hce_nthu
111年
資訊科學
第 30 題
The following is a joint probability table between two categorical variables, the education level and the exercise frequency.
| | 0-1 day per week | 2-4 days per week | 5-7 days per week |
| :--- | :--- | :--- | :--- |
| High school degree | 0.1 | 0.05 | 0.05 |
| Bachelor degree | 0.15 | 0.2 | 0.1 |
| Graduate degree | 0.15 | 0.15 | 0.05 |
Which of the following statements is correct?
| | 0-1 day per week | 2-4 days per week | 5-7 days per week |
| :--- | :--- | :--- | :--- |
| High school degree | 0.1 | 0.05 | 0.05 |
| Bachelor degree | 0.15 | 0.2 | 0.1 |
| Graduate degree | 0.15 | 0.15 | 0.05 |
Which of the following statements is correct?
- A If the table is from a large survey and can be treated as population data, we can say that the two variables are not independent
- B If the table is derived from sample data, we can test the independence of the two variables using $\chi^2$ test. The alternative hypothesis for the test is the independence of the two variables
- C If the table is derived from sample data, the degree of freedom of the null distribution of the $\chi^2$ test is 9
- D If the table is derived from sample data, the $\chi^2$ test statistic for testing the independence of the two variables is $\frac{(0.1-0.2\times0.4)^2}{0.2\times0.4}+\frac{(0.05-0.2\times0.4)^2}{0.2\times0.4}+\frac{(0.05-0.2\times0.2)^2}{0.2\times0.2}+\frac{(0.15-0.45\times0.4)^2}{0.45\times0.4}+\frac{(0.2-0.45\times0.4)^2}{0.45\times0.4}+\frac{(0.1-0.45\times0.2)^2}{0.45\times0.2}+\frac{(0.15-0.35\times0.4)^2}{0.35\times0.4}+\frac{(0.15-0.35\times0.4)^2}{0.35\times0.4}+\frac{(0.05-0.35\times0.2)^2}{0.35\times0.2}$
- E The $\chi^2$ test assumes that the samples categorized into the same cell are correlated
思路引導 VIP
若要判斷「教育程度」是否會影響「運動頻率」，在完全不看具體數字的情況下，如果這兩個變數「互不干涉（獨立）」，那麼我們應該如何利用各自出現的總機率，來推算出兩者同時發生的「理論機率」呢？試著將推算出的理論值與表中的實際數值做比對看看。
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AI 詳解
AI 專屬家教
太棒了！你能精準判斷出 (A) 為正確選項，代表你對聯合機率表（Joint Probability Table）以及統計獨立性的本質有很扎實的理解。這題的鑑別度在於區分「母體參數」與「樣本統計量」的應用。當我們將此表視為母體數據時，判斷獨立性的標準非常直觀：只需檢查所有格子的聯合機率 $P(A \cap B)$ 是否等於其邊際機率的乘積 $P(A) \times P(B)$。以「高中學歷且運動 0-1 天」為例，該格機率為 $0.1$，而邊際機率乘積為 $0.2 \times 0.4 = 0.08$，兩者不等，即證實兩變數不獨立。
統計檢定的常見陷阱
其餘選項則涉及了推論統計的細節。選項 (B) 的錯誤在於對立假設（$H_1$）應為「不獨立」；選項 (C) 的自由度應為 $(3-1) \times (3-1) = 4$；而選項 (D) 則忽略了 $\chi^2$ 檢定量必須基於「次數（Counts）」而非「機率」來計算，缺少了樣本總數 $n$。這題難度屬於中等，成功切入點在於不被複雜的計算公式迷惑，回歸獨立性的核心定義進行檢視。