Standard Deviation of a sample of a population's means

In summary, the standard deviation in a sample of a population can be derived from the mean and is equal to \sigma/\sqrt n when the population is normally distributed. This is due to the fact that the variance of the sample is equal to \sigma^2/n, making the standard deviation \sigma/\sqrt n. This derivation can also be applied if the variables in the population are not normally distributed under certain conditions.
  • #1
2^Oscar
45
0
Hey guys,

Just a question which has been puzzling me for some time.

I am told that means of samples of a population can be normally distributed with mean of [tex]\mu[/tex] and with standard deviation of [tex]\sigma[/tex]/[tex]\sqrt{}[/tex]n

Can someone please explain to me how the standard deviation is derived or is it ismply as a result of experimentation?

Thanks,
Oscar
 
Physics news on Phys.org
  • #2
It is derived. The given result follows if the population is normal.
 
  • #3
The mean is

[tex]
\overline X = \frac{\sum X}{n}
[/tex]

If all of the [tex] X [/tex] variables are independent, and identically distributed, then

[tex]
Var[X] = Var\left(\frac{\sum X}n\right) = \frac 1 {n^2} \sum Var(X) = \frac{n\sigma^2}{n^2} = \frac{\sigma} n
[/tex]

so the standard deviation is as you state.

If the [tex] X [/tex] values are themselves normally distributed, then the mean is as well (linear combinations of normally distributed normal random variables are normal)

If the [tex] X [/tex] values are not normally distributed, the distribution of [tex] \overline X [/tex] is approximately normal if certain conditions are satisfied.
 
  • #4
statdad said:
[tex]
Var[X] = Var\left(\frac{\sum X}n\right) = \frac 1 {n^2} \sum Var(X) = \frac{n\sigma^2}{n^2} = \frac{\sigma} n
[/tex]

You forgot the root in the denominator.

CS
 
  • #5
stewartcs said:
You forgot the root in the denominator.

CS
No, actually I forgot the square in the numerator, since I was writing out the variance. :frown:
Still a stupid mistake on my part.

The variance is

[tex]
\frac{\sigma^2} n
[/tex]

so the standard deviation is

[tex]
\frac{\sigma}{\sqrt n}
[/tex]
 
  • #6
statdad said:
No, actually I forgot the square in the numerator, since I was writing out the variance. :frown:
Still a stupid mistake on my part.

The variance is

[tex]
\frac{\sigma^2} n
[/tex]

so the standard deviation is

[tex]
\frac{\sigma}{\sqrt n}
[/tex]

Sorry...it appeared as if you were answering the OP's question about the standard deviation directly which is why I assumed you had finished the derivation all the way out to the standard deviation and not just the variance.

CS
 
  • #7
stewartcs; there is no need for you to apologize and I certainly did not mean to imply (in my earlier post) that there was. I am sorry if it seemed that way.
 
  • #8
Hey guys,

Thanks very much for showing me the derivation - it has helped make things a lot clearer :)

Thanks again,

Oscar
 

1. What is the purpose of calculating the Standard Deviation of a sample of a population's means?

The Standard Deviation of a sample of a population's means is a statistical measure that indicates how much the means of different samples from the same population vary from each other. It helps us understand the spread or variability of the data and is often used to make inferences about the entire population based on a smaller sample.

2. How is the Standard Deviation of a sample of a population's means calculated?

To calculate the Standard Deviation of a sample of a population's means, first find the mean of each sample. Then find the difference between each sample mean and the overall mean of all the samples. Square each of these differences, add them together, and divide by the number of samples. Finally, take the square root of this value to get the Standard Deviation.

3. What is the difference between Standard Deviation and Standard Error of the mean?

The Standard Deviation and Standard Error of the mean are both measures of variability, but they have different purposes. Standard Deviation measures the spread of data within a sample or population, while Standard Error of the mean measures the precision of the sample mean as an estimate of the population mean. Standard Error is typically used in hypothesis testing, while Standard Deviation is used in descriptive statistics.

4. How does the size of the sample affect the Standard Deviation of a population's means?

The larger the sample size, the more precise the estimate of the population mean will be. As the sample size increases, the Standard Deviation of the population's means will decrease, as the sample means will be closer to the population mean. This is known as the Central Limit Theorem, which states that as sample size increases, the distribution of sample means will approach a normal distribution.

5. Can Standard Deviation ever be negative?

No, Standard Deviation cannot be negative. By definition, it is the square root of the average of squared deviations from the mean, which will always result in a positive value. If you come across a negative value for Standard Deviation, it is likely due to an error in calculation or data entry.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
778
  • Set Theory, Logic, Probability, Statistics
Replies
5
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
943
  • Set Theory, Logic, Probability, Statistics
Replies
18
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
3
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
24
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
980
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
699
  • Set Theory, Logic, Probability, Statistics
Replies
28
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
9
Views
1K
Back
Top