I obtained two groups of data from my experiment and calculated the binomial distribution for each group, trying to see if they are significantly different from each other. However, the z sore I got was way smaller than -4, which was a little bit odd to me. Here is how I did my calculation:
The control group: # of trails n1=34; # of successes k1=2, success rate p1=2/34=5.9% The experimental group: # of trails n2=25; # of successes k2=2, success rate p2=2/25=8.0%
average p=6.8% standard error=0.0044 test statistic z=(5.9%-8%)/standard error=-4.77 It's a two-tailed test. I couldn't find a p-value corresponding to the z score I got (-4.77) on the table. Does that mean my p-value was just really small or I made mistakes here?
$$\frac{p_1-p_2}{\text{s.e.} (\bar{p})}$$
Where did you got the idea to use the difference in succesrate divided by the standard error of the average succes rate? (the test which seems most close is the 'z-test to compare the equivalence of two proportions' but it is using an estimate for the standard deviation instead of an estimate for the standard error, and also it does not work well with low number of expected counts)
– Sextus Empiricus Feb 25 '19 at 14:02