Differences

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--- sampling_distribution_in_r [2024/03/20 09:54] – [Sampling distribution in R e.g. 1] hkimscil
+++ sampling_distribution_in_r [2024/03/20 13:57] – [Sampling distribution in R e.g. 1] hkimscil
@@ Line 71: / Line 71: @@
 plot(h900, add = T, col="yellow")
-se4 <- sqrt(var(ajstu)/4)
-se25 <- sqrt(var(ajstu)/25)
-se100 <- sqrt(var(ajstu)/100)
-se400 <- sqrt(var(ajstu)/400)
-se900 <- sqrt(var(ajstu)/900)
-se1600 <- sqrt(var(ajstu)/1600)
-se2500 <- sqrt(var(ajstu)/2500)
-sss <- c(4,25,100,400,900,1600,2500)
+sss <- c(4,25,100,400,900,1600,2500) # sss sample sizes
-ses <- rep (NA, length(sss))
+ses <- rep (NA, length(sss)) # std errors
 for(i in 1:length(sss)){
   ses[i] = sqrt(var(ajstu)/sss[i])
 }
 ses
 se.1 <- ses
-se.2 <- 2*ses
+se.2 <- 2 * ses
-lower.part.2 <- mean(ajstu)-se.2
-upper.part.2 <- mean(ajstu)+se.2
+lower.s2 <- mean(ajstu)-se.2
-data.frame(cbind(sss, ses, lower.part.2, upper.part.2))
+upper.s2 <- mean(ajstu)+se.2
+data.frame(cbind(sss, ses, lower.s2, upper.s2))
 </code>
 <code>
 # n =1600 일 경우에
-# sample의 평균이 101.5보다 작을
+# sample의 평균이 100.15보다 작을
 # 확률은 어떻게 구해야 할까?
@@ Line 104: / Line 100: @@
 # 이것을 standard error라고 부른다
 # 따라서
-pnorm(101.5, mean(ajstu), sqrt(var(ajstu)/n.1600))
+pnorm(100.15, mean(ajstu), sqrt(var(ajstu)/n.1600))
 </code>