# Power analysis on chi-squared test with low cell counts

I am hoping to perform a chi-square test of independence on data in a 2x2 contingency table with the following values:

Group A: 627 successes, 28 failures
Group B:  59 successes,  2 failures


I understand that this is a small collection of observations, and I wish to determine the minimum number of observations required for a chi-squared test at 0.05 significance, 0.8 power. However, it seems that the normal approximation of the binomial distribution that is generally used for power analysis among comparisons of proportions is not appropriate here for at least two reasons:

1. p(success) is nowhere near 0.5 for the populations AND
2. the expected number for Group B Failures, (2+59)*(2+28)/716 = 2.56, is below the guideline of 5 for all cells in chi-square power analysis.

Any constructive guidance on how to perform this test of independence (or reference to other questions I may have overlooked) is greatly appreciated.

• By the time you get to a sample size as large as needed for your stated requirement, the normal approximation will be just fine. – rvl Feb 17 '15 at 18:19
• Thank you for your quick response, @rvl. Could you please clarify your meaning a bit? Specifically, are you saying that this sample size is large enough for use with power analysis? If so, how did you arrive at that conclusion? Thank you. – ArcticHare Feb 17 '15 at 18:42
• This is a FAQ: many, many people seem to confuse the rule of thumb that most expected counts should exceed 5 with a non-rule that most actual counts should exceed 5. Only the former is correct. You can find many references to this by searching our site. – whuber Feb 18 '15 at 16:24
• Sorry it took me so long to respond, but NO, I definitely do not mean that you already have enough data. I mean that if the sample size is made large enough to meet you power requirements based on the normal approximation, the result will likely be a big enough $n$ for the normal approximation to work well. You can of course verify this by examining the expected counts and seeing if they meet the rule of thumb. – rvl Feb 19 '15 at 20:35

Your lowest probability is $3.3\%$, and your lowest count is $6187$, meaning that the expected count would be $204\gg 5$. Thus, @rvl is right: you won't have to worry about that assumption. (You will have to worry about needing so much data to have reasonable power, I suspect.)