# How to determine statistical validity of results

I have two exclusive groups of people and a counter of how many events happened for each group.

Lets say group 1 has 7000 people and group 2 has 3000 people. group 1 had 50 events and group 2 had 40 events.

I'm calculating the event percentage for each group for example for group1 its 50/7000. for group 2 its 40/3000.

I want to calculate how much statistical validity these results have (in other words is the groups large enough or I need to collect more data). Probably in percentage (where >95% means its valid statistically)

Can someone point me to how to do it. I need to implement it in PHP code. I have little statistics knowledge. I think it involves square chi function but I'm not sure how to use the data with PHP chi square function http://www.php.net/manual/en/function.stats-cdf-chisquare.php

Thanks!

ADDITIONAL INFO: We're talking about visitors to a web store. I divide them randomly to 30% group B (test group) and 70% group A. I expose gruop A to a certain message. I compare the conversion rates of the groups (% of visitors who buy something). And I want to know when the samples are large enough to be statistically significant.

• Check wikipedia and then come back and clarify your question. It might be that you mean reliability. – John Mar 2 '11 at 13:00
• how were the groups chosen? What sort of events are we talking about? Are they "intrinsic" characteristics of the people sampled or choices they made on the particular day they were sampled or something else? Is each group homogeneous or heterogeneous? The answers to questions like these will provide a starting point to addressing your question. – cardinal Mar 2 '11 at 13:12
• Sounds like you're after a chi-squared test of independence. It would be much easier to do this in a statistics package than in PHP, especially as the PHP function isn't currently documented. – onestop Mar 2 '11 at 13:20

You are calculating the mean of a variable that is 0 if no event and 1 if there is an event. The sum of $N$ such (independent) binomial random variables has a variance $N\times p(1-p)$. The mean has a variance $p(1-p)/N$. We can use a two-sample difference in means test to see whether the difference in proportions between the groups is significant. Calculate:
$$\begin{equation*}\frac{p - q}{\sqrt{p(1-p)/N + q(1-q)/M}}\end{equation*}$$
where $p$ is the proportion from group 1, which has $N$ observations, and $q$ is the proportion from group 2 with $M$ observations. If this number is large in absolute value (bigger than 1.96 is a typical norm, giving a hypothesis test with a significance level of 5%), then you can reject the claim that the two groups have the same proportion of events.