# Explain how do we get standard error.

1. Mar 1, 2013

### Outrageous

1. The problem statement, all variables and given/known data
I know that standard error is to calculate how the sample deviate from the population.
But standard deviation divide √n ,then will get it.

2. Relevant equations
Standard error x sqrt(n) = Standard deviation

3. The attempt at a solution

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2. Mar 1, 2013

### haruspex

I'm not sure whether you understand that the "variance of T" is not the variance of a single sample of n; it's the variance of a random variable T each value of which is defined to be the sum of n independent trials of X.
I'm also unsure what you mean by standard error here. Standard error is usually used in reference to some statistic, such as an estimate of the mean. Your attachment doesn't mention standard error.
I assume that T/n is of interest because it could be used for estimates of the mean of X. The standard deviation of the r.v. T/n would then be the standard error of the mean so estimated.

3. Mar 1, 2013

### Outrageous

Not really understand.
Can you please teach me this first:
A sample of 5 students from 50 students.
The score of 5 students for math are recorded.
Mean of 5 students is 62.
sample σ from 5 students =27.44
The standard deviation of 50 students is 17.
Standard error = 17/√5.

What is the standard error here used for ?
This represents how well this sample deviate from the population?

So if I can get the mean of 50 students ,x
Then x-62 will be standard error?

4. Mar 1, 2013

### haruspex

If I'm reading that correctly, you happen to know (by some means) that the s.d. of the whole population is 17. If you attempt to estimate the mean of the whole population by taking a sample of 5, the standard error in the resulting number is 17/√5. That is, if you were to take lots of samples of 5 and look at the calculated means, their standard deviation would be about 17/√5.
But notice I have not made use of the 27.44. In practice, if you don't know the mean of the whole population then it's very unlikely that you know its s.d. So you'll be estimating both the mean and the s.d. from the sample.
No, standard error is the square root of the expected variance of the calculated statistic. Since the underlying population has a variance of 172, the sum of 5 samples has an expected variance of 5*172, and the average of 5 samples has an expected variance of (5*172)/52 = 172/5. The square root of that is 17/√5.

5. Mar 2, 2013

### Outrageous

Thank you.

This is totally same as what I uploaded, I don't understand why the average of 5 samples has an expected variance of (5*172)/52 = 172/5 ?

6. Mar 2, 2013

### haruspex

Average of N samples is ƩXi/N. Expected variance of this is E[(ƩXi/N)2] - (E[ƩXi/N])2 = (E[(ƩXi)2] - (E[ƩXi])2)/N2.
Leaving off the factor 1/N2 for now:
E[(ƩXi)2] - (E[ƩXi])2 = E[ƩXi2i≠jXiXj]-(ƩE[Xi])2 = E[ƩXi2]+E[Ʃi≠jXiXj]-(Nμ)2 = ƩE[Xi2]+Ʃi≠jE[XiXj]-(Nμ)2
= N(σ22)+N(N-1)μ2-(Nμ)2 = Nσ2
So expected variance of average is σ2/N

7. Mar 4, 2013

### Outrageous

Is that the definition? I mean we must define the variance in this way?

8. Mar 4, 2013

### haruspex

The variance of a set of numbers is the mean square difference from the mean: Ʃ(xi-μ)2/N, where μ = Ʃxi/N. It's not hard to show that this is the same as (Ʃxi2/N) - μ2. Or, writing in terms of expectations of a r.v., E[X2]-(E[X])2.

9. Mar 6, 2013

### Outrageous

I am not sure the expected is the mean or?
Can you please guide more .
Really thank

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