gal

I've tried to read up on this in my old statistics book, but most of my skills in this field are long gone..

I have 4 samples of batch Y and 2 samples of batch X. How can I test whether E(X) > E(Y)? I suspect I don't have enough observations to test this. I looked up some tables on signed-rank tests and the smallest observations they had were 4 to 5. The outcome should follow the normal distribution though so maybe I can use some other test..

Y: 10269, 16806, 14023, 8307
X: 15332, 20149

BAMF

I never took a statistical class, but I had one in probabilities. Calculate the average and the standard deviation for both X and Y. With average and the stdev you can plot a gaussian random variable for, lets say gX (for X) and gY (for Y).

The next step is something to do with ln, and then division and subtraction, and than an integration. But it is late and I cant remember. In the end you get the probability (between 0 and 1) that X is greater than Y.

I did this once in a research paper comparing the predicted resistance to the tested resistance of deep pile foundaitons. (When the predicted resistance is greater than the tested load capacity, then designed foundations could fail. The point was to evaluate the reliability of various prediction methods).

rsl12

I'll bump this one. I'm clueless, but I'm also interested. I sometimes run into problems with small data sets myself, and would like a statistician's POV on what to do.

My uneducated guess is just to compute means and leave it at that....

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sapiens

I seriously doubt that you would find a significant difference between the two groups regardless of what test you choose. Your sample sizes are miniscule.

BAMF

I have this written down on a piece of paper in my office. I will bring it home tommorow. From what I remember, the integral could not be solved analytically. So a numerical approximation is used at the end, probably gaussian quadrature. And sample size is irrelevent. All that is reqruied is the mean and standard deviation.

The biggest assumption is that the data will comply to a gaussian random variable. This is applicable for people heights, etc... However, it is not applicable for linear reandom (like rolling a dice).

And another thing, the result is not a test that X > Y. It is the probability that if you sampled X and Y one more time, the X result would be greater than the Y.

BAMF

Of course, the quality of the mean and standard deviation depend on the sample size. Maybe another problem would be finding the fractal dimension of the mean and stdev versus number of samples. But that was just an idea that came to me now; I do not know if there is a procedure in existance for determining how many samples need to be taken for quality data.

BAMF

Oh, gaussian and normal distributions are the same thing. Its the bell curve on the pdf graph.

lpj8

It depends on the nature of your variables, and what assumptions you can make about the data. It looks like you have one dependant variable, and one independent variable. Your dependant variables looks like it is continuous, and your independent variable looks nominal.
It seems like an Independent Samples T-test would answer your question. This would tell you if the difference of the means of Y and of X is statistically significant. However, due to your sample size (n=2, and n=4; N=6), you will not have enough statistical power to find any differences that may exist.

gal

Thanks all. The numbers are cycles to failure for two different batches of material, and the independend variable is the load level which is the same for all the outcomes here. The error is usually assumed to follow a normal distribution, but in this case I don't think I can say anything about the pdf. That's why I was looking into signed rank tests, but in this case I guess the sample size is too small to make any inferences.

Slavakion

You could use the t-score approximation. It's for estimating the mean of the population from small samples.

Statistically, there is a 90% chance that the population mean lies
Y: between 7879.97 and 16822.53
X: between 2533.25 and 32947.75

If you had a larger sample, it wouldn't be such a terrible range.

skier

If you compare the average mean from X (17740.5) and the average mean for Y (12351.25) you can see that X is larger than than Y. But it doesn't really mean very much with the small sample size and huge deviation (X is 3406.13 and Y 3800.49). There is huge room for error- as you can see in slavakion's t-score.

Looking at the p-valueof a 2 sample t-test it seems to me that there's a .75 probability that x is larger than y.

Of course, i'm pretending like I know this stuff- but i'm just taking the class this semester. So take from it what you will.

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Slavakion

Yeah, me too. But it was good practice for my next test.

BAMF

I did this recently, so im posting even though this is old.

Assuming X and Y and gaussian (normal/bell curve) Random Variables.
G = Y - X
Therefore G is a gaussian random variable.
the probability that G is less than 0 equals the probability that X > Y

Mean for G:
E[G] = E[Y] - E[X]

Var[X] = E[(X-E[X])^2] = stdev[X]^2
Var[Y] = E[(Y-E[Y])^2] = stdev[Y]^2

Var[G] = E[(G-E[G])^2]
Var[G] = E[(Y-X-(E[Y]-E[X]))^2]
Var[G] = E[((Y-E[Y])-(X-E[X]))^2]
Var[G] = E[((Y-E[Y])^2+(X-E[X])^2-2*(Y-E[Y])(X-E[X]))]
Var[G] = Var[Y] + Var[X] - 2*COV[X,Y]
If X and Y are independant then the covariance equals zero.

You now have the stdev and mean of G, and can use the characteristic function of a gaussian random variable to find the probability.

For the numbers you provided,
E[G] = mean of G = -5389.25
Var[G] = variance of G = 26045501
Stdev[G] = sqrt(var[G]) = 5103.479

Using E[G]/Stdev[G] = -1.056 and a table for characteristic function values, the probability that X > Y = 0.8531

In other words, 85%.

Correct me if you can.

Astrocloud

The thing that drives me nuts about these theoretical problems is that the problem is solved with more data. Right now we are talking about confidence levels of a weak correlation that Set X belonging in Set Y. How do we know that set Y doesn't represent a measure of cloud formations and Set X represent some measure of the mating patterns of sea horses? i.e. what exactly are we testing?

I have a very good friend who has a PhD in stats. I will see if he will entertain this.

BAMF

I assumed independant. Thus correlation coefficient is zero.

gal

Heh, nice to see I'm inspiring homework. I dunno about using the standard normal, BAMF. The t-test should be used for small samples, and back in the days I remember I had to use non-parametric methods for even smaller samples.. but I can't remember why.

Astrocloud

ANOVA test is what is typically done for normal sized samples. You MUST use the t-test but your degrees of freedom go through the roof. -from the prof.

BadNick

I think statistics are against me today. I came in here to see what these tests of men with small nipples might be all about.