b:head_first_statistics:estimating_populations_and_samples
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| b:head_first_statistics:estimating_populations_and_samples [2020/12/10 17:11] – [Expectation of samples proportions (Ps)] hkimscil | b:head_first_statistics:estimating_populations_and_samples [2022/11/17 12:47] (current) – [Exercise] hkimscil | ||
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| Line 228: | 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)/ | + | se <- sqrt((p*q)/ |
| - | 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) | ||
| </ | </ | ||
| < | < | ||
| + | > | ||
| > 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)/ | + | > se <- sqrt((p*q)/ |
| - | > 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 | ||
| </ | </ | ||
| Line 262: | 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 279: | 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} | + | & = \frac {1}{n^2} Var \left(X_{1} + X_{2} + . . . + X_{n} \right) \\ |
| - | & = & \frac{1}{n^2} | + | & = \frac{1}{n^2} (\sigma^2 + \sigma^2 + . . . + \sigma^2) \\ |
| - | & = & \frac{1}{n^2} | + | & = \frac{1}{n^2} n * (\sigma^2) \\ |
| - | & = & \frac{\sigma^2}{n} | + | & = \frac{\sigma^2}{n} |
| - | \end{eqnarray*} | + | |
| + | |||
| + | \end{align*} | ||
| Line 315: | Line 326: | ||
| ===== 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 | + | $X \sim B(n, p)$ 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: | ||
| $$\overline{X} \sim N(np, \; pq) $$ | $$\overline{X} \sim N(np, \; pq) $$ | ||
| Line 353: | Line 366: | ||
| </ | </ | ||
| discrepancy? | discrepancy? | ||
| + | < | ||
| + | > a <- sqrt(1/30) | ||
| + | > b <- 8.5-10 | ||
| + | > b/a | ||
| + | [1] -8.215838 | ||
| + | > pnorm(b/a) | ||
| + | [1] 1.053435e-16 | ||
| + | |||
| + | </ | ||
b/head_first_statistics/estimating_populations_and_samples.1607587915.txt.gz · Last modified: 2020/12/10 17:11 by hkimscil