Inferensi Regresi Linier Sederhana
Data
Data yang digunakan dalam praktikum ini adalah data Perusahaan toluca.
Perusahaan toluca ingin memprediksi berapa lama waktu yang dibutuhkan dalam bekerja berdasarkan banyaknya jumlah produksi yang dikerjakan.
## # A tibble: 6 × 2
## LotSize WorkHours
## <dbl> <dbl>
## 1 80 399
## 2 30 121
## 3 50 221
## 4 90 376
## 5 70 361
## 6 60 224
Persamaan Regresi
Persamaan Regresi menggunakan MKT dengan Rumus
toluca$xdif = toluca$LotSize-mean(toluca$LotSize)
toluca$ydif = toluca$WorkHours-mean(toluca$WorkHours)
toluca$crp = toluca$xdif*toluca$ydif
toluca$xsq = toluca$xdif^2
toluca$ysq=toluca$ydif^2
b1 <- sum(toluca$crp)/sum(toluca$xsq)
b1
## [1] 3.570202
## [1] 62.36586
Persamaan Regresi menggunakan fungsi lm
pada R
Estimasi paramter model regresi dapat diperoleh dengan juga menggunakan fungsi lm sebagai berikut,
##
## Call:
## lm(formula = WorkHours ~ LotSize, data = toluca)
##
## Residuals:
## Min 1Q Median 3Q Max
## -83.876 -34.088 -5.982 38.826 103.528
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 62.366 26.177 2.382 0.0259 *
## LotSize 3.570 0.347 10.290 4.45e-10 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 48.82 on 23 degrees of freedom
## Multiple R-squared: 0.8215, Adjusted R-squared: 0.8138
## F-statistic: 105.9 on 1 and 23 DF, p-value: 4.449e-10
Persaman Regresi dapat dituliskan sebagai berikut:
\(\hat{y}=b_{0}+b_{1}X\)
\(\hat{y}=62.366+3.570X\)
plot(toluca$LotSize,toluca$WorkHours,pch=16,xlab="LotSize",ylab="WorkHours")
abline(model2,col="red")
Fitted Value
## 1 2 3 4 5 6 7 8
## 347.9820 169.4719 240.8760 383.6840 312.2800 276.5780 490.7901 347.9820
## 9 10 11 12 13 14 15 16
## 419.3861 240.8760 205.1739 312.2800 383.6840 133.7699 455.0881 419.3861
## 17 18 19 20 21 22 23 24
## 169.4719 240.8760 383.6840 455.0881 169.4719 383.6840 205.1739 347.9820
## 25
## 312.2800
## [1] 347.9820 169.4719 240.8760 383.6840 312.2800 276.5780
Residuals (Sisaan/Galat)
## 1 2 3 4 5 6
## 51.0179798 -48.4719192 -19.8759596 -7.6840404 48.7200000 -52.5779798
## 7 8 9 10 11 12
## 55.2098990 4.0179798 -66.3860606 -83.8759596 -45.1739394 -60.2800000
## 13 14 15 16 17 18
## 5.3159596 -20.7698990 -20.0880808 0.6139394 42.5280808 27.1240404
## 19 20 21 22 23 24
## -6.6840404 -34.0880808 103.5280808 84.3159596 38.8260606 -5.9820202
## 25
## 10.7200000
## 1 2 3 4 5 6
## 51.01798 -48.47192 -19.87596 -7.68404 48.72000 -52.57798
Tabel ANOVA
## [1] 54825.46
## [1] 252377.6
## [1] 307203
## [1] 2383.716
## [1] 2383.716
## Analysis of Variance Table
##
## Response: WorkHours
## Df Sum Sq Mean Sq F value Pr(>F)
## LotSize 1 252378 252378 105.88 4.449e-10 ***
## Residuals 23 54825 2384
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Uji Hipotesis \(\beta_{1}\) dan \(\beta_{0}\)
## [1] 2383.716
## [1] 10.28959
sb0sqr<-KTG*(1/length(toluca$LotSize)+((mean(toluca$LotSize)^2)/sum((toluca$LotSize-mean(toluca$LotSize))^2)))
sb0<-sqrt(sb0sqr)
tb0<-b0/sb0
tb0
## [1] 2.382428
Selang Kepercayaan untuk \(\beta_{1}\) dan \(\beta_{0}\)
## 2.5 % 97.5 %
## (Intercept) 8.213711 116.518006
## LotSize 2.852435 4.287969
Selang Kepercayaan E{Yh}
#Selang Kepercayaan E{Yh}
xh1<-65
yh1<-b0+b1*xh1
syh1sqr<-KTG*(1/length(toluca$LotSize)+(((xh1-mean(toluca$LotSize))^2)/sum((toluca$LotSize-mean(toluca$LotSize))^2)))
syh1<-sqrt(syh1sqr)
Ttabel<-qt(0.975,length(toluca$LotSize)-2)
yh1_B<-yh1-Ttabel*syh1
yh1_A<-yh1+Ttabel*syh1
yh1_B
## [1] 273.9129
## [1] 314.9451
#Prediksi
predxc1<-predict(model2,data.frame(LotSize = c(xh1)),interval = "confidence", se.fit=FALSE,level = 0.95)
predxc1
## fit lwr upr
## 1 294.429 273.9129 314.9451
plot(toluca$LotSize, toluca$WorkHours)
abline(model2,col="blue")
points(xh1, predxc1[, "fit"],col="red",pch=15)
points(xh1, predxc1[, "lwr"], lty = "dotted",col="red")
points(xh1, predxc1[, "upr"], lty = "dotted",col="red")