# Analysing sample's

1. Aug 16, 2009

### Swatje

Hi,

I was wondering, if you have a certain sample set. How can you quantify the probability of a certain statistical fluctuation?

For example, let's say you have a poisson distribution of high numbers, up into the fifties. You have one peak... How do you determine the probability of this being a statistical fluctuation?

The simple answer would be for me, calculate the the mean, take the square root of the mean as the standard deviance of a poisson, and then see how much sigma's this is. You could then use the gauss tables, (for high numbers the poisson will look like a gauss), and see the probability.

Some things bother me with this, for example: the standard deviance is an estimation, therefor you should use the t-distribution. But the t-distribution requires the knowledge of the actual mean. Would it be better to use the t-distribution anyways, because the lack of knowledge of the standard deviance is a more weighing factor? How ever, if you would use the estimated mean in the t-distribution, you would still be wrong. Is there a distribution that gives values for both an estimated mean and an estimated standard deviation?

My second question lies in the fact, that if a poisson goes to a gauss for high numbers, would it mean you could estimate the standard deviation by using the square root of 1/(N-1) summation rule? Or do these meet at high numbers, so it's always better to use the poisson way?

EDIT: I forgot to ask this:

If you were to analyse this sample, would it be better to count the fluctation into your mean, or leave it out...? So do you consider it a part of your sample set, or do you take it away and see if it is consistent with the rest of your sample set...?

2. Aug 16, 2009

### EnumaElish

The one-sample t-statistic has the form (x - m)/s.e. where x is a normally-distributed random variable (e.g. sample average) and m is a deterministic value; and s.e. is the standard error of x (m is usually the expected value of x, although in regression analysis it's typically zero.)

In contrast, the two-sample t-statistic is (x - y)/s.e.p where x and y are both random variables and the denominator is the pooled standard error of x and y. In this formulation, the t-stat. does not include a deterministic parameter value. (Note that x and y are both normal, therefore x - y is normal.)

3. Aug 17, 2009

### Swatje

1) You have 10 numbers, and one hops out. What you do is you calculate s, you calculate the average, then you look at difference between the average and the "one hopping out", divide by s and then look at the t-value...

t = $$\frac{\bar{x}-x}{s}$$

With 9 d.o.f.

2) You have 10 numbers, and one hops out. But now you already know the actual mean value of your parent distribution, and again, you calculate the t-value in the same way...

t = $$\frac{\mu-x}{s}$$

With 10 d.o.f. (Because you know the mean.)

3) You have 10 numbers, and a different test of 40 numbers gave a certain average, but you do not know the s or sigma of that test, or any of the numbers to calculate it. You want to know if your mean is consistent with the other one, so you again:

t = $$\frac{\bar{x_{1}}-\bar{x_{2}}}{s_{1}/\sqrt{10}}$$

This one bothers me, but I see no other way to do it.

4) Same situation as 3, but now you know the s of the second sample.

(Two-sample pooled t-test, equal variances... A long formula which I'm not going to type out.)

Which methods are correct, and which are wrong?

4. Aug 23, 2009

### EnumaElish

(3) assumes equal variances (for the lack of a better solution), but (4) doesn't have to. Not so?