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b:head_first_statistics:estimating_populations_and_samples [2021/12/03 08:10] – [Exercise] hkimscilb:head_first_statistics:estimating_populations_and_samples [2022/11/17 12:47] (current) – [Exercise] hkimscil
Line 237: Line 237:
  
 <code> <code>
 +
 > p <- 0.25 > p <- 0.25
 > q <- 1-p > q <- 1-p
Line 245: Line 246:
 > o.c <- .4 - (1/(2*n)) > o.c <- .4 - (1/(2*n))
 > o.c  > o.c 
 +[1] 0.395
 > pnorm(o.c, p, se, lower.tail = F) > pnorm(o.c, p, se, lower.tail = F)
-[1] 0.0004313594+[1] 0.0004060586
 </code> </code>
  
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 \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)  \\
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 ===== Using CLT for the binomial distribution ===== ===== 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: 
 $$\overline{X} \sim N(np, \; pq) $$ $$\overline{X} \sim N(np, \; pq) $$
  
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 </code> </code>
 discrepancy? discrepancy?
 +<code>
 +> a <- sqrt(1/30)
 +> b <- 8.5-10
 +> b/a
 +[1] -8.215838
 +> pnorm(b/a)
 +[1] 1.053435e-16
 +
 +</code>
  
b/head_first_statistics/estimating_populations_and_samples.1638486640.txt.gz · Last modified: 2021/12/03 08:10 by hkimscil

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