b:head_first_statistics:estimating_populations_and_samples
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| b:head_first_statistics:estimating_populations_and_samples [2019/12/10 13:27] – [Estimating population variance] hkimscil | b:head_first_statistics:estimating_populations_and_samples [2022/11/17 12:47] (current) – [Exercise] hkimscil | ||
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| </ | </ | ||
| - | \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 ====== | ====== Estimating population variance ====== | ||
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| {{: | {{: | ||
| {{: | {{: | ||
| + | |||
| + | [[:Why N-1]] | ||
| < | < | ||
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| \end{eqnarray*} | \end{eqnarray*} | ||
| - | 이 때 각각의 시도에서의 (trial) proportion 기대값은 ($\hat{P}$): | + | 이 때 $n = 100$일때 각각의 시도에서의 (trial) proportion 기대값은 ($\hat{P}$): |
| - | \begin{eqnarray} | + | |
| - | \hat{P_{1}} & = & {X_{1}}/{100} = 0.3 \\ | + | \begin{align*} |
| - | \hat{P_{2}} & = & {X_{2}}/{100} = 0.7 \\ | + | n = 100, \\ |
| - | \hat{P_{3}} & = & {X_{3}}/{100} = 0.5 \\ | + | \hat{P_{1}} & = \frac{X_{1}}{n} = 0.34, (X_{1} = 34) \\ |
| - | \hat{P_{4}} & = & {X_{4}}/{100} = 0.4 \\ | + | \hat{P_{2}} & = \frac{X_{2}}{n} = 0.43, (X_{2} = 43) \\ |
| - | \cdots | + | \hat{P_{3}} & = \frac{X_{3}}{n} = 0.32, (X_{3} = 32) \\ |
| - | \hat{P_{k}} & = & {X_{k}}/{100} = 0.5 | + | \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 $P_{s} = \displaytype \frac{X}{n}$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, |
| {{: | {{: | ||
| Line 224: | 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 | ||
| </ | </ | ||
| </ | </ | ||
| - | ====== | + | ====== |
| <WRAP info 60%> | <WRAP info 60%> | ||
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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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| \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*} | ||
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| \end{eqnarray} | \end{eqnarray} | ||
| - | $$\text{standard error of the sample means} = \frac{\sigma}{\sqrt{n}}$$ | + | \begin{eqnarray*} |
| + | \text{standard error} & = & \text{standard deviation | ||
| + | & = & \frac{\sigma}{\sqrt{n}} | ||
| + | & = & \sqrt{\frac{\sigma^{2}}{n}} | ||
| + | \end{eqnarray*} | ||
| {{: | {{: | ||
| Line 307: | 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 345: | 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.1575952054.txt.gz · Last modified: 2019/12/10 13:27 by hkimscil