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--- t-test_summary [2026/04/12 11:48] – [rs.hypothesis.testing] hkimscil
+++ t-test_summary [2026/04/16 01:27] (current) – hkimscil
@@ Line 29: / Line 29: @@
 ################################
-sz <- 10
+sz <- 16
 iter <- 100000
-# n = 10 일 때의 p1에 대한 sampling dist 은 아래 시뮬레이션으로
+# n = 16 일 때의 p1에 대한 sampling dist 은 아래 시뮬레이션으로
 # 구해볼 수 있다.
 means <- rep(NA, iter)
@@ Line 50: / Line 50: @@
 sd.means
-# 위의 시뮬레이션으로 구한 sampling dist 대신에
+# 위 집합을 표준화하게 되면 그 집합의 평균과
-# 정확한 mean과 se 값을 갖는 집합 sdc를 만든다
+# 표준편차는 0, 1이 (분산도 1) 된다.
-sdc <- rnorm2(iter, m.means, sd.means)
+m.zmeans <- 0
-mean(sdc)
+ms.zmeans <- 1
-var(sdc)
+sd.zmeans <- 1
-sd(sdc)
-zsdc <- scale(sdc)
-m.zsdc <- mean(zsdc)
-ms.zsdc <- var(zsdc)
-sd.zsdc <- sd(zsdc)
-m.zsdc
-ms.zsdc
-sd.zsdc
-se2 <- c(sqrt(var(p2)/sz))
-z.p2 <- c((mean(p2)-mean(p1))/se2)
-sdc2 <- rnorm2(iter, mean(p2), se2)
-zsdc2 <- scale(sdc2)+z.p2
-mean(zsdc2)
-sd(zsdc2)
-curve(dnorm(x), from = -4, to = z.p2+4,
-      main = "normalized distribution of sample means from p1 and p2",
-      ylab = "Density", xlab = "t-value", col = "black", lwd = 2)
-curve(dnorm(x-(z.p2)), from = z.p2-3, to = z.p2+3, add = T,
-      main = "Distribution Curve",
-      ylab = "Density", xlab = "t-value", col = "blue", lwd = 2, lty=2)
-abline(v=mean(zsdc), col='black', lwd=2)
-abline(v=mean(zsdc2), col='blue', lwd=2)
-mean(zsdc2)
-text(x=mean(zsdc), y=.1, label=paste(round(mean(zsdc),4)), pos=4)
-text(x=mean(zsdc2), y=.1, label=paste(round(mean(zsdc2),4)), pos=4)
-###############
-################################
-sz <- 100
-iter <- 100000
-# n = 10 일 때의 p1에 대한 sampling dist 은 아래 시뮬레이션으로
-# 구해볼 수 있다.
-means <- rep(NA, iter)
-for (i in 1:iter) {
-  s1 <- sample(p1, sz, replace = T)
-  means[i] <- mean(s1)
-}
-mean(means)
-var(means)
-sd(means)
-# CLT에 의하면 위이 값은
-m.means <- mean(p1)
-ms.means <- c(var(p1)/sz)
-sd.means <- c(sqrt(var(p1)/sz))
-m.means
-ms.means
 sd.means
+# p2에서 구하는 샘플링 디스트리뷰션의 표준점수를
-# 위의 시뮬레이션으로 구한 sampling dist 대신에
+# p1의 표준점수에 (0, 1) 비교해서 배치하자면
-# 정확한 mean과 se 값을 갖는 집합 sdc를 만든다
+z.p2 <- c((mean(p2)-mean(p1))/sd.means) # 표준점수 평균
-sdc <- rnorm2(iter, m.means, sd.means)
+sd.means <- c(sqrt(var(p1)/sz)) # 표준편차
-mean(sdc)
-var(sdc)
-sd(sdc)
-zsdc <- scale(sdc)
-m.zsdc <- mean(zsdc)
-ms.zsdc <- var(zsdc)
-sd.zsdc <- sd(zsdc)
-m.zsdc
-ms.zsdc
-sd.zsdc
-se2 <- c(sqrt(var(p2)/sz))
-z.p2 <- c((mean(p2)-mean(p1))/se2)
-sdc2 <- rnorm2(iter, mean(p2), se2)
-zsdc2 <- scale(sdc2)+z.p2
-mean(zsdc2)
-sd(zsdc2)
 curve(dnorm(x), from = -4, to = z.p2+4,
-      main = "normalized distribution of sample \n means from p1 and p2 (n=100)",
+      main = "normalized distribution of sample \n means from p1 and p2 (n=10)",
       ylab = "Density", xlab = "z-value", col = "black", lwd = 2)
-curve(dnorm(x-(z.p2)), from = z.p2-3, to = z.p2+3, add = T,
+curve(dnorm(x-c(z.p2)), from = z.p2-3, to = z.p2+3, add = T,
       main = "",
       ylab = "Density", xlab = "z-value", col = "blue", lwd = 2, lty=2)
-abline(v=mean(zsdc), col='black', lwd=2)
+abline(v=0, col='black', lwd=2)
-abline(v=mean(zsdc2), col='blue', lwd=2)
+abline(v=z.p2, col='blue', lwd=2)
-mean(zsdc2)
+text(x=0, y=.1, label=paste(round(0, 4)), pos=4)
-text(x=mean(zsdc), y=.1, label=paste(round(mean(zsdc),4)), pos=4)
+text(x=z.p2, y=.1, label=paste(round(z.p2, 4)), pos=4)
-text(x=mean(zsdc2), y=.1, label=paste(round(mean(zsdc2),4)), pos=4)
-sz <- 10 # return to sample size 10
 #######################################
 #######################################
-#
 lo1 <- qnorm(.32/2)
 hi1 <- -lo1
@@ Line 169: / Line 96: @@
 text(x=hi3, y=.1, label=paste(round(hi3,3), "(3)", "\n","99%"), pos=4)
-mean.of.sample.a <- mean(sdc)+ 1.5*sd(sdc)
+mean.of.sample.a <- m.means+ 1.5*sd.means
 mean.of.sample.a
-diff <- (mean.of.sample.a - mean(sdc))
+diff <- (mean.of.sample.a - m.means)
 se.z <- sd(p1)/sqrt(sz)
 diff
@@ Line 193: / Line 120: @@
 # 새로운 UI로 게임을 하도록 한 후
-# UI점수를 10명에게 구했다고 가정하고
+# UI점수를 sz 명에게 구했다고 가정하고
 # 새로운 UI점수가 기존의 p1 paramter와
 # 다른지 테스트 해보라
@@ Line 201: / Line 128: @@
 # 하면 샘플의 평균과 p1의 평균은 다르다고 판단될 것이다.
 # 아래는 그럼에도 불구하고 실패하는 경우이다.
-set.seed(111)
+set.seed(110)
 smp <- sample(p2, sz, replace=T)
 m.smp <- mean(smp)
@@ Line 213: / Line 140: @@
 curve(dnorm(x), from = -4, to = z.p2+4,
-      main = "normalized distribution of sample means \n testing with a sample from p2 (failed)",
+      main = "normalized distribution of sample means
+      testing with a sample from p2 (failed)",
       ylab = "Density", xlab = "z-value", col = "black", lwd = 2)
 abline(v=0, col="black", lwd=2)
@@ Line 237: / Line 165: @@
 z.test(smp, mean(p1), sd(p1))
-z.p2 <- (mean(p2)-mean(p1))/se2
+z.p2 <- (mean(p2)-mean(p1))/se.z
 z.p2
 curve(dnorm(x), from = -5, to = z.p2+5,
@@ Line 259: / Line 187: @@
 # type i and type ii error
-z.p2 <- (mean(p2)-mean(p1))/se2
+two <- qnorm(.05/2)
-z.p2
+two
 curve(dnorm(x), from = -4.7, to = z.p2+4,
       main = "Distribution Curve",
@@ Line 268: / Line 197: @@
       ylab = "Density", xlab = "z-value", col = "blue", lwd = 2, lty=2)
 abline(v=0, col='black', lwd=2)
-z.cal1
-z.cal2
-two <- qnorm(.05/2)
-two
 abline(v=c(two, -two), col='black', lwd=2)
 abline(v=c(-z.cal1, z.cal1), col='red', lwd=2)
@@ Line 326: / Line 251: @@
 print(c(t.cal, df.smp, prob))
 print(c(m.smp+lo2*se.z, m.smp+hi2*se.z))
-cat("t =", t.cal, ", df =", round(df.smp,0), ", p-value =", prob,
+cat(" t =", t.cal, ", df =", round(df.smp,0), ", p-value =", prob,
 "\n", "95% confidence interval =", m.smp+lo2*se.z, m.smp+hi2*se.z)
 t.test(smp, mu=mean(p1))
@@ Line 386: / Line 311: @@
 t.test(group.a, group.b, var.equal = T)
 t.cal
-# t.cal=diff/se
-t.cal * se.s
-diff
-diff+lo2*se.s
-diff+hi2*se.s
-(t.cal+lo2)*se.s
-(t.cal+hi2)*se.s
 ######################
@@ Line 446: / Line 364: @@
 <tabbox ro.hypothesis.testing>
 <code>
->
 > rm(list=ls())
 > rnorm2 <- function(n,mean,sd){ mean+sd*scale(rnorm(n)) }
@@ Line 474: / Line 391: @@
 >
 > ################################
-> sz <- 10
+> sz <- 16
 > iter <- 100000
-> # n = 10 일 때의 p1에 대한 sampling dist 은 아래 시뮬레이션으로
+> # n = 16 일 때의 p1에 대한 sampling dist 은 아래 시뮬레이션으로
 > # 구해볼 수 있다.
 > means <- rep(NA, iter)
@@ Line 484: / Line 401: @@
 + }
 > mean(means)
-[1] 99.9946
+[1] 99.997
 > var(means)
-[1] 9.95743
+[1] 6.215719
 > sd(means)
-[1] 3.15554
+[1] 2.493134
 >
 > # CLT에 의하면 위이 값은
@@ Line 497: / Line 414: @@
 [1] 100
 > ms.means
-[1] 10
+[1] 6.25
 > sd.means
-[1] 3.162278
+[1] 2.5
 >
-> # 위의 시뮬레이션으로 구한 sampling dist 대신에
+> # 위 집합을 표준화하게 되면 그 집합의 평균과
-> # 정확한 mean과 se 값을 갖는 집합 sdc를 만든다
+> # 표준편차는 0, 1이 (분산도 1) 된다.
-> sdc <- rnorm2(iter, m.means, sd.means)
+> m.zmeans <- 0
-> mean(sdc)
+> ms.zmeans <- 1
-[1] 100
+> sd.zmeans <- 1
-> var(sdc)
-     [,1]
-[1,]   10
-> sd(sdc)
-[1] 3.162278
 >
-> zsdc <- scale(sdc)
-> m.zsdc <- mean(zsdc)
-> ms.zsdc <- var(zsdc)
-> sd.zsdc <- sd(zsdc)
-> m.zsdc
-[1] -2.40102e-17
-> ms.zsdc
-     [,1]
-[1,]    1
-> sd.zsdc
-[1] 1
->
-> se2 <- c(sqrt(var(p2)/sz))
-> z.p2 <- c((mean(p2)-mean(p1))/se2)
-> sdc2 <- rnorm2(iter, mean(p2), se2)
-> zsdc2 <- scale(sdc2)+z.p2
-> mean(zsdc2)
-[1] 1.897367
-> sd(zsdc2)
-[1] 1
->
-> curve(dnorm(x), from = -4, to = z.p2+4,
-+       main = "normalized distribution of sample means from p1 and p2",
-+       ylab = "Density", xlab = "t-value", col = "black", lwd = 2)
-> curve(dnorm(x-(z.p2)), from = z.p2-3, to = z.p2+3, add = T,
-+       main = "Distribution Curve",
-+       ylab = "Density", xlab = "t-value", col = "blue", lwd = 2, lty=2)
-> abline(v=mean(zsdc), col='black', lwd=2)
-> abline(v=mean(zsdc2), col='blue', lwd=2)
-> mean(zsdc2)
-[1] 1.897367
-> text(x=mean(zsdc), y=.1, label=paste(round(mean(zsdc),4)), pos=4)
-> text(x=mean(zsdc2), y=.1, label=paste(round(mean(zsdc2),4)), pos=4)
->
-</code>
-{{pasted:20260412-113853.png}}
-<code>
->
-> ################################
-> sz <- 100
-> iter <- 100000
-> # n = 10 일 때의 p1에 대한 sampling dist 은 아래 시뮬레이션으로
-> # 구해볼 수 있다.
-> means <- rep(NA, iter)
-> for (i in 1:iter) {
-+   s1 <- sample(p1, sz, replace = T)
-+   means[i] <- mean(s1)
-+ }
-> mean(means)
-[1] 100.0002
-> var(means)
-[1] 1.003404
-> sd(means)
-[1] 1.001701
->
-> # CLT에 의하면 위이 값은
-> m.means <- mean(p1)
-> ms.means <- c(var(p1)/sz)
-> sd.means <- c(sqrt(var(p1)/sz))
-> m.means
-[1] 100
-> ms.means
-[1] 1
 > sd.means
-[1] 1
+[1] 2.5
->
+> # p2에서 구하는 샘플링 디스트리뷰션의 표준점수를
-> # 위의 시뮬레이션으로 구한 sampling dist 대신에
+> # p1의 표준점수에 (0, 1) 비교해서 배치하자면
-> # 정확한 mean과 se 값을 갖는 집합 sdc를 만든다
+> z.p2 <- c((mean(p2)-mean(p1))/sd.means) # 표준점수 평균
-> sdc <- rnorm2(iter, m.means, sd.means)
+> sd.means <- c(sqrt(var(p1)/sz)) # 표준편차
-> mean(sdc)
-[1] 100
-> var(sdc)
-     [,1]
-[1,]    1
-> sd(sdc)
-[1] 1
->
-> zsdc <- scale(sdc)
-> m.zsdc <- mean(zsdc)
-> ms.zsdc <- var(zsdc)
-> sd.zsdc <- sd(zsdc)
-> m.zsdc
-[1] 3.271623e-17
-> ms.zsdc
-     [,1]
-[1,]    1
-> sd.zsdc
-[1] 1
->
-> se2 <- c(sqrt(var(p2)/sz))
-> z.p2 <- c((mean(p2)-mean(p1))/se2)
-> sdc2 <- rnorm2(iter, mean(p2), se2)
-> zsdc2 <- scale(sdc2)+z.p2
-> mean(zsdc2)
-[1] 6
-> sd(zsdc2)
-[1] 1
 >
 > curve(dnorm(x), from = -4, to = z.p2+4,
-+       main = "normalized distribution of sample \n means from p1 and p2 (n=100)",
++       main = "normalized distribution of sample \n means from p1 and p2 (n=10)",
 +       ylab = "Density", xlab = "z-value", col = "black", lwd = 2)
-> curve(dnorm(x-(z.p2)), from = z.p2-3, to = z.p2+3, add = T,
+> curve(dnorm(x-c(z.p2)), from = z.p2-3, to = z.p2+3, add = T,
 +       main = "",
 +       ylab = "Density", xlab = "z-value", col = "blue", lwd = 2, lty=2)
-> abline(v=mean(zsdc), col='black', lwd=2)
+> abline(v=0, col='black', lwd=2)
-> abline(v=mean(zsdc2), col='blue', lwd=2)
+> abline(v=z.p2, col='blue', lwd=2)
-> mean(zsdc2)
+> text(x=0, y=.1, label=paste(round(0, 4)), pos=4)
-[1] 6
+> text(x=z.p2, y=.1, label=paste(round(z.p2, 4)), pos=4)
-> text(x=mean(zsdc), y=.1, label=paste(round(mean(zsdc),4)), pos=4)
-> text(x=mean(zsdc2), y=.1, label=paste(round(mean(zsdc2),4)), pos=4)
 >
 </code>
-{{pasted:20260412-114016.png}}
+{{pasted:20260412-232126.png}}
 <code>
-> sz <- 10  # return to sample size 10
+> #######################################
->
+> #######################################
-> ####################################
-> ####################################
 > lo1 <- qnorm(.32/2)
 > hi1 <- -lo1
@@ Line 659: / Line 476: @@
 <code>
-> mean.of.sample.a <- mean(sdc)+ 1.5*sd(sdc)
+> mean.of.sample.a <- m.means+ 1.5*sd.means
 > mean.of.sample.a
-[1] 104.7434
+[1] 103.75
-> diff <- (mean.of.sample.a - mean(sdc))
+> diff <- (mean.of.sample.a - m.means)
 > se.z <- sd(p1)/sqrt(sz)
 > diff
-[1] 4.743416
+[1] 3.75
 > se.z
-[1] 3.162278
+[1] 2.5
 > z.score  <- diff / se.z
 > z.score
@@ Line 686: / Line 503: @@
 +                    "\n", "pnorm(-z.score)*2 =", round(prob,5)),
 +      pos=4, col='red')
 >
 </code>
 {{pasted:20260412-063608.png}}
@@ Line 700: / Line 517: @@
 > # 하면 샘플의 평균과 p1의 평균은 다르다고 판단될 것이다.
 > # 아래는 그럼에도 불구하고 실패하는 경우이다.
-> set.seed(111)
+> set.seed(110)
 > smp <- sample(p2, sz, replace=T)
 > m.smp <- mean(smp)
 > m.smp
-[1] 104.4742
+[1] 104.5958
 > diff <- m.smp - mean(p1)
 > se.z <- sqrt(var(p1)/sz)
@@ Line 710: / Line 527: @@
 > prob1 <- pnorm(abs(z.cal1), lower.tail = F)*2
 > print(c(z.cal1, sz, prob1))
-[1]  1.4148817 10.0000000  0.1571032
+[1]  2.906626913 40.000000000  0.003653487
 > z.test(smp, mean(p1), sd(p1))
- z value: 1.41488
+ z value: 2.90663
- p value: 0.1571032
+ p value: 0.00365349
- diff:    104.4742 - 100 = 4.474249
+ diff:    104.5958 - 100 = 4.595781
- se:      3.162278
+ se:      1.581139
-% CI:  93.80205 106.198>
+% CI:  96.90102 103.099>
 > curve(dnorm(x), from = -4, to = z.p2+4,
-+       main = "normalized distribution of sample means \n testing with a sample from p2 (failed)",
++       main = "normalized distribution of sample means
++       testing with a sample from p2 (failed)",
 +       ylab = "Density", xlab = "z-value", col = "black", lwd = 2)
 > abline(v=0, col="black", lwd=2)
@@ Line 728: / Line 546: @@
 +                    "\n", "pnorm(-z.cal1)*2 =", round(prob1,4)),
 +      pos=4, col='red')
+>
 >
 </code>
-{{pasted:20260412-063636.png}}
+{{pasted:20260412-233253.png}}
 <code>
 > # 같은 방법으로 했는데 성공한 경우
 > set.seed(211)
@@ Line 738: / Line 557: @@
 > m.smp <- mean(smp)
 > m.smp
-[1] 110.1154
+[1] 107.6795
 > diff <- m.smp - mean(p1)
 > se.z <- sqrt(var(p1)/sz)
@@ Line 744: / Line 563: @@
 > prob2 <- pnorm(abs(z.cal2), lower.tail = F)*2
 > print(c(z.cal2, sz, prob2))
-[1]  3.198763975 10.000000000  0.001380181
+[1] 4.856940e+00 4.000000e+01 1.192138e-06
 > z.test(smp, mean(p1), sd(p1))
- z value: 3.19876
+ z value: 4.85694
- p value: 0.00138018
+ p value: 1.19e-06
- diff:    110.1154 - 100 = 10.11538
+ diff:    107.6795 - 100 = 7.679496
- se:      3.162278
+ se:      1.581139
-% CI:  93.80205 106.198>
+% CI:  96.90102 103.099>
-> z.p2 <- (mean(p2)-mean(p1))/se2
+> z.p2 <- (mean(p2)-mean(p1))/se.z
 > z.p2
-[1] 1.897367
+         [,1]
+[1,] 3.794733
 > curve(dnorm(x), from = -5, to = z.p2+5,
 +       main = "normalized distribution of sample means \n testing with a sample from p2 (succeeded)",
@@ Line 760: / Line 580: @@
 > z.cal1
          [,1]
-[1,] 1.414882
+[1,] 2.906627
 > z.cal2
-         [,1]
+        [,1]
-[1,] 3.198764
+[1,] 4.85694
 > two <- qnorm(.05/2)
 > two
@@ Line 776: / Line 596: @@
 +      label=paste(round(-z.cal2,4)),
 +      col="darkgreen", cex=1, pos=2)
+>
 >
 </code>
-{{pasted:20260412-063652.png}}
+{{pasted:20260412-233208.png}}
 <code>
 > # type i and type ii error
-> z.p2 <- (mean(p2)-mean(p1))/se2
+> two <- qnorm(.05/2)
-> z.p2
+> two
-[1] 1.897367
+[1] -1.959964
+>
 > curve(dnorm(x), from = -4.7, to = z.p2+4,
 +       main = "Distribution Curve",
 +       ylab = "Density", xlab = "z-value", col = "black", lwd = 2)
-> curve(dnorm(x-(z.p2)), from = z.p2-3, to = z.p2+3, add = T,
+> curve(dnorm(x-c(z.p2)), from = z.p2-3, to = z.p2+3, add = T,
 +       main = "Distribution Curve",
 +       ylab = "Density", xlab = "z-value", col = "blue", lwd = 2, lty=2)
 > abline(v=0, col='black', lwd=2)
-> z.cal1
-         [,1]
-[1,] 1.414882
-> z.cal2
-         [,1]
-[1,] 3.198764
-> two <- qnorm(.05/2)
-> two
-[1] -1.959964
 > abline(v=c(two, -two), col='black', lwd=2)
 > abline(v=c(-z.cal1, z.cal1), col='red', lwd=2)
@@ Line 819: / Line 632: @@
 +      label=paste(round(-z.cal2,4)),
 +      col="darkgreen", cex=1, pos=2)
+>
 >
 </code>
-{{pasted:20260412-063709.png}}
+{{pasted:20260412-233411.png}}
 <code>
->
 > ############################
 > # one sample t-test
@@ Line 869: / Line 682: @@
 > print(c(m.smp+lo2*se.z, m.smp+hi2*se.z))
 [1] 102.5239 110.0970
-> cat("t =", t.cal, ", df =", round(df.smp,0), ", p-value =", prob,
+> cat(" t =", t.cal, ", df =", round(df.smp,0), ", p-value =", prob,
 + "\n", "95% confidence interval =", m.smp+lo2*se.z, m.smp+hi2*se.z)
-t = 3.488087 , df = 19 , p-value = 0.002460977
+ t = 3.488087 , df = 19 , p-value = 0.002460977
 % confidence interval = 102.5239 110.097> t.test(smp, mu=mean(p1))
@@ Line 884: / Line 697: @@
 mean of x
 .3104
 >
 > #################################
@@ Line 962: / Line 774: @@
 +      pos=4, col='red')
 >
 </code>
 {{pasted:20260412-063739.png}}
@@ Line 983: / Line 796: @@
 > t.cal
 [1] -3.070212
-> # t.cal=diff/se
-> t.cal * se.s
-[1] -8.871414
-> diff
-[1] -8.871414
-> diff+lo2*se.s
-[1] -14.68117
-> diff+hi2*se.s
-[1] -3.061661
-> (t.cal+lo2)*se.s
-[1] -14.68117
-> (t.cal+hi2)*se.s
-[1] -3.061661
 >
 > ######################
@@ Line 1063: / Line 863: @@
 +      col="red", pos=4)
 > text(x=t.cal, y=.2, label=c(round(t.cal,3)), col="red", pos=2)
+>
+> cat(t.cal, sz-1, prob)
+-3.88213 39 0.0003888961
+>
 </code>
 {{pasted:20260412-063758.png}}