토양에 녹아 있는 붕소(boron)의 양(Y)은 토양의 진흙의 비중(X1)과
토양의 pH(X2)에 관련되어 있다고 한다. 다음 자료는 20가지의 토양을
조사하여 붕소의 양, 진흙의 비중, pH 를 기록한 것이다.
1) 2차
다항회귀모형으로 적합시키고, 적합성 검토
2) 가설 H0 : B12=0, H1 :
B12 /= 0 을 유의수준 5% 에서 검정
=> 교호작용은 유의확률 0.46
이므로 0.05 보다 크므로 귀무가설 기각 안됨
Y = c(0.62, 0.69, 0.63, 0.61, 0.28, 0.33, 0.31, 0.37, 0.66, 0.70, 0.74, 0.63, 0.52, 0.47, 0.45, 0.42, 0.41, 0.42, 0.42, 0.41)
X1 = c(37,37,37,37,29,29,29,29,29,29,29,29,27,27,27,27,27,27,27,27)
X2 = c(5.3, 5.3, 5.5, 5.7, 5.6, 5.7, 5.7, 5.9, 6.0, 6.0, 6.0, 6.0, 5.5, 5.6, 5.6, 5.7, 5.5, 5.5, 5.5, 5.5)
str(X2)## num [1:20] 5.3 5.3 5.5 5.7 5.6 5.7 5.7 5.9 6 6 ...dataF = data.frame(Y, X1, X2)
dataF.lm = lm(Y ~ X1+X2+I(X1^2)+I(X2^2)+X1:X2, data=dataF)
summary(dataF.lm)##
## Call:
## lm(formula = Y ~ X1 + X2 + I(X1^2) + I(X2^2) + X1:X2, data = dataF)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.128912 -0.030792 0.002645 0.029003 0.079928
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 82.697074 20.398454 4.054 0.001184 **
## X1 -0.895735 0.224292 -3.994 0.001333 **
## X2 -24.635265 6.395311 -3.852 0.001760 **
## I(X1^2) 0.011673 0.002849 4.098 0.001087 **
## I(X2^2) 2.134101 0.476012 4.483 0.000515 ***
## X1:X2 0.029347 0.038728 0.758 0.461153
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.05755 on 14 degrees of freedom
## Multiple R-squared: 0.8807, Adjusted R-squared: 0.8381
## F-statistic: 20.67 on 5 and 14 DF, p-value: 5.191e-06다음 자료는 미국 와이오밍주에 있는 Snake River 에서 매년 4월1일의
눈에 포함된 수분량(X)과 4월부터 7월까지의 강수량(Y, 단위 inch)을
17년(1919~1935) 동안 측정한 것이다
1) X 와 Y의 산점도를 그리라
2) Xi - X(bar) 이용하여 반응변수 Y 에 대한 1차, 2차, 3차 다항회귀모형을
구하라
3) 2)의 3개의 다항회귀모형들 중에서 관측된 자료에 가장
적절한 다항회귀모형은 무엇인가?
X = c(23.1, 32.8, 31.8, 32.0, 30.4, 24.0, 39.5, 24.2, 52.5, 37.9, 30.5, 25.1, 12.4, 35.1, 31.5, 21.1, 27.6)
Y = c(10.5, 16.7, 18.2, 17.0, 16.3, 10.5, 23.1, 12.4, 24.9, 22.8, 14.1, 12.9, 8.8, 17.4, 14.9, 10.5, 16.1)
str(Y)## num [1:17] 10.5 16.7 18.2 17 16.3 10.5 23.1 12.4 24.9 22.8 ...dataF = data.frame(X,Y)
plot(dataF)
dataF.lm = lm(Y~X, data=dataF)
summary(dataF.lm)##
## Call:
## lm(formula = Y ~ X, data = dataF)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.1793 -1.5149 -0.3624 1.6276 3.1973
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.72538 1.54882 0.468 0.646
## X 0.49808 0.04952 10.058 4.63e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.743 on 15 degrees of freedom
## Multiple R-squared: 0.8709, Adjusted R-squared: 0.8623
## F-statistic: 101.2 on 1 and 15 DF, p-value: 4.632e-08plot(dataF.lm)
dataF.lm = lm(Y~X+I(X^2), data=dataF)
summary(dataF.lm)##
## Call:
## lm(formula = Y ~ X + I(X^2), data = dataF)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.1865 -1.5803 -0.3889 1.5699 3.1576
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.0937778 3.7751930 -0.025 0.9805
## X 0.5528327 0.2342043 2.360 0.0333 *
## I(X^2) -0.0008467 0.0035344 -0.240 0.8141
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.8 on 14 degrees of freedom
## Multiple R-squared: 0.8714, Adjusted R-squared: 0.853
## F-statistic: 47.43 on 2 and 14 DF, p-value: 5.818e-07dataF.lm = lm(Y~X+I(X^3), data=dataF)
summary(dataF.lm)##
## Call:
## lm(formula = Y ~ X + I(X^3), data = dataF)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.1694 -1.3339 -0.3754 1.4871 3.0753
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -5.786e-01 2.931e+00 -0.197 0.846357
## X 5.629e-01 1.326e-01 4.245 0.000816 ***
## I(X^3) -1.894e-05 3.578e-05 -0.529 0.604970
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.786 on 14 degrees of freedom
## Multiple R-squared: 0.8734, Adjusted R-squared: 0.8553
## F-statistic: 48.29 on 2 and 14 DF, p-value: 5.212e-07plot(dataF.lm)
dataF.lm = lm(Y~X+I(X^2)+I(X^3), data=dataF)
summary(dataF.lm)##
## Call:
## lm(formula = Y ~ X + I(X^2) + I(X^3), data = dataF)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.88800 -0.90941 -0.01824 0.73015 2.25675
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 21.1848945 6.8002641 3.115 0.00820 **
## X -1.8588156 0.7223336 -2.573 0.02315 *
## I(X^2) 0.0808675 0.0238868 3.385 0.00488 **
## I(X^3) -0.0008390 0.0002437 -3.442 0.00437 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.351 on 13 degrees of freedom
## Multiple R-squared: 0.9327, Adjusted R-squared: 0.9172
## F-statistic: 60.07 on 3 and 13 DF, p-value: 7.097e-08plot(dataF.lm)
## Warning in sqrt(crit * p * (1 - hh)/hh): NaNs produced
## Warning in sqrt(crit * p * (1 - hh)/hh): NaNs produced
다음은 미국 경제 자료로서 1940년부터 1950년까지의 가처분소득(Y)과 가계소비지출액(X)을 적어 놓은 것이다. 제2차 세계대전기간(1941~45) 동안의 기간을 I, 그리고 그 외의 기간을 0으로 하여 가처분소득과 가계소비지출액의 관계를 알아보라
Y = c(244.0, 277.9, 317.5, 332.1, 343.6, 338.1, 332.7, 318.8, 335.8, 336.8, 362.8)
X = c(299.9, 243.6, 241.1, 248.2, 255.2, 270.9, 301.0, 305.8, 312.2, 319.3, 337.3)
I = c(0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0)
I_d = factor(I, levels=c(0,1), label=c("Line0","Line1"))
str(Y)## num [1:11] 244 278 318 332 344 ...dataF = data.frame(Y,X,I_d)
dataF[c(1,2,3),]## Y X I_d
## 1 244.0 299.9 Line0
## 2 277.9 243.6 Line1
## 3 317.5 241.1 Line1plot(X,Y)
points(X[I_d=="Line1"], Y[I_d=="Line1"], pch=17, col="BLUE")
points(X[I_d=="Line0"], Y[I_d=="Line0"], pch=19, col="RED")
legend("bottomright", legend=levels(I_d), pch=c(19,17), col=c("RED","BLUE"))
dataF.lm = lm(Y ~ X+I_d, data=dataF)
summary(dataF.lm)##
## Call:
## lm(formula = Y ~ X + I_d, data = dataF)
##
## Residuals:
## Min 1Q Median 3Q Max
## -55.625 -9.711 8.852 15.233 31.150
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -225.0999 221.3456 -1.017 0.3389
## X 1.7497 0.7072 2.474 0.0385 *
## I_dLine1 106.3739 46.2084 2.302 0.0503 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 28.01 on 8 degrees of freedom
## Multiple R-squared: 0.4335, Adjusted R-squared: 0.2919
## F-statistic: 3.061 on 2 and 8 DF, p-value: 0.103dataF.lm = lm(Y ~ X+I_d+X:I_d, data=dataF)
summary(dataF.lm)##
## Call:
## lm(formula = Y ~ X + I_d + X:I_d, data = dataF)
##
## Residuals:
## Min 1Q Median 3Q Max
## -52.511 -8.650 9.795 14.882 33.994
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -301.8345 292.6733 -1.031 0.3367
## X 1.9952 0.9355 2.133 0.0704 .
## I_dLine1 291.0512 427.4711 0.681 0.5178
## X:I_dLine1 -0.6742 1.5503 -0.435 0.6768
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 29.54 on 7 degrees of freedom
## Multiple R-squared: 0.4484, Adjusted R-squared: 0.212
## F-statistic: 1.897 on 3 and 7 DF, p-value: 0.2185한 대학교에서 성별에 따라 교수들의 월급에 차이가 있는가를 보기 위하여
교수들의 월급액(SL), 직위(RK), 근속연수(YR), 최종학위(DG)들을 조사한
결과가 다음과 같다. 단, 직위변수에서는 1=조교수, 2=부교수, 3=정교수로,
최종학위변수에서는 0=석사, 1=박사로, 성별(SEX)변수에서는 1=여자,
0=남자로 기록하였다.
SEX=c(0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0, 0,1,0,0,1,0,0,1,1,0,1,0,0,0,0,0,0,1,0,0,1,1,0,1,1,1)
RK =c(3,3,3,3,3,3,3,3,3,3,3,2,3,2,3,3,3,2,2,3,1,2,3,3,2,3, 2,3,2,2,1,2,1,2,2,2,1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,1)
YR =c(25,13,10,7,19,16,0,16,13,13,12,15,9,9,9,7,13,11,10,6,16,8,7,8,9,5, 11,5,3,3,10,11,9,4,6,1,8,4,4,4,3,3,0,3,2,2,2,2,1,1,1,0)
DG =c(1,1,1,1,0,1,0,1,0,0,1,1,1,0,1,1,1,0,0,0,0,0,1,1,1,1, 1,1,0,0,0,0,0,0,0,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1)
SL =c(36350,35350,28200,26775,33696,28516,24900,31909,31850,32850,27025,24750,28200,23712,25748,29342,31114,24742,22906,24450,19175,20525,27959,38045,24832,25400,24800,25500,26182,23725,21600,23300,23713,20690,22450,20850,18304,17095,16700,17600,18075,18000,20999,17250,16500,16094,16150,15350,16244,16686,15000,20300)
dataF = data.frame(SEX,RK,YR,DG,SL)
dataF$SEX_D = factor(dataF$SEX, levels = c(0,1), label=c("male","female"))
dataF[c(1,2,3),]## SEX RK YR DG SL SEX_D
## 1 0 3 25 1 36350 male
## 2 0 3 13 1 35350 male
## 3 0 3 10 1 28200 male1)성별을 표시하는 기호를 이용하여 월급액과 근속연수에 대한 산점도를 그려라
plot(dataF$YR,dataF$SL)
points(dataF$YR[dataF$SEX_D=="male"], dataF$SL[dataF$SEX_D=="male"], pch=17, col="BLUE")
points(dataF$YR[dataF$SEX_D=="female"], dataF$SL[dataF$SEX_D=="female"], pch=19, col="RED")
2) 월급액에 대해서 근속연수와 성별을 설명변수로 하는 회귀모형을
구하라
dataf.lm = lm(SL~YR+SEX_D,data=dataF)
summary(dataF.lm)##
## Call:
## lm(formula = Y ~ X + I_d + X:I_d, data = dataF)
##
## Residuals:
## Min 1Q Median 3Q Max
## -52.511 -8.650 9.795 14.882 33.994
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -301.8345 292.6733 -1.031 0.3367
## X 1.9952 0.9355 2.133 0.0704 .
## I_dLine1 291.0512 427.4711 0.681 0.5178
## X:I_dLine1 -0.6742 1.5503 -0.435 0.6768
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 29.54 on 7 degrees of freedom
## Multiple R-squared: 0.4484, Adjusted R-squared: 0.212
## F-statistic: 1.897 on 3 and 7 DF, p-value: 0.2185dataf.lm2 = lm(SL~YR+SEX_D+RK,data=dataF)
summary(dataf.lm2)##
## Call:
## lm(formula = SL ~ YR + SEX_D + RK, data = dataF)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3166.5 -1570.1 -168.0 984.4 9037.0
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11011.76 966.95 11.388 3.03e-15 ***
## YR 393.86 74.53 5.285 3.04e-06 ***
## SEX_Dfemale 603.77 811.20 0.744 0.46
## RK 4747.18 452.58 10.489 5.18e-14 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2398 on 48 degrees of freedom
## Multiple R-squared: 0.8454, Adjusted R-squared: 0.8358
## F-statistic: 87.51 on 3 and 48 DF, p-value: < 2.2e-16dataf.lm3 = lm(SL~YR+SEX_D+RK+DG,data=dataF)
summary(dataf.lm3)##
## Call:
## lm(formula = SL ~ YR + SEX_D + RK + DG, data = dataF)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3261.8 -1618.4 -184.2 910.2 9075.6
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11101.27 1087.12 10.212 1.62e-13 ***
## YR 391.84 76.05 5.152 5.02e-06 ***
## SEX_Dfemale 608.10 819.80 0.742 0.462
## RK 4753.17 458.31 10.371 9.75e-14 ***
## DG -134.22 715.45 -0.188 0.852
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2422 on 47 degrees of freedom
## Multiple R-squared: 0.8455, Adjusted R-squared: 0.8324
## F-statistic: 64.33 on 4 and 47 DF, p-value: < 2.2e-16dataf.lm4 = lm(SL~YR+SEX+YR:SEX,data=dataF)
summary(dataf.lm4)##
## Call:
## lm(formula = SL ~ YR + SEX + YR:SEX, data = dataF)
##
## Residuals:
## Min 1Q Median 3Q Max
## -10904.0 -3150.2 -632.2 2896.8 13112.6
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 18222.6 1308.6 13.925 < 2e-16 ***
## YR 741.0 126.2 5.870 3.95e-07 ***
## SEX -570.8 2297.2 -0.248 0.805
## YR:SEX 169.1 387.0 0.437 0.664
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4342 on 48 degrees of freedom
## Multiple R-squared: 0.4932, Adjusted R-squared: 0.4615
## F-statistic: 15.57 on 3 and 48 DF, p-value: 3.323e-07dataf.lm5 = lm(SL~SEX,data=dataF)
summary(dataf.lm5)##
## Call:
## lm(formula = SL ~ SEX, data = dataF)
##
## Residuals:
## Min 1Q Median 3Q Max
## -8602.8 -4296.6 -100.8 3513.1 16687.9
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 24697 938 26.330 <2e-16 ***
## SEX -3340 1808 -1.847 0.0706 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5782 on 50 degrees of freedom
## Multiple R-squared: 0.0639, Adjusted R-squared: 0.04518
## F-statistic: 3.413 on 1 and 50 DF, p-value: 0.0706