Testing equality of two binomial proportions proportion (one near 100 %) I am testing equality of two binomial proportions 87/88 and 48/60. I use
this online calculator  and it noted that the Standard Normal approximation is not valid in this case. It seem quite strange to me. Is it some problem in testing equality of two binomial proportions if one of the proportions is very near to 100 %?
 A: If you trust wikipedia for rules of thumb for the validity of the normal approximation of the binomial distribution:

  
*
  
*One rule is that both $np$ and $n(1−p)$ must be greater than $5$.
  However, the specific number varies from source to source, and depends
  on how good an approximation one wants; some sources give $10$ which
  gives virtually the same results as the following rule for large ''n''
  until ''n'' is very large (ex: ''x=11, n=7752'').
  
*A second rule is that for $n > 5$ the normal approximation is
  adequate if
$$\left|(1/\sqrt{n})(\sqrt{(1-p)/p}-\sqrt{p/(1-p)})\right|<0.3$$
  
  
*
  
*Another commonly used rule holds that the normal approximation is
  appropriate only if everything within 3 standard deviations of its
  mean is within the range of possible values that is if
  
  
  $$ \mu \pm 3 \sigma = np \pm 3 \sqrt{np(1-p)} \in [0,n]. \,$$

All of those fail when $p$ is close to either $0$ or $1$. The intuitive idea is that then the distribution is:


*

*very skewed, and

*the normal approximation will be too significant outside the actual bounds of the binomial distribution, $[0, n]$.

A: +1 for @Jaime.  But as it happens in this case your null hypothesis is that both proportions equal a pooled figure of (87+48)/(88+60) = 0.91.  With your sample size this is in the acceptable area for approximations such as this z test or the equivalent chi square test.  See that the values in the "expected" (meaning expected under the null hypothesis of equal proportions) matrix below are all more than 5, usually accepted as an ok rule of thumb.
I would advocate as a simple solution a Chi square test with continuity correction - which agrees with you (low p value) that it is unlikely a common underlying proportion would produce these two observed sets of data.
> p <- (87+48)/(88+60)
> p
[1] 0.9121622
> obs <- matrix(c(87,1,48,12), nrow=2)
> obs
     [,1] [,2]
[1,]   87   48
[2,]    1   12
> expected <- rbind(p * margin.table(obs,2),(1-p) * margin.table(obs,2))
> expected
         [,1]     [,2]
[1,] 80.27027 54.72973
[2,]  7.72973  5.27027
> chisq.test(obs)

        Pearson's Chi-squared test with Yates' continuity correction

data:  obs 
X-squared = 13.5773, df = 1, p-value = 0.0002289

