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--- b:head_first_statistics:estimating_populations_and_samples [2019/11/28 08:44] – [Expectation of samples proportions (Ps)] hkimscil
+++ b:head_first_statistics:estimating_populations_and_samples [2024/11/11 08:23] (current) – [Recap] hkimscil
@@ Line 2: / Line 2: @@
 {{tablelayout?colwidth="350px"&rowsHeaderSource=1&rowsVisible=2&float=right}}
 |{{:b:head_first_statistics:pasted:20191125-101010.png}}|
-So how can we use the results of the sample taste test to tell us the mean
+So how can we use the results of the sample taste test to tell us the mean amount of time gumball flavor lasts for in the general gumball population?
-amount of time gumball flavor lasts for in the general gumball population?
 The answer is actually pretty intuitive. We assume that the mean flavor duration of the gumballs in the sample matches that of the population. In other words, we find the mean of the sample and use it as the mean for the population too.
-Here’s a sketch showing the distribution of the sample, and what you’d expect the distribution of the population to look like based on the sample. You’d expect the distribution of the population to be a similar shape to that of the sample, so you can assume that the mean of the sample and population have about the same value.
+Here's a sketch showing the distribution of the sample, and what you’d expect the distribution of the population to look like based on the sample. You’d expect the distribution of the population to be a similar shape to that of the sample, so you can assume that the mean of the sample and population have about the same value.
-$$\mu \;\;\;\; \hat\mu$$
+$$\mu \quad \quad \hat\mu$$
@@ Line 18: / 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 45: / Line 44: @@
 {{:b:head_first_statistics:pasted:20191125-103603.png}}
 {{:b:head_first_statistics:pasted:20191125-103510.png}}
+[[:Why N-1]]
 <code>
@@ Line 74: / Line 75: @@
 | $\sum{ds^2}$ |  |  | 62.32  |
 | $n-1$ |  |  | 9  |
+| $Var(x)$      |  |  | 6.924444   |
 ====== Estimating proportion ======
@@ Line 79: / Line 81: @@
 {{:b:head_first_statistics:pasted:20191125-105807.png}}
 {{:b:head_first_statistics:pasted:20191125-105856.png}}
+<fs large>Or</fs>
+\begin{eqnarray*}
+\Large{P_{s}} & = & \Large{\frac{\text{number of the same kind}}{\text{number in sample}}}
+\end{eqnarray*}
 p = 32/40 = 0.8
@@ Line 85: / Line 91: @@
 Mighty Gumball takes another sample of their super-long-lasting gumballs, and finds that in the sample, 10 out of 40 people prefer the pink gumballs to all other colors. What proportion of people prefer pink gumballs in the population? What’s the probability of choosing someone from the population who doesn’t prefer pink gumballs?
 </WRAP>
+핑크풍선검을 선호하는 사람의 비율
 \begin{eqnarray*}
 \hat{P} = P_{s} & = & \frac {10}{40}  \\
@@ Line 91: / Line 97: @@
 \end{eqnarray*}
+핑크가 아닌 풍선검을 선호하는 사람의 비율
 \begin{eqnarray*}
 \hat{P} = P_{s'} & = & 1 - \hat{P}  \\
@@ Line 112: / Line 119: @@
 population: gumball의 25%가 red라고 할 때,
 하나의 샘플을 뽑는다고 가정할 때의 기대값과 분산값은 무엇인가?
+<WRAP box>
 Bernoulli distribution에 따르면,
 하나의 검볼을 뽑을 때, 이것이 red인지 아닌지에 대한 기대값과 분산값은
@@ Line 123: / Line 130: @@
 위의 상황에서 100번 independent trial을 통해서 구한 평균과 분산값은:
 $X \sim B(100, 1/4)$의 분포를 따른다고 할 때,
+</WRAP>
+<WRAP box>
+혹은 위의 분포는 이항분포이므로 $X ~ B(n, p)$ 에서 $E(X) = np$; $V(X) = npq$ 이다.
+</WRAP>
 \begin{eqnarray*}
 E(X) & = & n * p = 100 * 1/4 = 25 \\
@@ Line 129: / Line 142: @@
 \end{eqnarray*}
-이 때 각각의 시도에서의 (trial) proportion 기대값은 ($\hat{P}$):
+위와 같이 $n = 100$ 일때 각각의 시도에서의 (trial) proportion 값은 ($\hat{P}$), 즉
-\begin{eqnarray}
+\begin{eqnarray*}
-\hat{P_{1}} & = & {X_{1}}/{100} = 0.3 \\
+X_{i} & = & \text{the number of red gumball,} \\
-\hat{P_{2}} & = & {X_{2}}/{100} = 0.7 \\
+n & = & 100
-\hat{P_{3}} & = & {X_{3}}/{100} = 0.5 \\
+\end{eqnarray*} 조건에서의 proportion (비율) 값은
-\hat{P_{4}} & = & {X_{4}}/{100} = 0.4 \\
+\begin{eqnarray*}
-\cdots & \cdots & \cdots              \\
+\hat{P_{1}} & = \frac{X_{1}}{n} = 0.34, (X_{1} = 34) \\
-\hat{P_{k}} & = & {X_{k}}/{100} = 0.5
+\hat{P_{2}} & = \frac{X_{2}}{n} = 0.23, (X_{2} = 23) \\
-\end{eqnarray}
+\hat{P_{3}} & = \frac{X_{3}}{n} = 0.22, (X_{3} = 22) \\
+\hat{P_{4}} & = \frac{X_{4}}{n} = 0.21, (X_{4} = 21) \\
+& \cdots \cdots \cdots \cdots \cdots \\
+\hat{P_{k}} & = \frac{X_{k}}{n} = 0.24, (X_{k} = 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).
+즉, $X \sim B(n, p)$ 일 때, sample의 비율은 $P_{s} = \dfrac{X}{n}$를 따른다 ($X$ = red gumball이 나온 갯수, $n$ = sample 크기).
 {{:b:head_first_statistics:pasted:20191126-073028.png}}
-위의 sampling을 계속한다면 (1)~(6)과 같은 결과를 의미한다 (아래 그림 참조).
+위의 sampling을 계속한다면 아래 그림과 같은 결과를 의미한다.
 {{:b:head_first_statistics:pasted:20191126-073652.png}}
-이렇게 계속 샘플링을 하여 그 확률(probability)를 구한다고 하면:
+이렇게 계속 샘플링을 하여 얻은 비율의 $P_{s}$ 기댓값을 ($E(P_{s})$) 구한다고 하면;
 \begin{eqnarray*}
@@ Line 154: / Line 171: @@
 & = & p
 \end{eqnarray*}
+아래는 위의 시뮬레이션이다.
+  * $X ~ B(100, 1/4)$의 이항분포에서 (n=100, p=1/4)
+  * random 하게 1000번의 (k=1000) 샘플링을 해서
+  * 얻는 Red gumball의 숫자
+<code>
+> set.seed(101)
+> k <- 1000
+> n <- 100
+> p <- 1/4
+> q <- 1-p
+# in order to clarify what we are doing
+# X~B(n,p) 일 때, 100개의 검볼을 샘플링해서
+# red gumball을 세봤더니
+> rbinom(1,n,p) # 24개 였다라는 뜻
+[1] 24
+# 아래는 이것을 1000번 (k번) 한 것
+> numbers.of.red.gumball <- rbinom(k, n, p)
+> numbers.of.red.gumball
+   [1] 18 27 27 22 23 26 23 26 25 30 27 28 32 24 26 29 22 24 18 27 33 22 27 31 29 19
+  [27] 24 24 27 24 23 21 21 25 31 21 29 16 31 24 24 28 23 24 22 19 31 28 20 19 24 27
+  [53] 28 24 28 27 25 27 26 29 29 26 36 29 27 16 23 30 32 22 32 26 29 29 22 18 22 27
+  [79] 33 27 28 28 34 15 32 23 24 20 16 27 31 27 21 22 29 24 22 19 18 20 17 24 30 27
+ [105] 23 19 17 28 37 20 18 26 30 30 34 30 25 23 26 24 20 19 25 22 29 25 25 27 19 27
+ [131] 23 22 23 26 25 25 32 25 27 32 22 32 23 30 21 25 27 17 24 21 24 26 33 20 22 26
+ [157] 28 25 30 33 27 30 26 23 39 23 31 18 26 27 34 25 28 31 35 28 29 32 27 31 28 25
+ [183] 22 23 15 22 20 26 21 22 16 23 22 31 24 27 31 21 24 26 26 22 22 34 19 30 22 28
+ [209] 25 24 29 25 25 16 27 23 25 32 18 22 25 25 24 24 21 32 20 28 29 22 23 22 25 21
+ [235] 27 22 24 29 24 22 30 22 21 17 25 23 21 27 22 22 25 22 29 24 26 32 28 20 22 22
+ [261] 27 26 22 24 31 18 27 29 28 17 27 33 23 33 25 32 26 23 19 21 20 23 15 19 23 26
+ [287] 27 28 23 24 35 27 30 23 25 24 31 23 20 22 22 26 21 22 26 28 26 23 21 13 29 27
+ [313] 21 34 28 24 19 26 27 25 23 27 25 19 29 18 28 21 27 28 28 22 22 20 20 25 27 17
+ [339] 16 27 32 23 18 28 31 29 21 27 27 30 21 25 20 25 26 30 26 21 15 29 22 21 16 25
+ [365] 25 27 26 27 28 21 27 24 25 24 39 24 28 33 20 26 24 27 20 31 27 27 20 21 31 25
+ [391] 22 22 30 34 27 23 21 25 20 24 29 19 30 27 33 22 29 30 22 29 26 24 18 26 36 26
+ [417] 23 24 22 32 33 16 24 28 24 25 29 31 28 28 29 26 24 25 28 27 24 31 25 31 33 26
+ [443] 26 24 33 28 20 23 22 23 22 30 25 25 23 27 27 23 24 28 24 28 23 22 26 30 26 27
+ [469] 21 23 23 27 26 23 25 30 25 24 22 28 18 23 18 16 27 26 18 25 27 22 20 19 27 25
+ [495] 31 27 22 21 24 24 26 23 23 29 27 23 25 20 21 21 27 25 22 29 28 21 21 24 27 24
+ [521] 28 19 14 32 27 22 24 35 26 28 28 26 25 25 19 26 24 20 19 28 25 25 24 21 30 27
+ [547] 30 20 22 26 31 26 20 20 27 25 26 18 30 20 29 16 38 26 22 29 22 30 26 19 27 24
+ [573] 29 29 25 19 23 24 24 23 25 31 18 24 33 27 25 27 29 28 24 23 24 28 20 24 30 24
+ [599] 21 20 25 24 24 30 22 26 23 25 21 21 24 27 18 20 22 30 25 23 27 26 23 23 28 18
+ [625] 29 27 25 32 26 15 22 24 21 34 23 23 18 29 23 27 28 23 37 20 17 25 11 21 28 22
+ [651] 28 25 22 25 21 18 20 27 30 24 28 23 30 31 24 23 37 19 27 32 25 27 28 29 22 26
+ [677] 26 20 22 25 24 19 27 21 32 27 31 29 24 24 29 29 25 22 34 23 18 33 18 23 24 26
+ [703] 18 20 23 30 28 26 34 17 33 30 32 30 22 28 19 19 23 23 20 23 21 31 30 20 24 23
+ [729] 23 28 26 34 27 33 31 20 25 12 25 20 20 25 27 24 29 26 22 30 26 28 28 27 23 18
+ [755] 28 22 21 27 22 26 21 22 27 24 19 27 29 37 30 27 25 30 19 22 22 28 32 22 33 26
+ [781] 20 31 23 24 24 26 24 30 17 21 20 22 20 17 24 22 24 23 23 24 23 16 16 17 23 27
+ [807] 29 26 16 21 34 19 25 25 28 32 17 22 26 23 23 24 22 22 14 30 25 33 26 25 31 28
+ [833] 30 21 19 17 19 21 16 21 26 21 29 27 31 32 19 22 24 25 25 24 23 30 21 22 19 20
+ [859] 21 20 21 28 19 26 28 26 29 28 26 21 31 32 31 22 23 25 27 26 22 27 30 24 25 23
+ [885] 27 25 24 24 30 29 26 32 29 23 24 20 26 26 22 22 19 23 33 18 27 26 28 18 26 24
+ [911] 24 26 27 17 26 23 27 25 32 20 22 23 25 25 24 28 20 19 22 20 22 24 17 19 22 17
+ [937] 19 27 27 28 29 18 24 30 26 34 26 24 25 24 29 28 29 23 24 21 24 23 23 29 19 29
+ [963] 30 33 25 30 32 23 30 27 17 20 21 24 36 21 26 30 26 25 22 21 38 21 24 21 25 21
+ [989] 32 20 29 24 19 21 32 26 27 18 21 20
+>
+</code>
+그런데 교재는 이 이항분포를 비율로 (proportion) 생각하므로, 같은 방식으로 Red gumball의 비율로 바꿔서 보면
+<code>
+> # 아래처럼 n으로 (100개의 검볼이 총 숫자이므로)
+> # 나눠주면 비율을 구할 수 있다
+> proportions.of.rg <- numbers.of.red.gumball/n
+> ps.k <- proportions.of.rg
+> ps.k
+   [1] 0.18 0.27 0.27 0.22 0.23 0.26 0.23 0.26 0.25 0.30 0.27 0.28 0.32 0.24 0.26
+  [16] 0.29 0.22 0.24 0.18 0.27 0.33 0.22 0.27 0.31 0.29 0.19 0.24 0.24 0.27 0.24
+  [31] 0.23 0.21 0.21 0.25 0.31 0.21 0.29 0.16 0.31 0.24 0.24 0.28 0.23 0.24 0.22
+  [46] 0.19 0.31 0.28 0.20 0.19 0.24 0.27 0.28 0.24 0.28 0.27 0.25 0.27 0.26 0.29
+  [61] 0.29 0.26 0.36 0.29 0.27 0.16 0.23 0.30 0.32 0.22 0.32 0.26 0.29 0.29 0.22
+  [76] 0.18 0.22 0.27 0.33 0.27 0.28 0.28 0.34 0.15 0.32 0.23 0.24 0.20 0.16 0.27
+  [91] 0.31 0.27 0.21 0.22 0.29 0.24 0.22 0.19 0.18 0.20 0.17 0.24 0.30 0.27 0.23
+ [106] 0.19 0.17 0.28 0.37 0.20 0.18 0.26 0.30 0.30 0.34 0.30 0.25 0.23 0.26 0.24
+ [121] 0.20 0.19 0.25 0.22 0.29 0.25 0.25 0.27 0.19 0.27 0.23 0.22 0.23 0.26 0.25
+ [136] 0.25 0.32 0.25 0.27 0.32 0.22 0.32 0.23 0.30 0.21 0.25 0.27 0.17 0.24 0.21
+ [151] 0.24 0.26 0.33 0.20 0.22 0.26 0.28 0.25 0.30 0.33 0.27 0.30 0.26 0.23 0.39
+ [166] 0.23 0.31 0.18 0.26 0.27 0.34 0.25 0.28 0.31 0.35 0.28 0.29 0.32 0.27 0.31
+ [181] 0.28 0.25 0.22 0.23 0.15 0.22 0.20 0.26 0.21 0.22 0.16 0.23 0.22 0.31 0.24
+ [196] 0.27 0.31 0.21 0.24 0.26 0.26 0.22 0.22 0.34 0.19 0.30 0.22 0.28 0.25 0.24
+ [211] 0.29 0.25 0.25 0.16 0.27 0.23 0.25 0.32 0.18 0.22 0.25 0.25 0.24 0.24 0.21
+ [226] 0.32 0.20 0.28 0.29 0.22 0.23 0.22 0.25 0.21 0.27 0.22 0.24 0.29 0.24 0.22
+ [241] 0.30 0.22 0.21 0.17 0.25 0.23 0.21 0.27 0.22 0.22 0.25 0.22 0.29 0.24 0.26
+ [256] 0.32 0.28 0.20 0.22 0.22 0.27 0.26 0.22 0.24 0.31 0.18 0.27 0.29 0.28 0.17
+ [271] 0.27 0.33 0.23 0.33 0.25 0.32 0.26 0.23 0.19 0.21 0.20 0.23 0.15 0.19 0.23
+ [286] 0.26 0.27 0.28 0.23 0.24 0.35 0.27 0.30 0.23 0.25 0.24 0.31 0.23 0.20 0.22
+ [301] 0.22 0.26 0.21 0.22 0.26 0.28 0.26 0.23 0.21 0.13 0.29 0.27 0.21 0.34 0.28
+ [316] 0.24 0.19 0.26 0.27 0.25 0.23 0.27 0.25 0.19 0.29 0.18 0.28 0.21 0.27 0.28
+ [331] 0.28 0.22 0.22 0.20 0.20 0.25 0.27 0.17 0.16 0.27 0.32 0.23 0.18 0.28 0.31
+ [346] 0.29 0.21 0.27 0.27 0.30 0.21 0.25 0.20 0.25 0.26 0.30 0.26 0.21 0.15 0.29
+ [361] 0.22 0.21 0.16 0.25 0.25 0.27 0.26 0.27 0.28 0.21 0.27 0.24 0.25 0.24 0.39
+ [376] 0.24 0.28 0.33 0.20 0.26 0.24 0.27 0.20 0.31 0.27 0.27 0.20 0.21 0.31 0.25
+ [391] 0.22 0.22 0.30 0.34 0.27 0.23 0.21 0.25 0.20 0.24 0.29 0.19 0.30 0.27 0.33
+ [406] 0.22 0.29 0.30 0.22 0.29 0.26 0.24 0.18 0.26 0.36 0.26 0.23 0.24 0.22 0.32
+ [421] 0.33 0.16 0.24 0.28 0.24 0.25 0.29 0.31 0.28 0.28 0.29 0.26 0.24 0.25 0.28
+ [436] 0.27 0.24 0.31 0.25 0.31 0.33 0.26 0.26 0.24 0.33 0.28 0.20 0.23 0.22 0.23
+ [451] 0.22 0.30 0.25 0.25 0.23 0.27 0.27 0.23 0.24 0.28 0.24 0.28 0.23 0.22 0.26
+ [466] 0.30 0.26 0.27 0.21 0.23 0.23 0.27 0.26 0.23 0.25 0.30 0.25 0.24 0.22 0.28
+ [481] 0.18 0.23 0.18 0.16 0.27 0.26 0.18 0.25 0.27 0.22 0.20 0.19 0.27 0.25 0.31
+ [496] 0.27 0.22 0.21 0.24 0.24 0.26 0.23 0.23 0.29 0.27 0.23 0.25 0.20 0.21 0.21
+ [511] 0.27 0.25 0.22 0.29 0.28 0.21 0.21 0.24 0.27 0.24 0.28 0.19 0.14 0.32 0.27
+ [526] 0.22 0.24 0.35 0.26 0.28 0.28 0.26 0.25 0.25 0.19 0.26 0.24 0.20 0.19 0.28
+ [541] 0.25 0.25 0.24 0.21 0.30 0.27 0.30 0.20 0.22 0.26 0.31 0.26 0.20 0.20 0.27
+ [556] 0.25 0.26 0.18 0.30 0.20 0.29 0.16 0.38 0.26 0.22 0.29 0.22 0.30 0.26 0.19
+ [571] 0.27 0.24 0.29 0.29 0.25 0.19 0.23 0.24 0.24 0.23 0.25 0.31 0.18 0.24 0.33
+ [586] 0.27 0.25 0.27 0.29 0.28 0.24 0.23 0.24 0.28 0.20 0.24 0.30 0.24 0.21 0.20
+ [601] 0.25 0.24 0.24 0.30 0.22 0.26 0.23 0.25 0.21 0.21 0.24 0.27 0.18 0.20 0.22
+ [616] 0.30 0.25 0.23 0.27 0.26 0.23 0.23 0.28 0.18 0.29 0.27 0.25 0.32 0.26 0.15
+ [631] 0.22 0.24 0.21 0.34 0.23 0.23 0.18 0.29 0.23 0.27 0.28 0.23 0.37 0.20 0.17
+ [646] 0.25 0.11 0.21 0.28 0.22 0.28 0.25 0.22 0.25 0.21 0.18 0.20 0.27 0.30 0.24
+ [661] 0.28 0.23 0.30 0.31 0.24 0.23 0.37 0.19 0.27 0.32 0.25 0.27 0.28 0.29 0.22
+ [676] 0.26 0.26 0.20 0.22 0.25 0.24 0.19 0.27 0.21 0.32 0.27 0.31 0.29 0.24 0.24
+ [691] 0.29 0.29 0.25 0.22 0.34 0.23 0.18 0.33 0.18 0.23 0.24 0.26 0.18 0.20 0.23
+ [706] 0.30 0.28 0.26 0.34 0.17 0.33 0.30 0.32 0.30 0.22 0.28 0.19 0.19 0.23 0.23
+ [721] 0.20 0.23 0.21 0.31 0.30 0.20 0.24 0.23 0.23 0.28 0.26 0.34 0.27 0.33 0.31
+ [736] 0.20 0.25 0.12 0.25 0.20 0.20 0.25 0.27 0.24 0.29 0.26 0.22 0.30 0.26 0.28
+ [751] 0.28 0.27 0.23 0.18 0.28 0.22 0.21 0.27 0.22 0.26 0.21 0.22 0.27 0.24 0.19
+ [766] 0.27 0.29 0.37 0.30 0.27 0.25 0.30 0.19 0.22 0.22 0.28 0.32 0.22 0.33 0.26
+ [781] 0.20 0.31 0.23 0.24 0.24 0.26 0.24 0.30 0.17 0.21 0.20 0.22 0.20 0.17 0.24
+ [796] 0.22 0.24 0.23 0.23 0.24 0.23 0.16 0.16 0.17 0.23 0.27 0.29 0.26 0.16 0.21
+ [811] 0.34 0.19 0.25 0.25 0.28 0.32 0.17 0.22 0.26 0.23 0.23 0.24 0.22 0.22 0.14
+ [826] 0.30 0.25 0.33 0.26 0.25 0.31 0.28 0.30 0.21 0.19 0.17 0.19 0.21 0.16 0.21
+ [841] 0.26 0.21 0.29 0.27 0.31 0.32 0.19 0.22 0.24 0.25 0.25 0.24 0.23 0.30 0.21
+ [856] 0.22 0.19 0.20 0.21 0.20 0.21 0.28 0.19 0.26 0.28 0.26 0.29 0.28 0.26 0.21
+ [871] 0.31 0.32 0.31 0.22 0.23 0.25 0.27 0.26 0.22 0.27 0.30 0.24 0.25 0.23 0.27
+ [886] 0.25 0.24 0.24 0.30 0.29 0.26 0.32 0.29 0.23 0.24 0.20 0.26 0.26 0.22 0.22
+ [901] 0.19 0.23 0.33 0.18 0.27 0.26 0.28 0.18 0.26 0.24 0.24 0.26 0.27 0.17 0.26
+ [916] 0.23 0.27 0.25 0.32 0.20 0.22 0.23 0.25 0.25 0.24 0.28 0.20 0.19 0.22 0.20
+ [931] 0.22 0.24 0.17 0.19 0.22 0.17 0.19 0.27 0.27 0.28 0.29 0.18 0.24 0.30 0.26
+ [946] 0.34 0.26 0.24 0.25 0.24 0.29 0.28 0.29 0.23 0.24 0.21 0.24 0.23 0.23 0.29
+ [961] 0.19 0.29 0.30 0.33 0.25 0.30 0.32 0.23 0.30 0.27 0.17 0.20 0.21 0.24 0.36
+ [976] 0.21 0.26 0.30 0.26 0.25 0.22 0.21 0.38 0.21 0.24 0.21 0.25 0.21 0.32 0.20
+ [991] 0.29 0.24 0.19 0.21 0.32 0.26 0.27 0.18 0.21 0.20
+>
+</code>
+위의 비율의 기댓값을 (평균을) 구한다는 것이 교재가 하는 이야기
+<code>
+> mean.ps.k <- mean(ps.k)
+> mean.ps.k
+[1] 0.24893
+>
+</code>
+위의 결과를 histogram으로 그려보면
+<code>
+hist(ps.k)
+</code>
+이는 평균이 0.25에 (p값에) 근접하는 값이 된다. 교재의 p값이 되는 것은 k가 무한대로 큰 값을 가질 때의 이야기.
+아래는 k를 1000번이 아닌 1000000번 (백만번일 때의 이야기). 평균비율이 0.25가 된다.
+<code>
+> set.seed(101)
+> k <- 1000000
+> n <- 100
+> p <- 1/4
+> q <- 1-p
+> numbers.of.red.gumball <- rbinom(k, n, p)
+> # 아래처럼 n으로 (100개의 검볼이 총 숫자이므로)
+> # 나눠주면 비율을 구할 수 있다
+> proportions.of.rg <- numbers.of.red.gumball/n
+> ps.k <- proportions.of.rg
+> mean.ps.k <- mean(ps.k)
+> mean.ps.k
+[1] 0.2500217
+>
+</code>
+{{:b:head_first_statistics:pasted:20241106-081710.png}}
 ^ references  ^
@@ Line 162: / Line 346: @@
 ===== What about variance =====
+그렇다면 위의 분포에서의 분산값은 얼마가 될까? 그리고 표준편차값은 얼마가 될까?
 \begin{eqnarray*}
-Var(\text{probability of sample proportions}) & = & Var(P_{s}) \\
+\text{Variance of sample proportions} & = & Var(P_{s}) \\
 & = & Var\left(\frac{X}{n}\right) \\
 & = & \frac {Var(X)}{n^{2}} \\
 & = & \frac {npq}{n^{2}} \\
-& = & \frac {pq}{n}
+& = & \frac {pq}{n} \\
-\end{eqnarray*}
-\begin{eqnarray*}
 \text{Standard deviation of sample proportions} & = & \sqrt{\frac{pq}{n}} \\
 & = & \text{Standard error of sample proportions}
 \end{eqnarray*}
+우리는 위의 Standard deviation of sample proportions를 특별하게 standard error라고 부른다.
-이를 종합하면, Sample proportions 들에 대한 기대값과 분산은 각각 아래와 같다 (그림 참조).
+종합하면, Sample proportions 들에 대한 기대값과 분산은 각각 아래와 같다 (그림 참조).
 $$E(P_{s}) = p \qquad\qquad\qquad Var(P_{s}) = \displaystyle \frac{pq}{n}$$
@@ Line 183: / Line 366: @@
 continuity correction: $$\pm \frac{1}{2n}$$
+R에서의 simulation을 계속해서 보면
+<code>
+> # variance?
+> var.cal <- var(ps.k)
+> var.value <- (p*q)/n
+> var.cal
+[1] 0.001869001
+> var.value
+[1] 0.001875
+>
+> # standard deviation
+> sd.cal <- sqrt(var.cal)
+> sd.value <- sqrt(var.value)
+> sd.cal
+[1] 0.04323195
+> sd.value
+[1] 0.04330127
+> se <- sd.value
+> # 우리는 standard deviation of sample
+> # proportions 를 standard error라고
+> # 부른다
+>
+</code>
+위의 se는 standard deviation의 일종이므로 그 특성을 갖는다 (68, 95, 99%). 따라서 Red gumball의 비율이 1/4임을 알고 있을 때, n=100개의 gumball을 샘플링하면 (한번), red gumball의 비율은 p를 (0.25) 중심으로 위아래도 2*se 범위의 값이 나올 확률이 95%임을 안다는 것이 된다. 위에서 계산해보면;
+<code>
+# 위의 histogram 에서 mean 값은 이론적으로
+p
+# standard deviation값은
+se
+# 우리는 평균값에서 +- 2*sd.cal 구간이 95%인줄 안다.
+se2 <- se * 2
+# 즉, 아래 구간이
+lower <- p-se2
+upper <- p+se2
+lower
+upper
+hist(ps.k)
+abline(v=lower, col=2, lwd=2)
+abline(v=upper, col=2, lwd=2)
+</code>
+즉 아래의 그래프에서
+{{:b:head_first_statistics:pasted:20241106-084520.png}}
+lower: 0.1633975와 (16.33975%) upper: 0.3366025 사이에서 (33.66025%) red gumaball의 비율이 나올 확률이 95%라는 이야기.
+그렇다면 만약에 30% 이상이 red gumball일 확률은 무엇이라는 질문이라면
+우리는 X ~ B(100, 1/4)에서 도출되는
+X ~ N(p, se) 에서 P(X>_0.3)을 구하는 질문이므로
+-pnorm(0.295, p, se) 가 답이 되겠다.
+-pnorm(0.295, p, se)
+[1] 0.1493488
 ===== Exercise =====
@@ Line 224: / Line 462: @@
 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>
 </WRAP>
-====== How many gumballs? -- Probability of sample means ======
+====== Sampling distribution of sample mean ======
 <WRAP info 60%>
@@ Line 258: / Line 504: @@
 \overline{X} = \frac{X_{1} + X_{2} + . . . + X_{n}}{n}
 \end{eqnarray*}
+Let $ X_{1}, X_{2}, X_{3}, . . . X_{n} $ be a random sample of size of $n$.
+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)  \\
 & = & \frac{1}{n}\: E \left(X_{1} + X_{2} + . . . + X_{n}\right) \\
@@ Line 267: / Line 516: @@
 & = & \mu
 \end{eqnarray*}
+헷갈릴까봐 정리
+|                |   봉지 ID  ||||||
+|  | bag 1   | bag 2   | bag 3   | . . . .  | bag n-1  | bag n  |
+|  | 9   | 10  | 12  | . . . .  | 8   | 7   |
+|  | 5   | 12  | 9   | . . . .  | 12  | 10  |
+|  | 11  | 8   | 10  | . . . .  | 10  | 9   |
+| . . .  | ..  | ..  | ..  | . . . .  | ..  | ..  |
+| mean of $\overline{X}s = E(\overline{X})$      | mean of \\ bag 1 \\ $ X_{1} = 10 $   | mean of \\ bag 2 \\ $ X_{2} = 10 $     | mean of \\ bag 3 \\ $ X_{3} = 10 $    | . . . .  | mean of \\ bag n-1 \\ $ X_{n-1} = 10 $     | mean of \\ bag n \\ $ X_{n} = 10 $     |
 ===== Variance of sample means =====
@@ Line 275: / Line 535: @@
 \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 290: / Line 552: @@
 \end{eqnarray}
-$$\text{standard error of the sample means} = \frac{\sigma}{\sqrt{n}}$$
+\begin{eqnarray*}
+\text{standard error} & = & \text{standard deviation of sample means} \\
+& = & \frac{\sigma}{\sqrt{n}} \\
+& = & \sqrt{\frac{\sigma^{2}}{n}}
+\end{eqnarray*}
 {{:b:head_first_statistics:pasted:20191126-093924.png}}
@@ Line 307: / Line 573: @@
 ===== 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:
 $$\overline{X} \sim N(np, \; pq) $$
@@ Line 331: / Line 599: @@
 $$\overline{X} \sim N(10, \frac{1}{30})$$
+$P (\overline{X} < 8.5)$ 을 묻는 문제이므로
+\begin{eqnarray*}
+z & = & \frac{8.5-10}{\sqrt{\frac{1}{30}}} \\
+& = & -8.22
+\end{eqnarray*}
+따라서, 위의 문제는 $P(Z < z) = P(Z < -8.22)$를 묻는 문제
+<code>
+> pnorm(-8.22)
+[1] 1.017516e-16
+> pnorm(8.5, 10, sqrt(1/30))  # or . . . .
+[1] 1.053435e-16
+>
+</code>
+discrepancy?
+<code>
+> a <- sqrt(1/30)
+> b <- 8.5-10
+> b/a
+[1] -8.215838
+> pnorm(b/a)
+[1] 1.053435e-16
+</code>
+====== Recap ======
+Distribution of **Sample** <fc #ff0000>**P**</fc>roportion<fc #ff0000>**s**</fc>, <fc #ff0000>$Ps$</fc>,
+when sampling n entities (repeatedly) from a population whose proportion is p.
+\begin{eqnarray*}
+Ps & \sim & N(p,  \frac{pq}{n}) \\
+\text{hence, } \\
+\text{standard deviation of} \\
+\text{sample proportions} & = & \sqrt{\frac{pq}{n}}
+\end{eqnarray*}
+Distribution of **Sample** <fc #ff0000>Means, $\overline{X}$</fc>
+when sampling a sample whose size is n from a population whose mean is $\mu$ and variance is $\sigma^2$.
+\begin{eqnarray*}
+\overline{X} & \sim & N(\mu,  \frac{\sigma^2}{n}) \\
+\text{hence, } \\
+\text{standard deviation of} \\
+\text{sample means} & = &  \sqrt{\frac{\sigma^2}{n}} \\
+& = &  \frac{\sigma}{\sqrt{n}}
+\end{eqnarray*}