[問題] 一題兩獨立母體比例差的假說檢定

[libitum (libitum)]

想請教遇到的一題問題，題目如下


In one country, 52% of the population are males, and 48% are female.

In another country, 47% are male, and 53% are females.

Suppose that a random sample of 100 people are picked from each country

(total number of observations: 200).


What is the probability that a sample of this size would be

able to reject the null hypothesis that the proportion of males

in the first population are greater than the second population

at an alpha level of 0.05?

(note: round this probability to two decimal places)


我的問題主要是在 the probability of rejecting the null hypothesis?

不是根據alpha value去決定那個機率是否reject null hypothesis嗎

但感覺又沒有那麼簡單...

謝謝


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→ andrew43: 應該是要你算power的意思
→ libitum: 我來試試看 謝謝
想確認一下

power is the probability of rejecting H0 when H0 is false

所以

p1 = the probability of male shows in the first group (0.52)
p2 = the probaiblity of male shows in the second group (0.47)

H0: p1 <= p2
H1: P1 > p2

是這樣嗎 謝謝

※ 編輯: libitum (72.88.166.22 美國), 05/11/2020 11:53:41
→ andrew43: 就是算實際上p=0.47的抽樣分配可拒絕不正確的H0:p=0.52
→ yhliu: 是在 p1=0.52, p2=0.47 的情況下計算 reject H0 的機率.
→ yhliu: "Reject H0" 是在問題所定顯著水準下所決定的事件.
→ andrew43: 抱歉說錯。謝謝老師。H0應該是p1-p2<=0，且應關心p1-p2
→ andrew43: 的抽樣分配。
→ andrew43: 解題應該就從比較兩樣本比例的檢驗查起。
→ andrew43: 常見教材應該會寫到以CLT近似的解法。
→ andrew43: 更正 H0 p1-p2 gt 0
→ libitum: H0:p1-p2 >= 0 ?
→ libitum: 所以要算Pr(p1-p2<0)?
→ yhliu: H1 是 p1 > p2, H0 是 p1=p2 或 p1≦p2. 在 p1=p2 之下根據
→ yhliu: 顯著水準 0.05 決定 reject H0 的規則, 然後在 p1=0.52,
→ yhliu: p2 = 0.47 之下計算 reject H0 這一規則成立的機率.