Why square root of n for standard error?

#1
Hi,

I am trying to understand the formula for the standard error of the mean. OK so far, but:
I don't understand why it is the square root of n instead of n which is used in the
denominator - and I couldnt find a stats book which explains the formula...

any help & comments are greatly appreciated!

Nina
 
#3
Hey Nina!

I was just trying to work out that same question a few weeks ago. I think I may have found the answer. I am pretty confident in my thinking. Take a look at what I wrote down and see if you see the answer to your question! There are some basic algebra tricks and basic stat concepts that should reveal the answer.

Hopefully, if anyone sees any flaws in my workings they could point them out.

Another thing: The reason so many people become confused at this stuff is because it is EXTREMELY easy to mix up which variables and distributions you are dealing with...even at the most basic level. Some of the equations will make no sense at all if you confuse something as simple as...the "distribution of the variable under consideration" such as heart rate, versus the "sampling distribution of heart rate (distribution of sample means)."

I think it is easiest to understand why by starting with understanding the Population Variance. So I basically derived the expression of Standard Error of the Mean from taking population variance and dividing by n, then just doing some basic algebra. I did it in practically 2 steps, "1) and 2)" as shown in the picture.

Let me know if you have any other questions, I hope I could help!

Take a look:

 
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Dragan

Super Moderator
#4
Hi,

I am trying to understand the formula for the standard error of the mean. OK so far, but:
I don't understand why it is the square root of n instead of n which is used in the
denominator - and I couldnt find a stats book which explains the formula...

any help & comments are greatly appreciated!

Nina

If X_1, X_2,....,X_n are independent observations from a population with mean Mu and standard deviation Sigma, then the variance of the Total is:
T = (X_1 + X_2 + ...+ X_n) = n*Sigma^2.

The variance of T/n must be (1/n^2)*n*Sigma^2 = Sigma^2/n.

The standard deviation of T/n must be Sigma/Sqrt[n], where T/n is the sample mean.
 
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#5
Hi,

I am trying to understand the formula for the standard error of the mean. OK so far, but:
I don't understand why it is the square root of n instead of n which is used in the
denominator - and I couldnt find a stats book which explains the formula...

any help & comments are greatly appreciated!

Nina
Variability (or difference) can be measured in different ways..sometimes we use SS (sums of squares, sometimes variance, sometimes standard deviation). They all measure variability in different units of measurement. We tend to use standard deviation when we compare means since those are in the same units of measurement. (ie we compare means when we do t-tests..the numerator of the formula)

So for your question...you can use s2 (variance) divided by n then take the square root..or sample standard deviation (s) over the square root of n. They both mean the same thing.
 
#6
Hi,

I am trying to understand the formula for the standard error of the mean. OK so far, but:
I don't understand why it is the square root of n instead of n which is used in the
denominator - and I couldnt find a stats book which explains the formula...

any help & comments are greatly appreciated!

Nina
I have always had that very same question; but never got a satisfactory answer. I don't see any answer explaining WHY. It makes no sense to me.
 

Dason

Ambassador to the humans
#7
I have always had that very same question; but never got a satisfactory answer. I don't see any answer explaining WHY. It makes no sense to me.
Let's turn it around. Can you give a satisfactory answer as to why it should be 'n' instead?

On a related note: If I increase the area of a square by a factor of 4 how much did I increase the side length by?