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퀴즈 1 문제 중

모집단 Mean = 180; SD = 20, 정규분포일때

49. N=16의 샘플을 추출할 때 샘플들의 평균 분포가 갖는 표준편차 값은?
standard error 값을 묻는 질문이므로
se = sigma / sqrt(n) = 20 / 4 = 5

50. n=400일 때 샘플평균들의 분포가 갖는 표준편차 값은?
1

51. n=100 의 크기의 샘플을 취한다고 할 때 이 샘플의 평균값이 나올 구간을 99퍼센트의 확신성을 가지고 구하시오.
52. 위와합산
se값이 2이고 99퍼센트의 구간은 se값을 위로 3 밑은 3 포함하는 구간이 되므로 180-6, 180+6 이 정답

위의 문제는 모두 샘플평균들을 모아 놓은 집합의 평균과 표준편차를 (표준오차) 구하는 문제이다.
즉, 무 = 180, 시그마 = 20 일 때,

샘플평균들의 표준편차는 아래처럼 각각 5, 1, 2가 된다.

> rm(list=ls())
> rnorm2 <- function(n,mean,sd){ 
+   mean+sd*scale(rnorm(n)) 
+ }
> ss <- function(x) {
+   sum((x-mean(x))^2)
+ }
> 
> mu = 180
> sigma = 20
> var = 400
> 
> n.a <- 16
> n.b <- 400
> n.c <- 100
> 
> se.a <- sigma/sqrt(n.a)
> se.b <- sigma/sqrt(n.b)
> se.c <- sigma/sqrt(n.c)
> se.a
[1] 5
> se.b
[1] 1
> se.c
[1] 2

Recap

Distribution of Sample Means – mu = 40, sigma = 4 (hence var = 16) 인 모집단에서 n = n 사이즈의 샘플링을 무한 반복할 때 그 샘플평균들이 모인 집합

rscript01

rm(list=ls())
rnorm2 <- function(n,mean,sd){ 
  mean+sd*scale(rnorm(n)) 
}
ss <- function(x) {
  sum((x-mean(x))^2)
}

mu <- 40
sigma <- 4
iter <- 1000000
sz <- 16
se <- sigma/sqrt(16)
################################
means <- rnorm2(iter, mu, se)
hist(means, breaks=50,
     xlim = c(mu-6*se, mu+6*se),
     main = paste("sampling distribution"))
abline(v=mu, col='black', lwd=2)
lo1 <- mu - se*1
hi1 <- mu + se*1
lo2 <- mu - se*2
hi2 <- mu + se*2
lo3 <- mu - se*3
hi3 <- mu + se*3

abline(v=c(lo1, lo2, lo3, hi1, hi2, hi3),
       col=c("green","blue", "black"),
       lwd=2)

print(c(lo2, hi2))

m.samp <- 37
p.val <- pnorm(m.samp, mu, se)*2
p.val
z.cal <- (m.samp-mu)/se
z.cal
p.val <- pnorm(z.cal)*2
p.val

zmeans <- scale(means)
hist(zmeans, breaks=50, 
     xlim = c(0-10*1, 0+10*1), 
     main=("normalized distribution\nof sample means"))
abline(v=0, col="black", lwd=2)
abline(v=z.cal, col='blue', lwd=2)
abline(v=-z.cal, col="green", lwd=2)
text(x=-6, y=50000, 
     label=paste("z.cal =", z.cal),
     pos = 1,
     col="blue", cex=1)
text(x=4, y=50000, 
     label=paste(-z.cal), 
     pos=1,
     col="green", cex=1)
text(x=-6, y=30000, 
     label=paste("pnorm(z.cal)*2 =", "\n", 
                 round(p.val,3)),
     pos = 1,
     col="red", cex=.8)

hist(zmeans, breaks=50, 
     xlim = c(0-10*1, 0+10*1), 
     main=("normalized distribution\nof sample means"))
abline(v=0, col="black", lwd=2)
abline(v=c(-1,-2,-3,1,2,3), 
       col=c("green", "blue", "black"), lwd=2)

z.cal
p.val
#####
# 위의 아이디어로는 z.cal 점수가 
# +-2 밖에 있는지 보면 된다. 즉, 
# 이는 prob가 0.05보다 작은지 
# 보면 되는 것이다.
#####
# +-2 는 정확한 숫자가 아니고
# qnorm(.05/2) 에 해당하는 숫자
# 가 정확한 숫자
two.minus.exact <- qnorm(.05/2)
two.plus.exact <- qnorm(1-(.05/2))
c(two.minus.exact, two.plus.exact)
#####
# 그러나 R 사용시에는 z 점수로 
# 판단하기 보다는 
# 직접 구하는 prob.로 판단
pnorm(z.cal)*2
p.val
#####
# 위에서 그룹 간의 차이를
# standard error로 나누는 것에 주의
# 


################
m.samp <- 43
sd.samp <- 4
sz <- 16
samp <- rnorm2(sz, m.samp, sd.samp)
diff <- m.samp - mu
se <- sd.samp / sqrt(sz)
t.cal <- diff/se
df <- sz-1
p.val <- pt(t.cal, df=df, lower.tail = F)*2
t.cal
df
p.val
t.test(samp, mu=mu)

out01

> rm(list=ls())
> rnorm2 <- function(n,mean,sd){ 
+   mean+sd*scale(rnorm(n)) 
+ }
> ss <- function(x) {
+   sum((x-mean(x))^2)
+ }
> 
> mu <- 40
> sigma <- 4
> iter <- 1000000
> sz <- 16
> se <- sigma/sqrt(16)
> ################################
> means <- rnorm2(iter, mu, se)
> hist(means, breaks=50,
+      xlim = c(mu-6*se, mu+6*se),
+      main = paste("sampling distribution"))
> abline(v=mu, col='black', lwd=2)
> lo1 <- mu - se*1
> hi1 <- mu + se*1
> lo2 <- mu - se*2
> hi2 <- mu + se*2
> lo3 <- mu - se*3
> hi3 <- mu + se*3
> 
> abline(v=c(lo1, lo2, lo3, hi1, hi2, hi3),
+        col=c("green","blue", "black"),
+        lwd=2)
> 
> print(c(lo2, hi2))
[1] 38 42
> 
> m.samp <- 37
> p.val <- pnorm(m.samp, mu, se)*2
> p.val
[1] 0.002699796
> z.cal <- (m.samp-mu)/se
> z.cal
[1] -3
> p.val <- pnorm(z.cal)*2
> p.val
[1] 0.002699796
> 
> zmeans <- scale(means)
> hist(zmeans, breaks=50, 
+      xlim = c(0-10*1, 0+10*1), 
+      main=("normalized distribution\nof sample means"))
> abline(v=0, col="black", lwd=2)
> abline(v=z.cal, col='blue', lwd=2)
> abline(v=-z.cal, col="green", lwd=2)
> text(x=-6, y=50000, 
+      label=paste("z.cal =", z.cal),
+      pos = 1,
+      col="blue", cex=1)
> text(x=4, y=50000, 
+      label=paste(-z.cal), 
+      pos=1,
+      col="green", cex=1)
> text(x=-6, y=30000, 
+      label=paste("pnorm(z.cal)*2 =", "\n", 
+                  round(p.val,3)),
+      pos = 1,
+      col="red", cex=.8)
> 
> hist(zmeans, breaks=50, 
+      xlim = c(0-10*1, 0+10*1), 
+      main=("normalized distribution\nof sample means"))
> abline(v=0, col="black", lwd=2)
> abline(v=c(-1,-2,-3,1,2,3), 
+        col=c("green", "blue", "black"), lwd=2)
> 
> z.cal
[1] -3
> p.val
[1] 0.002699796
> #####
> # 위의 아이디어로는 z.cal 점수가 
> # +-2 밖에 있는지 보면 된다. 즉, 
> # 이는 prob가 0.05보다 작은지 
> # 보면 되는 것이다.
> #####
> # +-2 는 정확한 숫자가 아니고
> # qnorm(.05/2) 에 해당하는 숫자
> # 가 정확한 숫자
> two.minus.exact <- qnorm(.05/2)
> two.plus.exact <- qnorm(1-(.05/2))
> c(two.minus.exact, two.plus.exact)
[1] -1.959964  1.959964
> #####
> # 그러나 R 사용시에는 z 점수로 
> # 판단하기 보다는 
> # 직접 구하는 prob.로 판단
> pnorm(z.cal)*2
[1] 0.002699796
> p.val
[1] 0.002699796
> #####
> # 위에서 그룹 간의 차이를
> # standard error로 나누는 것에 주의
> # 
> 
> 
> ################
> m.samp <- 43
> sd.samp <- 4
> sz <- 16
> samp <- rnorm2(sz, m.samp, sd.samp)
> diff <- m.samp - mu
> se <- sd.samp / sqrt(sz)
> t.cal <- diff/se
> df <- sz-1
> p.val <- pt(t.cal, df=df, lower.tail = F)*2
> t.cal
[1] 3
> df
[1] 15
> p.val
[1] 0.008972737
> t.test(samp, mu=mu)

	One Sample t-test

data:  samp
t = 3, df = 15, p-value = 0.008973
alternative hypothesis: true mean is not equal to 40
95 percent confidence interval:
 40.86855 45.13145
sample estimates:
mean of x 
       43 

>

rscript02

#####
# 
m.a <- 5.8
m.b <- 6.3
sd.a <- .5
sd.b <- .5
sz.a <- 16
sz.b <- 16
df.a <- sz.a-1
df.b <- sz.b-1
df <- df.a + df.b
a <- rnorm2(sz.a, m.a, sd.a)
b <- rnorm2(sz.b, m.b, sd.b)
diff <- m.a - m.b
pv <- (ss(a)+ss(b))/(df.a+df.b)
se <- sqrt(pv/sz.a+pv/sz.b)
t.cal <- diff / se
p.val <- pt(t.cal, df=df)*2

diff
se
t.cal
df
p.val
t.test(a,b, var.equal = T)
diff - se*2
diff + se*2
lo <- qt(.05/2,df)
lo
hi <- -lo
diff + se*lo
diff + se*hi

#####
# t-test repeated measre
#####
m.t1 <- 103
m.t2 <- 111
sd.t1 <- 10
sd.t2 <- 10
sz <- 16
t1 <- rnorm2(sz, m.t1, sd.t1)
t2 <- rnorm2(sz, m.t2, sd.t2)
t1
t2
mdiff <- m.t1-m.t2
diff <- t1-t2
sd.diff <- sd(diff)
se <- sd.diff/sqrt(sz)
t.cal <- mdiff/se
p.val <- pt(t.cal, df=sz-1)*2
t.cal
sz-1
p.val
t.test(t1,t2, paired=T)
two <- qt(.05/2, df=sz-1)
two
lo <- se*two
hi <- -lo
c(lo, hi)
c(mdiff+lo, mdiff+hi)

rout02

> #####
> # 
> m.a <- 5.8
> m.b <- 6.3
> sd.a <- .5
> sd.b <- .5
> sz.a <- 16
> sz.b <- 16
> df.a <- sz.a-1
> df.b <- sz.b-1
> df <- df.a + df.b
> a <- rnorm2(sz.a, m.a, sd.a)
> b <- rnorm2(sz.b, m.b, sd.b)
> diff <- m.a - m.b
> pv <- (ss(a)+ss(b))/(df.a+df.b)
> se <- sqrt(pv/sz.a+pv/sz.b)
> t.cal <- diff / se
> p.val <- pt(t.cal, df=df)*2
> 
> diff
[1] -0.5
> se
[1] 0.1767767
> t.cal
[1] -2.828427
> df
[1] 30
> p.val
[1] 0.008257336
> t.test(a,b, var.equal = T)

	Two Sample t-test

data:  a and b
t = -2.8284, df = 30, p-value = 0.008257
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.8610262 -0.1389738
sample estimates:
mean of x mean of y 
      5.8       6.3 

> diff - se*2
[1] -0.8535534
> diff + se*2
[1] -0.1464466
> lo <- qt(.05/2,df)
> lo
[1] -2.042272
> hi <- -lo
> diff + se*lo
[1] -0.8610262
> diff + se*hi
[1] -0.1389738
> 
> #####
> # t-test repeated measre
> #####
> m.t1 <- 103
> m.t2 <- 111
> sd.t1 <- 10
> sd.t2 <- 10
> sz <- 16
> t1 <- rnorm2(sz, m.t1, sd.t1)
> t2 <- rnorm2(sz, m.t2, sd.t2)
> t1
           [,1]
 [1,]  89.58295
 [2,]  97.20986
 [3,] 100.82700
 [4,] 120.11867
 [5,] 103.06410
 [6,] 117.36762
 [7,]  98.82191
 [8,] 111.72472
 [9,] 100.06093
[10,] 114.58757
[11,] 105.99472
[12,]  84.34803
[13,]  94.63867
[14,]  94.49667
[15,] 106.03514
[16,] 109.12144
attr(,"scaled:center")
[1] 0.08912759
attr(,"scaled:scale")
[1] 0.9759765
> t2
           [,1]
 [1,] 114.76609
 [2,] 111.81937
 [3,] 102.93248
 [4,] 122.85959
 [5,] 105.68180
 [6,] 110.43890
 [7,] 115.34844
 [8,]  97.39180
 [9,] 117.00475
[10,]  98.63924
[11,] 118.87807
[12,] 107.55519
[13,] 128.46569
[14,]  93.50094
[15,] 107.15280
[16,] 123.56487
attr(,"scaled:center")
[1] 0.2000755
attr(,"scaled:scale")
[1] 0.8946962
> mdiff <- m.t1-m.t2
> diff <- t1-t2
> sd.diff <- sd(diff)
> se <- sd.diff/sqrt(sz)
> t.cal <- mdiff/se
> p.val <- pt(t.cal, df=sz-1)*2
> t.cal
[1] -2.2741
> sz-1
[1] 15
> p.val
[1] 0.03808083
> t.test(t1,t2, paired=T)

	Paired t-test

data:  t1 and t2
t = -2.2741, df = 15, p-value = 0.03808
alternative hypothesis: true mean difference is not equal to 0
95 percent confidence interval:
 -15.4981736  -0.5018264
sample estimates:
mean difference 
             -8 

> two <- qt(.05/2, df=sz-1)
> two
[1] -2.13145
> lo <- se*two
> hi <- -lo
> c(lo, hi)
[1] -7.498174  7.498174
> c(mdiff+lo, mdiff+hi)
[1] -15.4981736  -0.5018264
> 
>