Thanks for the reference to library(exactRankTests). That seems like a reasonable alternative to "prop.test" with small samples.
However, aren't "exact tests" and the related bootstrap methodology what Deming called "enumerative techniques", more relating to describing a fixed finite population than "enumerative techniques" for describing more general processes that will likely generate similar samples in the future? Don't "exact tests" and bootstraps answer different ("enumerative") questions from those posed by standard ("analytic") parametric procedures? (I know that the chi-square distribution is only an approximation to the distribution of the contingency table chi-square; however, that is a different issue from the question of enumerative vs. analytic studies.)
Thanks again for this and your many other interesting contributions to r-help. Spencer Graves
Torsten Hothorn wrote:
Hello,
I'm looking for some guidance with the following problem:
I've 2 samples A (111 items) and B (10 items) drawn from the same unknown population. Witihn A I find 9 "positives" and in B 0 positives. I'd like to know if the 2 samples A and B are different, ie is there a way to find out whether the number of "positives" is significantly different in A and B?
I'm currently using prop.test, but unfortunately some of my data contains
less than 5 items in a group (like in the example above), and the test
statistics may not hold:
The statistic is fine, the approximation to its null distribution may be questionable :-)
prop.test(c(9,0), c(111,10))2-sample test for equality of proportions with continuity correction
data: c(9, 0) out of c(111, 10) X-squared = 0.0941, df = 1, p-value = 0.759 alternative hypothesis: two.sided 95 percent confidence interval: -0.02420252 0.18636468 sample estimates: prop 1 prop 2 0.08108108 0.00000000
Warning message: Chi-squared approximation may be incorrect in: prop.test(c(9, 0), c(111, 10))
Do you have suggestions for an alternative test?
you may consider a permutation test for two independent samples:
R> library(exactRankTests) R> x = c(rep(1, 9), rep(0, 102)) R> y = rep(0, 10) R> mean(x) [1] 0.08108108 R> mean(y) [1] 0 R> perm.test(y, x, exact = TRUE)
2-sample Permutation Test
data: y and x T = 0, p-value = 0.6092 alternative hypothesis: true mu is not equal to 0
Best,
Torsten
many thanks for your help, +kind regards,
Arne
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