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Openai/691718ed-7ba4-8006-96d5-a66e0d9678fe
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=== Assistant: 對!你觀察得非常精準 👏 === 對!你觀察得非常精準 👏 這其實來自二項分布的變異數公式: Var(X)=np0(1−p0),σ=np0(1−p0)\text{Var}(X) = n p_0 (1-p_0), \quad \sigma = \sqrt{n p_0 (1-p_0)}Var(X)=np0(1−p0),σ=np0(1−p0) ==== 為什麼 ==== 𝑝 0 p 0 越靠近 0.5 誤差越大? * 當 p0=0.5p_0 = 0.5p0=0.5: - 變異數最大 → σ=n×0.25=12n\sigma = \sqrt{n \times 0.25} = \tfrac{1}{2}\sqrt{n}σ=n×0.25=21n - 所以臨界比例差最大。 * 當 p0p_0p0 偏離 0.5(例如 0.2 或 0.8): - p0(1−p0)\,p_0(1-p_0)p0(1−p0) 變小 → 標準差變小 → 信賴區間縮小。 - 所以在相同 nnn 下,檢定「更敏感」。 ==== 舉例(n = 1000, α = 0.05, 雙尾) ==== 臨界比例差 ≈ ±1.96p0(1−p0)/n\pm 1.96 \sqrt{p_0 (1-p_0) / n}±1.96p0(1−p0)/n * p0=0.5p_0 = 0.5p0=0.5: - 誤差 ≈ ±3.1% * p0=0.2p_0 = 0.2p0=0.2: - 誤差 ≈ ±2.5% * p0=0.1p_0 = 0.1p0=0.1: - 誤差 ≈ ±1.9% ==== 直觀比喻 ==== 想像「擲硬幣」: * 公平硬幣 (p=0.5) → 結果最難判斷是不是偏差,因為波動最大。 * 偏硬幣 (p=0.2 或 0.8) → 分布天生就比較「穩定」,少量差異就比較容易被偵測。 要不要我幫你畫一張圖,橫軸 p0p_0p0(0 到 1),縱軸顯著水準下的「臨界誤差範圍」(±%),讓你直觀看到「在 0.5 最寬、越靠邊越窄」的效果?
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