不等機率抽樣(PPS) - 考試
By Charlotte
at 2013-08-25T10:48
at 2013-08-25T10:48
Table of Contents
在儲全滋教授書中提及不等機率抽樣法中:
Zi = Ni/ΣNi
當群集的平均無顯著差異時,則可以用變量取代,i.e. Zi = Xi/ΣXi
只是我看到老師書上舉例時,我對符號有感到疑惑不知道是數據打錯還是我理解有誤:
Xij Xi Xi_bar
N1 2,18 10 5
N2 1,2,3,10,10 25 5
Xi不是把Xij作加總得到,對N1與N2來說應是20與26,為何是10與25?
也想請教到底如何判斷平均無顯著差異與每個抽樣單位Ni差異不大?
因在閱讀黃文隆老師與鄭光甫教授書後仍無法體會,煩請說明.謝謝!
--
一個人澈悟的程度
恰等于他所受痛苦的深度
~~林語堂
--
Zi = Ni/ΣNi
當群集的平均無顯著差異時,則可以用變量取代,i.e. Zi = Xi/ΣXi
只是我看到老師書上舉例時,我對符號有感到疑惑不知道是數據打錯還是我理解有誤:
Xij Xi Xi_bar
N1 2,18 10 5
N2 1,2,3,10,10 25 5
Xi不是把Xij作加總得到,對N1與N2來說應是20與26,為何是10與25?
也想請教到底如何判斷平均無顯著差異與每個抽樣單位Ni差異不大?
因在閱讀黃文隆老師與鄭光甫教授書後仍無法體會,煩請說明.謝謝!
--
一個人澈悟的程度
恰等于他所受痛苦的深度
~~林語堂
--
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考試
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