표 4.4 데이터에 대한 선형회귀 모형을 적합하고 다음에 답하시오
<표 4.4>
| 환자번호 | 콜레스트롤(mg/100ml) | 몸무게(kg) | 나이(year) |
| 1 | 354 | 84 | 46 |
| 2 | 190 | 73 | 20 |
| 3 | 405 | 65 | 52 |
| 4 | 263 | 73 | 30 |
| 5 | 451 | 76 | 57 |
| 6 | 302 | 69 | 25 |
| 7 | 288 | 63 | 28 |
| 8 | 385 | 72 | 36 |
| 9 | 402 | 79 | 57 |
| 10 | 365 | 75 | 44 |
| 11 | 209 | 27 | 24 |
| 12 | 290 | 89 | 31 |
| 13 | 346 | 65 | 52 |
| 14 | 254 | 57 | 23 |
| 15 | 395 | 59 | 60 |
| 16 | 434 | 69 | 48 |
| 17 | 220 | 60 | 34 |
| 18 | 374 | 79 | 51 |
| 19 | 308 | 75 | 50 |
| 20 | 220 | 82 | 34 |
| 21 | 311 | 59 | 46 |
| 22 | 181 | 67 | 23 |
| 23 | 274 | 85 | 37 |
| 24 | 303 | 55 | 40 |
| 25 | 244 | 63 | 30 |
(a) 가정에 대한 검토로 잔차를 이용하여 등분산성을 점검하시오
> cholesterol = data.frame(
+ 환자번호 = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25),
+ 콜레스테롤 = c(354, 190, 405, 263, 451, 302, 288, 385, 402, 365, 209, 290, 346, 254, 395, 434, 220, 374, 308, 220, 311, 181, 274, 303, 244),
+ 몸무게 = c(84, 73, 65, 70, 76, 69, 63, 72, 79, 75, 27, 89, 65, 57, 59, 69, 60, 79, 75, 82, 59, 67, 85, 55, 63),
+ 나이 = c(46, 20, 52, 30, 57, 25, 28, 36, 57, 44, 24, 31, 52, 23, 60, 48, 34, 51, 50, 34, 46, 23, 37, 40, 30)
+ )
> print(cholesterol)
환자번호 콜레스테롤 몸무게 나이
1 1 354 84 46
2 2 190 73 20
3 3 405 65 52
4 4 263 70 30
5 5 451 76 57
6 6 302 69 25
7 7 288 63 28
8 8 385 72 36
9 9 402 79 57
10 10 365 75 44
11 11 209 27 24
12 12 290 89 31
13 13 346 65 52
14 14 254 57 23
15 15 395 59 60
16 16 434 69 48
17 17 220 60 34
18 18 374 79 51
19 19 308 75 50
20 20 220 82 34
21 21 311 59 46
22 22 181 67 23
23 23 274 85 37
24 24 303 55 40
25 25 244 63 30
> cholesterol.lm = lm(콜레스테롤 ~ 몸무게 + 나이, data = cholesterol)
> plot(cholesterol.lm$fitted.values, resid(cholesterol.lm), col=4)
> abline(h = 0, col = "red")
(b) 가정에 대한 검토로 잔차를 이용하여 정규성을 점검하시오
> residuals=residuals(cholesterol.lm)
> qqnorm(residuals,pch = 19, col = "blue")
> qqline(residuals, col = "red", lwd = 2)
(c) 지렛대점이 있는지 점검하시오
> hat_values = hatvalues(cholesterol.lm)
> n=nrow(cholesterol)
> p=length(coef(cholesterol.lm))-1
> threshold = 2 * (p + 1) / n
> plot(hat_values, type = "h", main = "Hat Value")
> abline(h = threshold, col = "red", lwd = 2)
(d) 이상점이 있는지 검토하시오
> standardized_residuals = rstandard(cholesterol.lm)
> outliers_residuals = which(abs(standardized_residuals) > 2)
> outliers_residuals
8
> plot(standardized_residuals, main = "표준화 잔차", ylab = "표준화 잔차", xlab = "데이터 인덱스")
> abline(h = c(-2, 2), col = "red", lwd = 2)
(e) 영향점이 있는지 검토하시오
> cooks_distance = cooks.distance(cholesterol.lm)
> influential_points = which(cooks_distance > 0.5)
> outliers_cooks
named integer(0)
> plot(cooks_distance, type = "h", main = "쿡의 거리",xlab = "데이터 인덱스", ylab = "쿡의 거리")
> abline(h = 0.5, col = "red", lty = 2)
(f) 설명변수들과 반응변수 간의 회귀식이 적합한지 진단하시오
> summary(swiss.lm)
Call:
lm(formula = Fertility ~ ., data = swiss)
Residuals:
Min 1Q Median 3Q Max
-15.2743 -5.2617 0.5032 4.1198 15.3213
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 66.91518 10.70604 6.250 1.91e-07
Agriculture -0.17211 0.07030 -2.448 0.01873
Examination -0.25801 0.25388 -1.016 0.31546
Education -0.87094 0.18303 -4.758 2.43e-05
Catholic 0.10412 0.03526 2.953 0.00519
Infant.Mortality 1.07705 0.38172 2.822 0.00734
(Intercept) ***
Agriculture *
Examination
Education ***
Catholic **
Infant.Mortality **
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 7.165 on 41 degrees of freedom
Multiple R-squared: 0.7067, Adjusted R-squared: 0.671
F-statistic: 19.76 on 5 and 41 DF, p-value: 5.594e-10
