hce_nthu
112年
資訊科學
第 28 題
From the general knowledge, the population occurrence rate of asthma is 0.1 for children who visit hospitals. A pediatrician moved his clinic to a new place and he found 9 children with asthma among 50 appointments in his first day at the new place. If the 50 appointments are considered as Bernoulli random variables with 0 and 1 outcomes, the total number of asthma children, indicated as X, should follow a Binomial distribution with probability 0.1 under the general condition. He then makes the following guesses. Which one is a correct inference for the occurrence rate at the new place?
- A If the occurrence rate holds the same as other places, the standard deviation of each Bernoulli random variable in this problem is $\sqrt{0.1 \times (1 - 0.1)} = 0.3$. The standard deviation of X should be $50 \times 0.3 = 15$
- B Since $9/50 = 0.18$ is greater than 0.1, there must be an increase of the occurrence rate
- C If the occurrence rate is greater than 0.1, we should also see the proportion of asthma children to be greater than 0.1 in the next day
- D Since $P(X=9) = 0.033$ is smaller than 0.05, we will decide this as an unusual situation, and the evidence supports an increased occurrence rate. The threshold 0.05 can be applied when the number of appointments is 70 in a day
- E Since $P(X\ge 9) = 0.058$ is greater than 0.05, we do not have to worry too much about the situation, and it might be a random fluctuation from day to day. The threshold 0.05 can be applied when the number of appointments is 70 in a day
思路引導 VIP
想像你在丟一枚公平的硬幣，如果連續出現了幾次正面，你會立刻斷定這枚硬幣有問題嗎？在統計決策中，如果我們想判斷一個觀測結果是否「極不尋常」，我們應該只計算「剛好發生那個次數」的機率，還是要考慮「包含那個次數以及更極端情況」的總機率呢？
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AI 詳解
AI 專屬家教
恭喜你精確地辨識出統計推論中的核心觀念！這題考驗的是你對「二項分布」（Binomial Distribution）性質與「顯著性檢定」邏輯的綜合判斷能力。
二項分布與統計決策的平衡
在統計學中，我們不能僅憑觀測比例（如本題的 $9/50 = 0.18$）高於母體機率（0.1）就草率斷定發生率上升。正確的做法是計算在原假設（$p=0.1$）下，發生「如此極端或更極端結果」的機率，也就是 P值 (p-value)。選項 (E) 提到 $P(X \ge 9) = 0.058$，由於這個數值大於常用的顯著水準 $0.05$，代表在統計上我們無法排除這只是日常隨機波動的可能性。相較之下，選項 (A) 的標準差計算錯誤，二項分布的標準差應為 $\sqrt{np(1-p)} = \sqrt{50 \times 0.1 \times 0.9} \approx 2.12$，而非 $15$。
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