Review of Statistics

相關主題 統計之基礎理論與觀念 基礎統計在品管與製程上之應用(含可靠度) 進階統計之一 ~ 實驗計劃法與田口式品質工程 進階統計之一 ~ 實驗計劃法與田口式品質工程 進階統計之二 ~ 反應曲面技術 進階統計之三 ~ 迴歸分析 進階統計之四 ~ 時間序列 進階統計之五 ~ 多變量分析

機率 Vs. 統計 抽樣 母體 樣本 推論 母體參數 樣本統計量

敘述統計(一)-統計圖表 次數分配圖 and Stem-and-Leaf Diagram 交叉列表 長條圖 (Histogram) 長方圖 圓形圖 Box Plot 時間序列圖 (Time Sequence Plots) 圖表之誤解/誤用

敘述統計(二)-統計量 集中趨勢 - 平均值(m ), 中位數(Me), 眾數(Mo) 位置統計量 - 百分位, 四分位 離散統計量 - 全距(R), 變異數(s 2), 標準差(s ), 變異係數(CV) 形狀統計量 - 偏度(Skewness), 峰度(Kurtosis) 經驗法則 常態分配 => 68.3%, 95.4%, 99.7% 其他 => 柴比雪夫不等式(Chebyshev’s Inequality)

常用之機率公式 若P(AB) = P(A) * P(B), 則事件A與B獨立 P(AB) = P(A) + P(B) - P(A  B) 條件機率 貝氏定理 (Bayesian)

期望值與變異數之公式 母體平均數(m ) = 隨機變數之期望值 E(X) 母體變異數(s 2) = 隨機變數之變異數 V(X)

期望值與變異數之公式

機率分配 離散型變數 二項分配 超幾何分配 波以松分配 連續型變數 常態分配 指數分配 韋伯分配

常態分配(Normal) 機率函數: E(X) = m, V(X) = s 2 P(m-s<X<m+s) = 0.683 相加性: 標準化公式:

Standardization

Central Limit Theorem

Criteria for Point Estimator Unbiased Minimum Variance Absolute Efficiency Relative Efficiency

假設檢定(Hypothesis Testing) “A person is innocent until proven guilty beyond a reasonable doubt.” 在沒有充分證據證明其犯罪之前, 任何人皆是清白的. 假設檢定 H0: m = 50 cm/s H1: m  50 cm/s Null Hypothesis (H0) Vs. Alternative Hypothesis (H1) One-sided and two-sided Hypotheses A statistical hypothesis is a statement about the parameters of one or more populations.

Errors in Hypothesis Testing 檢定結果可能為 Type I Error(a): Reject H0 while H0 is true. Type II Error(b): Fail to reject H0 while H0 is false.

Hypothesis Testing on m - Variance Known

Construction of the C.I. From Central Limit Theory, Use standardization and the properties of Z,

Summary Table of Influence Procedures for a Single Sample (I)

Summary Table of Influence Procedures for a Single Sample (II)

Goodness-of-Fit Test (II) If the population follows the hypothesized distribution, X02 has approximately a chi-square distribution with k-p-1 d.f., where p represents the number of parameters of the hypothesized distribution estimated by sample statistics. That is, Reject the hypothesis if

Contingency Table Test - The Problem Formulation (I) There are two classifications, one has r levels and the other has c levels. (3 pension plans and 2 type of workers) Want to know whether two methods of classification are statistically independent. (whether the preference of pension plans is independent of job classification) The table:

Contingency Table Test - The Problem Formulation (II) Let pij be the probability that a random selected element falls in the ijth cell, given that the two classifications are independent. Then pij = uivj, where the estimator for ui and vj are Therefore, the expected frequency of each cell is Then, for large n, the statistic has an approximate chi-square distribution with (r-1)(c-1) d.f.