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JMP Blog

A blog for anyone curious about data visualization, design of experiments, statistics, predictive modeling, and more
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變異數分析 (ANOVA) 與兩兩比較的思考脈絡與分析方法

在之前的文章中,我們介紹了組間比較的基本操作,並在上期文章中詳細介紹了t 檢定在JMP中的實現。 t 檢定是用於檢驗兩組均值差異的統計方法,本篇文章將帶您詳細瞭解什麼是變異數分析(ANOVA)、使用變異數分析前須考量哪些條件、以及如何使用JMP進行變異數分析(ANOVA)與兩兩比較。


在本文中,我們以圖1的資料為例進行講解。

 

Michelle_Wu_0-1632714609445.png

圖1 範例資料

 

什麼是變異數分析(ANOVA)? 

 

變異數分析,是把全部觀察值的總變異分解成組間變異和誤差變異,然後將組間變異與隨機誤差進行比較,從而判斷總體均數間的差別是否具有統計學意義。

使用變異數分析須滿足3大條件

 

變異數分析是 t 檢定更一般性的推廣,t 檢定可以看做是變異數分析的特例。

因此使用變異數分析的前提條件與 t 檢定一致:
①各個樣本是相互獨立的;
②各組資料均為常態分佈;
③各組間的變異數相等。

 

意思是,進行變異數分析前,我們需要進行常態檢定和變異數同質性檢定 (Homogeneity of variance test),由於變異數分析只能得出「組間是否有差異」的結論,具體哪幾組之間有差異,仍需要進一步統計分析,這時就需要用到兩兩比較方法,常見的兩兩比較方法有 Bonferroni 法、Tukey’ HSD 法和 Dunnett 法。

 

Bonferroni法

 

Bonferroni 法,是在進行兩兩比較時調整檢驗水準。通常組間比較以 0.05 作為檢驗水準,但在兩兩比較時,每次比較就會有 5% 的概率發生 I 類錯誤。

使用 Bonferroni 法的思路,是通過將 0.05 除以要比較的次數,降低檢驗水準,從而減少假陽性錯誤。如 4 組兩兩比較共需比較 6 次,則兩兩比較的檢驗水準需調整為 0.05/6=0.0083,即認為 p<0.0083 才算有統計學差異。但是該方法在比較次數較多時不太適合使用,因為校正後的檢驗水準會過小。

Tukey'HSD法

 

Tukey 法,是常見的兩兩比較方法,該方法曾經只能用於各組例數相等的情形,後來提出了改進的 Tukey 法,可用於各組例數不等的情形。 JMP 提供的就是改進的 Tukey 法,該方法可作為兩兩比較的首選方法。

Dunnett法

 

Dunnett t 檢定,專門用於比較 1 個對照組和多個試驗組間的差異,試驗組之間不做比較。


使用JMP進行變異數分析(ANOVA)

 

在 (圖1) 資料中,若比較不同心功能分級患者的軀體健康評分是否存在差異,心功能分級分為一到四級,因此這是一個四組之間的比較,不能直接用 t 檢定,而應考慮多組比較的方法。首先通過點選JMP菜單「分析→以X擬合Y」(如圖2),進入組間差異比較的界面。

Michelle_Wu_1-1632714609197.png

圖2  變異數分析操作——菜單選擇

本例中軀體健康評分為結果,心功能分級為分組,因此在對話框中將軀體健康評分放入「Y,響應」,將心功能分級放入「X,因子」(圖3)。

Michelle_Wu_2-1632714609456.png

圖3 變異數分析操作——變量選擇

 

進入結果畫面後,我們需要結合常態檢定和變異數同質性檢定 (Homogeneity of variance test) 的結果,選擇合適的統計方法,分析結果如下:

> 常態檢定結果顯示各組資料均為常態分佈。

> 變異數同質性檢定 (Homogeneity of variance test)結果見 (圖4)。

 

多組資料的變異數同質性檢定 (Homogeneity of variance test)多用 Bartlett 檢定和 Levene檢定。至於兩種檢定的使用情況略有不同,Bartlett 檢定主要用於常態分佈的資料,Levene檢定多用於資料不滿足常態分佈的情形。

 

閱讀文章〈在JMP中進行常態檢定與變異數同質性檢定〉瞭解如何進行該檢驗。


在這個範例,資料為常態分佈,因此我們採用Bartlett 檢定,分析結果顯示為 P=0.0224,為方差不齊。

Michelle_Wu_3-1632714609496.png

圖4 變異數同質性檢定 (Homogeneity of variance test)結果

當資料為常態分佈但不滿足變異數同質性檢定 (Homogeneity of variance test)時,採用Welch變異數分析,方法選擇可參考文章〈實現資料的所有組間比較的強大功能 -- 以X擬合Y〉。

Welch 變異數分析結果見變異數同質性檢定 (Homogeneity of variance test) 結果的最後一部分(圖5)。結果顯示四組間軀體健康評分的差異有統計學意義(F=40.2951,P<0.0001)。

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圖5 Welch變異數分析輸出結果

 

如果資料滿足常態且方差齊,則可直接採用變異數分析,儘管從條件來看,本例資料應該用 Welch 檢定,但做為範例,我們同時也介紹一下變異數分析的結果如何輸出。點擊「心功能分級-軀體健康評分」單因子分析旁邊的紅色三角形按鈕,在下拉菜單中選擇「均值/變異數分析」,如圖6。

Michelle_Wu_5-1632714609297.png

 

圖6 變異數分析操作——方法選擇

 

輸出結果見 (圖7),變異數分析結果表明四組的軀體健康評分差異有統計學意義 (F=16.0080,P<0.0001)。



Michelle_Wu_6-1632714609352.png

 

圖7 變異數分析輸出結果
從上述分析結果可以看出,變異數分析的 F 值與 Welch 檢定結果有一定的差異。因此對於連續變量的組間比較一定要綜合考慮其常態性與變異數同值性。


JMP中的兩兩比較

 

如果總的變異數分析結果,顯示無統計學差異,提示各組間均無統計學差異,就不要再做兩兩比較;不過如果總的組間比較結果顯示四組的差異有統計學意義,那麼通常還需要進行組間兩兩比較,以明確具體是哪兩組之間有差異。

JMP中常態資料的兩兩比較比較操作在「比較均值」的選項中完成操作,由於我們要比較任意兩組之間的差異,可選擇 Tukey 法,操作見圖8。



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圖8 兩兩比較操作

 

點擊「心功能分級-軀體健康評分」單因子分析旁邊的紅色三角形按鈕,在下拉菜單中選擇「比較均值→所有對,Tukey HSD」。

 

輸出結果見 (圖9),結果顯示除了心功能分級3和4間無差異,其它組之間都有統計學差異。本例分析結果表明,不同心功能分級人群的軀體健康評分差異有統計學意義 (F=16.0080,P<0.0001),除了心功能分級3和4間無差異,其它心功能分級之間的差異都有統計學意義。

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圖9 兩兩比較輸出結果圖

以上就是本期我們為大家帶來的實用分享,立即下载 JMP 試試以上的操作吧!

 

原文:干货 | 方差分析及两两比较的思路与实现

 

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Last Modified: Dec 19, 2023 2:50 PM