X is a random variable. with any distribution. Central limit theorem The sum of X (many many independent and identically distributed X) will become normal as the number of X goes larger. rule of thumb: when n>=30 the sum of X is already close to normal. 记. Sampling distribution: distribution of $\bar{X}$, $\hat{P}$ For $\bar{X}$ sample mean. sample mean sample proportion. 1) assume sample size $n \geq 30$. 2) assume sample size $n < 0.1N$ . Because $n < 0.1N$. . we have independent. sampling. Because $n \geq 30$ we can use C.L.T. Y = $X_1$ + ... + $X_n \sim N$ Because of C.L.T. $\bar{X}$ = $\frac{Y}{n} \sim N$ $E(\bar{X}) = E(\frac{Y}{n})$ = $\frac{1}{n} E(Y)$ = $\frac{1}{n} \cdot n E(X)$ Sample mean is normal because of CLT. individual's distribution is unknown in Sample mean's Center of distribution population center of distribution shape. Std($\bar{X}$) = Std($\frac{Y}{n}$) = $\frac{1}{n}$ Std(Y) = $\frac{1}{n} \cdot \sqrt{n}$ Std(X) = $\frac{Std(X)}{\sqrt{n}}$ 当 population 为正态时: $X_1$,...,$X_n$ 天然为正态. $Y = X_1+...+X_n$ 也为正态. 无须 $n \geq 30$. $\bar{X} = \frac{Y}{n} \sim N$. 如何可知 population 为正态 ① Sample 无 skewness, 无 outlier |常考. ② use Q-Q plot. |很少考. ③. Shapiro-Wilk & Kolmogorov-Smirnov. 不考. PBL用. For Sample proportion. $\hat{P}$ $\hat{P} = \frac{Y}{n}$. Y is Binomial distribution Sum of Bernoulli distribution 已知当 $np \geq 10$ 且 $n(1-p) \geq 10$ 时. $Y \sim Binom(n, p) \implies Y \sim Normal(np, \sqrt{np(1-p)})$ $\hat{P} \sim Normal$. $E(\hat{P}) = E(\frac{Y}{n}) = \frac{1}{n} E(Y) = P$. mean of Sample proportion. population proportion Prob. Std($\hat{P}$) = Std($\frac{Y}{n}$) = $\frac{1}{n}$ Std(Y) = $\frac{1}{n} \sqrt{np(1-p)}$. = $\sqrt{\frac{P(1-P)}{n}}$ $n \uparrow \implies Std(\hat{P}) \downarrow$