RStudio : Bonferroni Test

Bartlettテストで等分散性が棄却されなかったのでBonferroniの多重比較を実施します。

１．データインポート

> score_value=c(11,11.15384615,11.56521739,10.63636364,11.13157895,10.55555556,9.833333333,9.571428571,9.838709677,7.678571429,9.176470588,7.333333333,6.909090909,10.38461538,10.69230769,11.7826087,10.72727273,11.15789474,11,9.666666667,9.214285714,9.612903226,7.607142857,8.352941176,7.333333333,7,11.38461538,11.92307692,11.7826087,10.61363636,10.57894737,10.55555556,9,9.571428571,9.387096774,7.464285714,8.352941176,8.111111111,6.545454545,10.76923077,10.61538462,11.73913043,10.40909091,10.28947368,10,9.166666667,8.285714286,9.096774194,7.035714286,7.823529412,6.555555556,6.545454545)

２．グループ作成

> group=factor(rep(c('awareness', 'decision', 'communication', 'leadership'), c(13, 13, 13, 13)))

３．検定実行

> pairwise.t.test(score_value, group, p.adjust.method='bonferroni',paired=TRUE)

結果

 Pairwise comparisons using paired t tests 

data:  score_value and group 

              awareness communication decision
communication 1.0000    -             -       
decision      1.0000    1.0000        -       
leadership    0.0010    0.0227        0.0054  

P value adjustment method: bonferroni 

leadershipが他の要素と有意差があることがわかった。 

RStudio : Boxplot

RStudioでBoxplotを書きます。

１．データのインポート

> awareness_mean1=c(11,11.15384615,11.56521739,10.63636364,11.13157895,10.55555556,9.833333333,9.571428571,9.838709677,7.678571429,9.176470588,7.333333333,6.909090909)
> decision_mean1=c(10.38461538,10.69230769,11.7826087,10.72727273,11.15789474,11,9.666666667,9.214285714,9.612903226,7.607142857,8.352941176,7.333333333,7)
> comunication_mean1=c(11.38461538,11.92307692,11.7826087,10.61363636,10.57894737,10.55555556,9,9.571428571,9.387096774,7.464285714,8.352941176,8.111111111,6.545454545)
> leadership_mean1=c(10.76923077,10.61538462,11.73913043,10.40909091,10.28947368,10,9.166666667,8.285714286,9.096774194,7.035714286,7.823529412,6.555555556,6.545454545)

２．グラフの描画

> boxplot(awareness_mean1,decision_mean1,comunication_mean1,leadership_mean1, main='BoxPlot for mean of each category',names = c('Awareness', 'Decision', 'Communica','Leadership'))

もう一つ例

> awareness_variance1=c(1.976331361,3.146683673,3.167533819,3.244897959,2,2.851211073,0.805555556,2.082644628,1.135802469,1.538461538,1.685950413,1.114265928,0.767485822)
> decision_variance1=c(1.905325444,5.095663265,4.172736733,3.168367347,1.777777778,2.46366782,1.222222222,2.727272727,1.111111111,2.236686391,1.652892562,0.922437673,0.691871456)
> comunication_variance1=c(0.071005917,5.677295918,4.172736733,3.673469388,2.987654321,3.404844291,1.333333333,2.975206612,1.135802469,1.621301775,1.646177686,0.822714681,0.517958412)
> leadership_variance1=c(10.23668639,6.605867347,5.635796046,4.918367347,3.135802469,2.851211073,2.472222222,1.884297521,1.777777778,1.562130178,1.559917355,1.100415512,0.801512287)
> boxplot(awareness_variance1,decision_variance1,comunication_variance1,leadership_variance1, main='BoxPlot for variance of each category',names = c('Awareness', 'Decision', 'Communica','Leadership'))

RStudio : Bartlett Test

RStudioを使って多重比較のための等分散性の検定を実施します。

帰無仮説：全てのカテゴリの母分散が等しい
対立仮説：少なくとも一つの母分散が他の母分散と異なる

１．データを入力

> score=c(11,10.38461538,11.38461538,10.76923077,11.15384615,10.69230769,11.92307692,10.61538462,11.56521739,11.7826087,11.7826087,11.73913043,10.63636364,10.72727273,10.61363636,10.40909091,11.13157895,11.15789474,10.57894737,10.28947368,10.55555556,11,10.55555556,10,9.833333333,9.666666667,9,9.166666667,9.571428571,9.214285714,9.571428571,8.285714286,9.838709677,9.612903226,9.387096774,9.096774194,7.678571429,7.607142857,7.464285714,7.035714286,9.176470588,8.352941176,8.352941176,7.823529412,7.333333333,7.333333333,8.111111111,6.555555556,6.909090909,7,6.545454545,6.545454545)

２． ラベルと標本数

> group=factor(rep(c("Awareness", "Decision", "Comunication", "Leadership"), c(13, 13, 13, 13)))

３．テスト実施 
> bartlett.test(score ~ group)

 Bartlett test of homogeneity of variances

data:  score by group
Bartlett's K-squared = 5.9198, df = 3, p-value = 0.1156

４．結果
p-value = 0.1156なので各群の母分散は等しくないと言えない（等しいとして次の検定を実施）。この場合は Bonferroniを実施

もう１件例を書いておく

> score_v=c(1.976331361,1.905325444,0.071005917,10.23668639,3.146683673,5.095663265,5.677295918,6.605867347,3.167533819,4.172736733,4.172736733,5.635796046,3.244897959,3.168367347,3.673469388,4.918367347,2,1.777777778,2.987654321,3.135802469,2.851211073,2.46366782,3.404844291,2.851211073,0.805555556,1.222222222,1.333333333,2.472222222,2.082644628,2.727272727,2.975206612,1.884297521,1.135802469,1.111111111,1.135802469,1.777777778,1.538461538,2.236686391,1.621301775,1.562130178,1.685950413,1.652892562,1.646177686,1.559917355,1.114265928,0.922437673,0.822714681,1.100415512,0.767485822,0.691871456,0.517958412,0.801512287)
> group=factor(rep(c("Awareness", "Decision", "Comunication", "Leadership"), c(13, 13, 13, 13)))
> bartlett.test(score_v ~ group)

 Bartlett test of homogeneity of variances

data:  score_v by group
Bartlett's K-squared = 40.782, df = 3, p-value = 7.274e-09

今回は有意差ありなので、等分散性は棄却される。この場合は  Steel–Dwass nonparametric testを実施。

参考
http://bio-info.biz/tips/r_bartlett_test.html