Diff for "FAQ/semsd" - CBU statistics Wiki
location: Diff for "FAQ/semsd"
Differences between revisions 7 and 9 (spanning 2 versions)
Revision 7 as of 2013-04-10 09:49:33
Size: 1711
Editor: PeterWatson
Comment:
Revision 9 as of 2013-04-10 13:44:30
Size: 2217
Editor: PeterWatson
Comment:
Deletions are marked like this. Additions are marked like this.
Line 12: Line 12:
The variance of the mean = 1/N x variance of the response = 1/N x The variance of N randomly sampled responses from a parent population = 1/ N^2 x (Nsigma^2) The variance of the mean = 1/N x variance of the response = 1/N x The variance of N randomly sampled responses from a parent population = 1/ N^text{2} x (Nsigma^2)
Line 19: Line 19:
So it follows that the one sample z-test statistic = mean / s.e.(mean) will increase with increasing sample size because the mean stays the same and the s.e.(mean) decreases. So it follows that the one sample t-test statistic = mean / s.e.(mean) will increase with increasing sample size because the mean stays the same and the s.e.(mean) decreases. This also follows from the fact that as the sample size increases a t random variable converges to a value having a standard Normal distribution (z-value). If any z-value is squared it becomes a value having a chi-square distribution (with one degree of freedom) and chi-squares increase with sample size (see e.g. Howell p.157) so the square of the mean divided by its standard error goes up with N.
Line 21: Line 21:
__Reference__ __References__
Line 25: Line 25:
Howell, D.C. (1979) Statistical Methods for Psychologists. Fourth Edition. Wadsworth:Belmont,CA.

How do I obtain the standard deviation from the standard error of the mean (s.e.m.) and how does this and the mean vary with sample size?

$$\frac{\mbox{The standard deviation}}{\sqrt{\mbox{sample size}}}$$= standard error of the mean

i.e.

standard deviation = $$\sqrt{\mbox{sample size}} \mbox{multiplied by the standard error of the mean}$$

It follows from this (e.g. p.218 of Babbie) that the standard error fo the mean decreases with sample size, N.

This follows since: The variance of the mean = 1/N x variance of the response = 1/N x The variance of N randomly sampled responses from a parent population = 1/ Ntext{2} x (Nsigma2)

= sigma2/N where sigma2 is the (unobserved true) variance of the (parent population of the) response. Since this involves a term 1/N the variance (or its square root = standard error) of the mean decreases with sample size.

The mean (= to the midpoint or median in a Normal distribution) on the other hand is not proportional to sample size so is uneffected by N. One can see this easily by considering an example: suppose we have a sample of size 3 of a responses = 1 2 3 then the mean is 2. Suppose I take a sample of size 7 say of the same response and get values of 1 1 2 2 2 3 3 then the mean = 2 there since it is symmetric about the (hypothesised true) mean of 2 (which follows from sampling from a response following a normal distribution).

So it follows that the one sample t-test statistic = mean / s.e.(mean) will increase with increasing sample size because the mean stays the same and the s.e.(mean) decreases. This also follows from the fact that as the sample size increases a t random variable converges to a value having a standard Normal distribution (z-value). If any z-value is squared it becomes a value having a chi-square distribution (with one degree of freedom) and chi-squares increase with sample size (see e.g. Howell p.157) so the square of the mean divided by its standard error goes up with N.

References

Babbie, E. (2008). The Basics of Social Research. Fourth Edition. Thomson Wadsworth: Belmont.CA.

Howell, D.C. (1979) Statistical Methods for Psychologists. Fourth Edition. Wadsworth:Belmont,CA.

None: FAQ/semsd (last edited 2013-08-30 09:34:03 by PeterWatson)