Confidence integrals when n is small

[TaliskerBA]

Should Say Intervals.. I'm tired...

I am probably going wrong somewhere but I am running into problems with understanding this. My understanding of a 95% confidence interval is that in a sample of n the sample mean is 95% likely to be within 1.96 standard errors of the actual mean. I have a problem because I think I have an example where this isn't true.

I play a bit of online poker and have been playing around with my results to help me grasp some of these concepts and it is with my poker results I get the contradiction.

In 425 games I have won 58 for $51.50 profit, come 2nd in 53 for $24.50 profit, come 3rd in 41 for $11 profit and not cashed in 273 for -$16. With these figures I work out:

sample mean = $0.87
sample variance = 614
standard deviation = $24.77

So, here is my problem. Say I play another tournament (ie. n=1), based on this sample there is a 58/425 = 13.6% chance that I win $51.50 but a 95% confidence interval states that:

Pr(0.87 - 1.96*24.77 < X < 0.87 + 1.96*24.77) = 0.95

Pr(-47.7<X<49.4) = 0.95

So it suggests that I am 95% likely to have a result that yields me less than $49.40 even though I already know I am 13.6% likely to win $51.50...

I presume it's all to do with n being small but my notes don't give any acknowledgment that confidence intervals aren't completely sound when n is small. Where am I going wrong?

 

[pwsnafu]

Say I play another tournament (ie. n=1)

No no no. You need to talk about http://en.wikipedia.org/wiki/Prediction_interval" for your example.

 

[Hurkyl]

Is X really an (approximately) normally distributed random variable?

 

[TaliskerBA]

No no no. You need to talk about http://en.wikipedia.org/wiki/Prediction_interval" for your example.

That is useful to know, but the notation in this example still claims that Pr(-47.7<X<49.4) = 0.95 which clearly doesn't work when n=1. My notes state that over many repetitions of sampling then 95% of intervals will include X, but what if all samples were of size n=1? Or are the samples sizes themselves random as well (in which case this would most likely work)..?

* Edit. I think I understand now, thanks for your help.

Is X really an (approximately) normally distributed random variable?

I think I get your point now. It's not a normally distributed RV because I have dictated what the sample size should be. Right?

Thanks for your help.

 

[SW VandeCarr]

I think I get your point now. It's not a normally distributed RV because I have dictated what the sample size should be. Right?

Thanks for your help.

No. Sample size is almost always stated a priori. To calculate confidence intervals, you must assume the random sample came from a normally distributed population. Confidence intervals for small samples (typically less than 30) may be estimated from the t score. The standard deviation estimate is based on the number of degrees of freedom.

The standard deviation estimate for the t distribution is $$\sqrt{\frac {n}{n - 2}}$$. You can use a t test table to get the relation between the standard deviation (or just n) and the confidence limits. As with the Z score, confidence intervals extending 2 sd from the mean are considered acceptable for most applications.

EDIT: More specifically to your question the CI is $$\bar X \pm_{Z_{\alpha/2}} \sigma/ \sqrt {n}$$ if n is at least 8.