# Thread: Standard error - why sqrt of n?

1. ## Standard error - why sqrt of n?

I happened to be browsing through some standard explanations of confidence intervals and came upon the formula for standard error...

t x (stddev / sqrt of n)

I couldn't quite grasp why we would use sqrt of the sample size. Yes, a stats 101 question, but my brain can't seem to go that far back. Can someone explain this in simple terms? Thank you.

2. Originally Posted by jawon
I happened to be browsing through some standard explanations of confidence intervals and came upon the formula for standard error...

t x (stddev / sqrt of n)

I couldn't quite grasp why we would use sqrt of the sample size. Yes, a stats 101 question, but my brain can't seem to go that far back. Can someone explain this in simple terms? Thank you.
stddev is the standard deviation of one observation from the population.
But then you take average of n observations(sample mean), standard deviation sample mean will smaller than stddev
(sample mean should vary less from sample-to-sample, when sample size is large we will get approximately same sample mean every time )
and it will be
stddev / sqrt of n

if you need more understandingon this, do a simulation study.

3. Xm, nice one to read

>
http://yusung.********.com/2008/09/s...deviation.html

********=blog spot !!! Since when that's banned?

4. is the ******** means blog spot ?

5. Ha, I thought the ban only worked for hyperlinks

6. Hmmm. Thanks for the replies. I think my question was not very clear. Let me try this...

For the sample size, why do you square-root?

I of course understand mathematically what the result of the formulas are, but I'm trying to get a more general explanation why you would square root. Sorry if this is a silly naive question... may not be worth any more of anyone's time!

7. Originally Posted by jawon
Hmmm. Thanks for the replies.

I think my question was not very clear. Let me try this...

For the sample size, why do you square-root?

... I'm trying to get a more general explanation why you would square root.

Let X1,...Xi,...,XN form a random sample from a population with variance Sigma^2. Let the sample mean be XBar=(X1+...+Xi+...+XN)/N.

The variance of XBar is: (Sum of each variance of Xi + Sum of all covariances between Xi and Xj)/N^2.

Because each pair of Xi and Xj are uncorrelated, each covariance is zero.
So the variance of XBar is the sum of all the variances of Xi/N^2.

Each component Xi of the sample has the same variance Sigma^2,

the variance of XBar is Sigma^2/N.

Thus, the standard error of XBar is Sigma/Sqrt[N].

How's that?

8. when you square root n, is n the sample mean or the sample of the problem. For example finding the standard error when the standard deviation is 18, sample mean is 45 and the sample is 100. What would you square root?

9. Originally Posted by tiffnnie01
when you square root n, is n the sample mean or the sample of the problem. For example finding the standard error when the standard deviation is 18, sample mean is 45 and the sample is 100. What would you square root?
18 divided by 10 ( sqrt of 100) (multiplied by t) is the standard error, I think.

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