1 개인적인 방법

2 ks.test(콜모고로프-스미르노프 검정)

3 chisq.test(카이제곱 검정)

4 참고

• ks.test --> 같은 분포를 가진 샘플인가?
  
• chisq.test --> 각각의 bin에 차이가 있나?
  
• run.test --> 차이가 정규분포인가? 지속적으로 밑으로 깔려있거나 위에 있으면 이것도 문제
  
  

x1 <- rnorm(50)
x2 <- rnorm(50, -1)

compare <- function(x, y) {
  n <- length(x); m <- length(y)
  w <- c(x, y)
  o <- order(w)
  z <- cumsum(ifelse(o <= n, m, -n))
  i <- which.max(abs(z))
  w[o[i]]
}
u <- compare(x1,x2)
e.x <- ecdf(x1)
e.y <- ecdf(x2)
abs(e.x(u) - e.y(u))
ks.test(x1,x2)$statistic

plot(e.x, col="Blue", main="ECDF", xlab="Value", ylab="Probability", xlim=range(c(x1,x2)))
plot(e.y, add=TRUE, col="Red")
lines(c(u,u), c(0,1), col="Gray")
lines(c(u,u), c(e.x(u), e.y(u)), lwd=2)
text(u*1.04, abs(e.x(u)-e.y(u)) * 1.5, label="D")

x1 <- rnorm(50)
x2 <- rnorm(50, -1)

library("rgr")
gx.ks.test(x1, x2)

> gx.ks.test(x1, x2)

	Two-sample Kolmogorov-Smirnov test

data:  x1 and x2
D = 0.3, p-value = 0.02171
alternative hypothesis: two-sided

plot(ecdf(x1), do.points = FALSE, verticals=T, xlim=range(x1, x2))
lines(ecdf(x2), lty=3, do.points = FALSE, verticals=T)
ks.test(x1, x2, alternative="two.sided")

chisq.test(x1, p=x2/sum(x2))

> chisq.test(x1, p=x2/sum(x2))

	Chi-squared test for given probabilities

data:  x1
X-squared = 169.74, df = 49, p-value = 3.196e-15

Warning message:
In chisq.test(x1, p = x2/sum(x2)) :
  카이제곱 approximation은 정확하지 않을수도 있습니다

• http://stackoverflow.com/questions/27233738/finding-location-of-maximum-d-statistic-from-ks-test
  
• https://rpubs.com/mharris/KSplot