> summary(cars)
speed dist
Min. :
4.0 Min. :
2.00
1st Qu.:12.0
1st Qu.: 26.00
Median :15.0
Median : 36.00
Mean
:15.4 Mean : 42.98
3rd Qu.:19.0
3rd Qu.: 56.00
Max.
:25.0 Max. :120.00
> str(cars)
'data.frame': 50 obs. of
2 variables:
$ speed: num
4 4 7 7 8 9 10 10 10 11 ...
$ dist : num
2 10 4 22 16 10 18 26 34 17 ...
> head(cars)
speed dist
1 4
2
2 4
10
3 7
4
4 7
22
5 8
16
6 9
10
> lm(speed~dist,
cars)
Call:
lm(formula
= speed ~ dist, data = cars)
Coefficients:
(Intercept) dist
8.2839 0.1656
>
> m<-lm(dist~speed,
cars)
> summary(m)
Call:
lm(formula
= dist ~ speed, data = cars)
Residuals:
Min
1Q Median 3Q
Max
-29.069 -9.525
-2.272 9.215 43.201
Coefficients:
Estimate Std. Error t value
Pr(>|t|)
(Intercept)
-17.5791 6.7584 -2.601
0.0123 *
speed 3.9324 0.4155
9.464 1.49e-12 ***
---
Signif.
codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05
‘.’ 0.1 ‘ ’ 1
Residual
standard error: 15.38 on 48 degrees of freedom
Multiple
R-squared: 0.6511, Adjusted R-squared: 0.6438
F-statistic:
89.57 on 1 and 48 DF, p-value: 1.49e-12
>
> # Intercept
= 절편
> #
coefficients(회귀식의 계수) 종속변수 = 독립변수*계수 + 절편
> #
dist=speed*3.932-17.579
> # 계수가 양수이므로 양의 상관관계를 지님
>
>
>
> # 추정해보기(신뢰구간)
>
> predict(m,
data.frame(speed=3), interval="confidence")
fit lwr
upr
1
-5.781869 -17.02659 5.462853
>
> # fit = 예측값
> # lwr, upr =
신뢰구간
> # speed가 3일 때, dist의 평균값은 -17.03~5.46 사이에 있음
>
>
> plot(cars)
> abline(coef(m))
#coef는 회선의 계수
> coef(m)
(Intercept) speed
-17.579095
3.932409
>
>
> [출처] [R프로그래밍] RStudio 선형 회귀 (1) 단순 선형 회귀|작성자 아트
http://chloe-ynlee.me/221302458526
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cars
summary(cars)
str(cars)
head(cars)
lm(speed~dist, cars)
m<-lm(dist~speed, cars)
summary(m)
# Intercept = 절편
# coefficients(회귀식의 계수) 종속변수 = 독립변수*계수 + 절편
# dist=speed*3.932-17.579
# 계수가 양수이므로 양의 상관관계를 지님
# 추정해보기(신뢰구간)
predict(m, data.frame(speed=3), interval="confidence")
# fit = 예측값
# lwr, upr = 신뢰구간
# speed가 3일 때, dist의 평균값은 -17.03~5.46 사이에 있음
plot(cars)
abline(coef(m)) #coef는 회선의 계수
coef(m)
[출처] [R프로그래밍] RStudio 선형 회귀 (1) 단순 선형 회귀|작성자 아트
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