User Tools

Site Tools


b:head_first_statistics:estimating_populations_and_samples

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revisionPrevious revision
Next revision
Previous revision
Next revisionBoth sides next revision
b:head_first_statistics:estimating_populations_and_samples [2020/12/10 16:52] – [Expectation of samples proportions (Ps)] hkimscilb:head_first_statistics:estimating_populations_and_samples [2022/11/15 23:44] – [Sampling distribution of sample mean] hkimscil
Line 132: Line 132:
    
 이 때 $n = 100$일때 각각의 시도에서의 (trial) proportion 기대값은 ($\hat{P}$): 이 때 $n = 100$일때 각각의 시도에서의 (trial) proportion 기대값은 ($\hat{P}$):
-\begin{eqnarray+ 
-\hat{P_{1}} & = {E(X_{1})}/{n} = 0.34 \\ +\begin{align*} 
-\hat{P_{2}} & = {E(X_{2})}/{n} = 0.43 \\ +n = 100, \\ 
-\hat{P_{3}} & = {E(X_{3})}/{n} = 0.32 \\ +\hat{P_{1}} & = \frac{X_{1}}{n} = 0.34, (X_{1} = 34) \\ 
-\hat{P_{4}} & = {E(X_{4})}/{n} = 0.42 \\ +\hat{P_{2}} & = \frac{X_{2}}{n} = 0.43, (X_{2} = 43) \\ 
-\cdots \cdots \cdots              \\ +\hat{P_{3}} & = \frac{X_{3}}{n} = 0.32, (X_{3} = 32) \\ 
-\hat{P_{k}} & = {E(X_{k})}/{n} = 0.24  +\hat{P_{4}} & = \frac{X_{4}}{n} = 0.42, (X_{4} = 42) \\ 
-\end{eqnarray}+\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). 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).
Line 226: Line 228:
 q <- 1-p q <- 1-p
 n <- 100 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>
  
 <code> <code>
 +
 > p <- 0.25 > p <- 0.25
 > q <- 1-p > q <- 1-p
 > n <- 100 > 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> </code>
  
Line 260: Line 270:
 \overline{X} = \frac{X_{1} + X_{2} + . . . + X_{n}}{n}  \overline{X} = \frac{X_{1} + X_{2} + . . . + X_{n}}{n} 
 \end{eqnarray*} \end{eqnarray*}
 +위는 풍선검 봉지 30개로 이루어진 샘플의 평균을 이야기하고 
 +아래는 이 평균을 계속 모았을 때의 평균을 이야기한다. 
 \begin{eqnarray*} \begin{eqnarray*}
 E(\overline{X}) & = & E\left(\frac{X_{1} + X_{2} + . . . + X_{n}}{n}\right)  \\ E(\overline{X}) & = & E\left(\frac{X_{1} + X_{2} + . . . + X_{n}}{n}\right)  \\
Line 277: Line 288:
 \end{eqnarray*} \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*}
  
  
b/head_first_statistics/estimating_populations_and_samples.txt · Last modified: 2022/11/17 12:47 by hkimscil

Donate Powered by PHP Valid HTML5 Valid CSS Driven by DokuWiki