimator.

Let $\hat{\sigma^{2}} = \frac {1}{n} \sum^{n}_{i=1} (X_{i} - \bar{X})^{2}$.

Then $\hat{\sigma^{2}}$ is biased estimator of $\sigma^{2}$

In other words, $E(\hat{\sigma^{2}}) \neq \sigma^{2}$ (The expected value is actually $\frac {n-1}{n} \ \sigma^{2}$ .)

$S^{2} = \frac {1}{n-1} \sum^{n}_{i=1} (X_{i} - \bar{X})^{2}$ , on the other hand, is an unbiased estimator of $\sigma^{2}$ because $E(S^{2}) = \sigma^{2}$

3. Dear Random Variable,

Will you be little bit more explanatory in simplistic way?

Kind Regards,

Gulam Dastgir

4. Originally Posted by dastgir
Dear Random Variable,

Will you be little bit more explanatory in simplistic way?

Kind Regards,

Gulam Dastgir
$\frac {1}{n} \sum^{n}_{i=1} (X_{i} - \bar{X})^{2}$ tends to underestimate $\sigma^{2}$ because a sample (at least not a very large one) tends not to have data points from the fringes of the distribution.

5. Dear Random Variable,
The sample variance and population variance are the same the but factor
∑(X-Xm)^2/(n). My instrest is where this factor come from? What if do not
include this? How this relationship is arrived at?

Kind Regards,

Gulam Dastgir