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--- b:head_first_statistics:estimating_populations_and_samples [2019/12/10 13:42] – [Variance of sample means] hkimscil
+++ b:head_first_statistics:estimating_populations_and_samples [2022/11/15 23:52] – [Using CLT for the binomial distribution] hkimscil
@@ Line 17: / Line 17: @@
 </WRAP>
-\begin{eqnarray*}
+\begin{align*}
-\overline{X} & = & \frac {\sum{X}}{n} \\
+\overline{X} & = \frac {\sum{X}}{n} \\
-& = & \frac {\sum_{i=1}^{n} X_{i}}{n} \\
+& = \frac{ \sum_{i=1}^{n} X_{i} } {n} \\
-& = & \hat{\mu}
+& = \hat{\mu}
-\end{eqnarray*}
+\end{align*}
 ====== Estimating population variance ======
@@ Line 44: / Line 44: @@
 {{:b:head_first_statistics:pasted:20191125-103603.png}}
 {{:b:head_first_statistics:pasted:20191125-103510.png}}
+[[:Why N-1]]
 <code>
@@ Line 130: / Line 132: @@
 이 때 $n = 100$일때 각각의 시도에서의 (trial) proportion 기대값은 ($\hat{P}$):
-\begin{eqnarray}
-\hat{P_{1}} & = & {E(X_{1})}/{n} = 0.34 \\
-\hat{P_{2}} & = & {E(X_{2})}/{n} = 0.43 \\
-\hat{P_{3}} & = & {E(X_{3})}/{n} = 0.32 \\
-\hat{P_{4}} & = & {E(X_{4})}/{n} = 0.42 \\
-\cdots & \cdots & \cdots              \\
-\hat{P_{k}} & = & {E(X_{k})}/{n} = 0.24
-\end{eqnarray}
-즉, $X \sim B(n, p)$ 일 때, sample의 확률 $P_{s} = \displaytype \frac{X}{n}$를 따른다 (X = red gumball이 나온 갯수, n = sample 크기).
+\begin{align*}
+n = 100, \\
+\hat{P_{1}} & = \frac{X_{1}}{n} = 0.34, (X_{1} = 34) \\
+\hat{P_{2}} & = \frac{X_{2}}{n} = 0.43, (X_{2} = 43) \\
+\hat{P_{3}} & = \frac{X_{3}}{n} = 0.32, (X_{3} = 32) \\
+\hat{P_{4}} & = \frac{X_{4}}{n} = 0.42, (X_{4} = 42) \\
+\cdots \cdots \cdots \\
+\hat{P_{k}} & = \frac{X_{k}}{n} = 0.24, (X_{1} = 24) \\
+\end{align*}
+즉, $X \sim B(n, p)$ 일 때, sample의 확률 $P_{s} = \dfrac{X}{n}$를 따른다 ($X$ = red gumball이 나온 갯수, $n$ = sample 크기).
 {{:b:head_first_statistics:pasted:20191126-073028.png}}
@@ Line 224: / Line 228: @@
 q <- 1-p
 n <- 100
-var <- (p*q)/(n-1)
+var <- (p*q)/(n)
-se  <- sqrt((p*q)/(n-1))
+se  <- sqrt((p*q)/(n))
-pnorm(.395, p, se, lower.tail = F)
+o <- .4
+o.c <- .4 - (1/(2*n))
+o.c
+pnorm(o.c, p, se, lower.tail = F)
 </code>
 <code>
+>
 > p <- 0.25
 > q <- 1-p
 > n <- 100
-> var <- (p*q)/(n-1)
+> var <- (p*q)/(n)
-> se  <- sqrt((p*q)/(n-1))
+> se  <- sqrt((p*q)/(n))
-> pnorm(.395, p, se, lower.tail = F)
+> o <- .4
-[1] 0.0004313594
+> o.c <- .4 - (1/(2*n))
+> o.c
+[1] 0.395
+> pnorm(o.c, p, se, lower.tail = F)
+[1] 0.0004060586
 </code>
@@ Line 258: / Line 270: @@
 \overline{X} = \frac{X_{1} + X_{2} + . . . + X_{n}}{n}
 \end{eqnarray*}
+위는 풍선검 봉지 30개로 이루어진 샘플의 평균을 이야기하고
+아래는 이 평균을 계속 모았을 때의 평균을 이야기한다.
 \begin{eqnarray*}
 E(\overline{X}) & = & E\left(\frac{X_{1} + X_{2} + . . . + X_{n}}{n}\right)  \\
@@ Line 275: / Line 288: @@
 \end{eqnarray*}
-\begin{eqnarray*}
+\begin{align*}
-Var(\overline{X}) & = & Var \left(\frac{X_{1} + X_{2} + . . . + X_{n}}{n}\right) \\
+Var(\overline{X}) & = Var \left(\frac{X_{1} + X_{2} + . . . + X_{n}}{n}\right) \\
-& = & \frac{1}{n^2} \ Var \left({X_{1} + X_{2} + . . . + X_{n}\right) \\
+& = \frac {1}{n^2} Var \left(X_{1} + X_{2} + . . . + X_{n} \right) \\
-& = & \frac{1}{n^2} \ (\sigma^2 + \sigma^2 + . . . + \sigma^2) \\
+& = \frac{1}{n^2} (\sigma^2 + \sigma^2 + . . . + \sigma^2) \\
-& = & \frac{1}{n^2} \ n * (\sigma^2) \\
+& = \frac{1}{n^2} n * (\sigma^2) \\
-& = & \frac{\sigma^2}{n}
+& = \frac{\sigma^2}{n}
-\end{eqnarray*}
+\end{align*}
@@ Line 311: / Line 326: @@
 ===== Using CLT for the binomial distribution =====
-$X \sim B(n, p)$, n이 30이 넘는 조건에서, $\mu = np$, $\sigma^2 = npq$ 이므로 이를 $\overline{X} \sim N(\mu, \frac{\sigma^2}{n})$에 대입해 보면:
+$X \sim B(n, p)$ 에서 $\mu = np$, $\sigma^2 = npq$ 이고,
+n이 30이 넘는 조건에서 이항분포가 정상분포를 이룬다고 하므로
+$\overline{X} \sim N(\mu, \frac{\sigma^2}{n})$에 대입해 보면:
 $$\overline{X} \sim N(np, \; pq) $$