r.220.149.117.26/teaching/r/anova.pdf · one-way anova ¯‰À˘@–“À˘ 1˝,...

$: R.220.149.117.26/teaching/r/anova.pdf · one-way ANOVA ¯‰À˘@–“À˘ 1˝, ¯‰À˘XÑŁ(factor level)t2˝ t`x‰° l„Ý— 0x ÝÌ‹„D ˜X(t ‹ﬂÀ• \ä. ì0˝ l„Ý@¯‰À˘\$
R과데이터분석분산분석

양창모

청주교육대학교컴퓨터교육과

2015년여름

양창모 (청주교육대학교컴퓨터교육과) Data Analysis using R 2015년여름 1 / 30

개요

분산분석analysis of variance, ANOVA

Factor(범주형변수,분류형변수)가독립변수에포함될때종속변수의분산을설명하는독립변수의평균차이의유의성을알아보는방법

종속변수 -측정형변수,독립변수 -분류형변수

세개이상의평균비교에대한검정방법

분산의비율은 F분산을따른다.

독립변수,종속변수모두가측정형이면회귀분석

독립변수,종속변수모두가분류형이면교차분석


개요

일원분산분석one-way ANOVA

독립변수와종속변수가각각 1개,독립변수의집단(factor level)이 2개이상인경우

예

가구소득에따른식료품소비정도의차이가있는지검정한다.여기서가구소득은독립변수로가구소득집단의구분-저소득,중산층,고소득층등으로 2개이상이다.


개요

이원분산분석two-way ANOVA

독립변수가두개이상일때집단간차이가유의한지를검증하기위하여

사용한다.

예

학력및성별에따른휴대폰요금의차이를분석하고자한다.학력,성별은독립변수이고종속변수는휴대폰요금이된다.


개요

다변량분산분석MANOVA

단순한분산분석을확장하여두개이상의종속변수가서로관계된상황에

적용시킨것

둘이상의집단간차이를검증할수있다.


개요

일원분산분석

oneway.test(x ∼ f)

범주형벡터 textitf의값에따라수치형벡터 x의평균이유의하게차이가있는지검정

귀무가설 :그룹의평균이같음

oneway.test(x ∼ f, var.equal=TRUE), aov(x ∼ f)

범주형벡터 f의값에따라수치형벡터 x의평균이유의하게차이가있는지검정

범주형벡터 f의값에따라수치형벡터 x의분산이같은경우귀무가설 :그룹의평균이같음


개요

일원분산분석의예

자동차브랜드에따른연비분석

> mpg = c(34, 35, 34.3, 35.5, 35.8, 35.3, 36.5, 36.4,+ 37, 37.6, 33.3, 34, 34.7, 33, 34.9)> brand = c("A", "A", "A", "A", "A", "B", "B", "B",+ "B", "B", "C", "C", "C", "C", "C")> mileage <- data.frame(mpg, brand)> head(mileage)

mpg brand1 34.0 A2 35.0 A3 34.3 A4 35.5 A5 35.8 A6 35.3 B


개요



> boxplot(mpg ~ brand)


개요



독립변수 -브랜드,종속변수 -연비분산분석에서독립변수는범주형변수 factor이어야한다.

> head(brand)[1] "A" "A" "A" "A" "A" "B"> str(brand)chr [1:15] "A" "A" "A" "A" "A" "B" "B" "B" "B" ...

> brand <- factor(brand)> str(brand)Factor w/ 3 levels "A","B","C": 1 1 1 1 1 2 2 2 2 2 ...

> head(brand)[1] A A A A A BLevels: A B C


개요



독립변수 - brand종속변수 - mpgformula - mpg brand

> xtabs(mpg~brand)brand

A B C174.6 182.8 169.9


개요



p − value = 0.00498 < 0.05이므로평균이같다는귀무가설을기각한다.

> xtabs(mpg~brand)brand

A B C174.6 182.8 169.9

> oneway.test(mpg~brand)

One-way analysis of means (not assuming equal variances)

data: mpg and brandF = 11.077, num df = 2.0000, denom df = 7.9827, p-value = 0.00498


개요



p − value = 0.001076 < 0.05이므로평균이같다는귀무가설을기각한다.

> oneway.test(mpg~brand, var.equal = T)

One-way analysis of means

data: mpg and brandF = 12.742, num df = 2, denom df = 12, p-value = 0.001076

> a <- aov(mpg~brand)> summary(a)

Df Sum Sq Mean Sq F value Pr(>F)brand 2 17.049 8.525 12.74 0.00108 **Residuals 12 8.028 0.669---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


개요


어떤그룹의평균의쌍이유의미하게다른가?

사후검정(Post-Hoc test)또는다중비교(multiple comparison)

Fisher’s Least Significant DifferenceTukey W procedure -가장일반적Student-Newman-Keuls procedure ...


개요



Tukey사후검정

> TukeyHSD(a)Tukey multiple comparisons of means95% family-wise confidence level

Fit: aov(formula = mpg ~ brand)

$branddiff lwr upr p adj

B-A 1.64 0.2599123 3.0200877 0.0204273C-A -0.94 -2.3200877 0.4400877 0.2056606C-B -2.58 -3.9600877 -1.1999123 0.0008518


개요



Tukey사후검정

diff평균의차이

lwr, upr :신뢰구간p adj : B-A와 C-B의경우귀무가설을기각한다(평균차이가유의미하다)

$branddiff lwr upr p adj

B-A 1.64 0.2599123 3.0200877 0.0204273C-A -0.94 -2.3200877 0.4400877 0.2056606C-B -2.58 -3.9600877 -1.1999123 0.0008518


개요

일원분산분석의예 2

호수의 ppm차이분석

> lake <- c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2,2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3)

> observ <- c(0, 2, 1, 3, 1, 2, 3, 4, 1, 5, 1, 3, 4, 6, 8, 7, 5,3, 4, 5, 14, 26, 25, 18, 19, 22, 21, 16, 20, 30)

> lk <- factor(lake)


개요



> oneway.test(observ~lk)

One-way analysis of means (not assuming equal variances)

data: observ and lkF = 66.719, num df = 2.000, denom df = 16.231, p-value = 1.48e-08


개요



> r <- aov(observ~lk)> summary(r)

Df Sum Sq Mean Sq F value Pr(>F)lk 2 2117.4 1059 105.5 1.73e-13 ***Residuals 27 270.9 10---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


개요



> TukeyHSD(r)Tukey multiple comparisons of means95% family-wise confidence level

Fit: aov(formula = observ ~ lk)

$lkdiff lwr upr p adj

2-1 2.4 -1.112265 5.912265 0.22571133-1 18.9 15.387735 22.412265 0.00000003-2 16.5 12.987735 20.012265 0.0000000


개요

이원분산분석

하나의종속변수가두개의독립변수에영향을받는지검정

곡물의품종과비료의종류가수확량에영향을미치는정도

교사의강의경력과교육방법이교육효과에영향을미치는정도


개요

이원분산분석

aov(x ∼ f1+f2)

summary(aov(x ∼ f1+f2))

범주형벡터 f1과 f2의값에따라수치형벡터 x의평균이유의하게차이가있는지검정

귀무가설 :그룹의평균이같음,즉,영향을주지않음


개요

이원분산분석의예

지역과비료의종류에따라토마토생산량에차이가있는가?

A B C A B C A B C A B C

1 1 1 2 2 2 3 3 3 4 4 4

42.8 52.3 48.2 38.6 43.5 40.3 50.2 58.7 53.5 48.2 50.8 51.2


개요



> fer <- c("A", "B", "C", "A", "B", "C", "A", "B", "C",+ "A", "B", "C")> loc <- c(1,1,1,2,2,2,3,3,3,4,4,4)> earn <- c(42.8, 52.3, 48.2, 38.6, 43.5, 40.3, 50.2, 58.7,+ 53.5, 48.2, 50.8, 51.2)> f <- factor(fer)> l <- factor(loc)


개요



> aov(earn ~ f + l)Call:

aov(formula = earn ~ f + l)

Terms:f l Residuals

Sum of Squares 81.35167 280.90917 18.46833Deg. of Freedom 2 3 6

Residual standard error: 1.754439Estimated effects may be unbalanced


개요



> r <- aov(earn ~ f + l)> summary(r)

Df Sum Sq Mean Sq F value Pr(>F)f 2 81.35 40.68 13.21 0.006333 **l 3 280.91 93.64 30.42 0.000502 ***Residuals 6 18.47 3.08---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


개요



> TukeyHSD(r)Tukey multiple comparisons of means95% family-wise confidence level

Fit: aov(formula = earn ~ f + l)

$fdiff lwr upr p adj

B-A 6.375 2.5685737 10.1814263 0.0051348C-A 3.350 -0.4564263 7.1564263 0.0791552C-B -3.025 -6.8314263 0.7814263 0.1104202


개요



$ldiff lwr upr p adj

2-1 -6.966667 -11.925545 -2.0077883 0.01113713-1 6.366667 1.407788 11.3255450 0.01701274-1 2.300000 -2.658878 7.2588783 0.44146403-2 13.333333 8.374455 18.2922117 0.00036314-2 9.266667 4.307788 14.2255450 0.00264304-3 -4.066667 -9.025545 0.8922117 0.1043733


개요

이원분산분석의예 2

교수법과성별에따라학업성취도에차이가?

전통(1) 시청각(2) 개인(3)

남자(1)

1 3 4

4 5 6

4 7 8

1 8

9

여자(2)

3 5 11

4 7 11

5 9 10

6 12

> met = c(1,1,1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3)> sex = c(1,1,1,1,2,2,2,2,1,1,1,2,2,2,1,1,1,1,1,2,2,2,2)> sco = c(1,4,4,1,3,4,5,6,3,5,7,5,7,9,4,6,8,8,9,11,11,10,12)> m <- factor(met)> s <- factor(sex)


개요



> summary(aov(sco~방법+성별))Df Sum Sq Mean Sq F value Pr(>F)

방법 2 118.36 59.18 21.04 1.52e-05 ***성별 1 44.12 44.12 15.69 0.000838 ***Residuals 19 53.44 2.81---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


개요



> TukeyHSD(aov(sco~방법+성별))Tukey multiple comparisons of means95% family-wise confidence level

Fit: aov(formula = sco ~ 방법 + 성별)

$방법diff lwr upr p adj

2-1 2.500000 0.1990994 4.800901 0.03189743-1 5.277778 3.2075761 7.347980 0.00000953-2 2.777778 0.5323287 5.023227 0.0141466

$성별diff lwr upr p adj

2-1 2.768519 1.303325 4.233712 0.0008495


r.220.149.117.26/teaching/r/anova.pdf · one-way anova ¯‰À˘@–“À˘ 1˝,...

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