0.0.1 Data input
data <- read.csv("Student_Performance.csv")
install_load('DT')
datatable(data, filter = 'top',
options = list(pageLength = 5))y <- data$Performance.Index
n <- nrow(data)
#Matriks X
X <- cbind(X0 = rep(1, n),
X1 = data$Hours.Studied,
X2 = data$Previous.Scores,
X3 = data$Sleep.Hours,
X4 = data$Sample.Question.Papers.Practiced)1 1. Matriks Penduga Beta
\[ \mathbf{\hat{\beta}} = (\mathbf{X}'\mathbf{X})^{-1}*\mathbf{X}'\mathbf{y} \]
b <- solve(t(X) %*% X) %*% t(X) %*% y #mencari matriks beta = (X'X)^-1*X'y
cat('Nilai dari Penduga Beta adalah : \n')## Nilai dari Penduga Beta adalah :b## [,1]
## X0 -33.7637261
## X1 2.8534292
## X2 1.0185835
## X3 0.4763330
## X4 0.1951983Dari nilai parameter \(\hat{\beta}\) diatas, didapatkan model :\[\hat{Y_{i}}= -33.7637261 + 2.8534292X_{1}-1.0185835X_{2} + 0.4763330 X_{3} + 0.1951983 X_{4} ε_i\]
2 2. Selang kepercayaan bagi b0, b1, b2, b3, dan b4 dengan \(\alpha= 5\%\)
Formula Selang kepercayaan :
\[\mathbf{\beta_i} \pm t_{\left(n-p;\frac{\alpha}{2} \right)} s \sqrt{ (\mathbf{X}^t\mathbf{X})^-1}_{(i,i)} \]
2.1 Cari Bahan Formula
2.1.1 Simpangan Baku
\[ s = \sqrt{s^2} = \sqrt{\frac{(y-\mathbf{X}^t\mathbf{b})^t (y-\mathbf{X}^t\mathbf{b})}{n-p}} \]
#Ragam
p <- nrow(b) #banyak beta pada model
s <- sqrt( (t(y-(X %*% b)) %*%( y-(X %*% b)) ) / (n-p) )
cat("Simpangan baku (s) :", s)## Simpangan baku (s) : 2.0608982.1.2 Matriks \((\mathbf{X}^t\mathbf{X})^-1\)
tX.X.inv <- solve(t(X) %*% X); tX.X.inv## X0 X1 X2 X3 X4
## X0 3.787989e-03 -7.518191e-05 -2.302961e-05 -2.251988e-04 -5.293624e-05
## X1 -7.518191e-05 1.492372e-05 2.793098e-08 -2.847483e-08 -2.366096e-07
## X2 -2.302961e-05 2.793098e-08 3.325813e-07 -2.016101e-08 -1.626058e-08
## X3 -2.251988e-04 -2.847483e-08 -2.016101e-08 3.477641e-05 -8.066048e-08
## X4 -5.293624e-05 -2.366096e-07 -1.626058e-08 -8.066048e-08 1.216887e-052.1.3 Nilai \(t_{\left(n-p;\frac{\alpha}{2} \right)}\)
Dengan \(\alpha = 5\%\), maka selang Kepercayaan \(95\%\).
alpha <- 5/100
t <- qt(1 - alpha/2, n-p)
cat('t :', t)## t : 1.9602012.2 Selang kepercayaan \(95\%\) bagi \(\beta\)
2.2.1 Selang kepercayaan \(95\%\) bagi \(\beta_0\)
cat(" Selang kepercayaan 95% bagi beta_0\n",
"Batas Bawah :", b[1] - t *s *sqrt(tX.X.inv[1,1]), "\n",
"Batas Atas :", b[1] + t *s *sqrt(tX.X.inv[1,1])
)## Selang kepercayaan 95% bagi beta_0
## Batas Bawah : -34.01236
## Batas Atas : -33.51509Didapatkan hasil :
\[\mathbf{\beta_0} \pm t_{\left(n-p;\frac{\alpha}{2} \right)} s \sqrt{ (\mathbf{X}^t\mathbf{X})^-1}_{(0,0)} = [-34.01236 ; -33.51509] \]
2.2.2 Selang kepercayaan \(95\%\) bagi \(\beta_1\)
cat(" Selang kepercayaan 95% bagi beta_1\n",
"Batas Bawah :", b[2] - t *s *sqrt(tX.X.inv[2,2]), "\n",
"Batas Atas :", b[2] + t *s *sqrt(tX.X.inv[2,2])
)## Selang kepercayaan 95% bagi beta_1
## Batas Bawah : 2.837823
## Batas Atas : 2.869035Didapatkan hasil :
\[\mathbf{\beta_1} \pm t_{\left(n-p;\frac{\alpha}{2} \right)} s \sqrt{ (\mathbf{X}^t\mathbf{X})^-1}_{(1,1)} = [2.837823 ; 2.869035] \]
2.2.3 Selang kepercayaan \(95\%\) bagi \(\beta_2\)
cat(" Selang kepercayaan 95% bagi beta_2\n",
"Batas Bawah :", b[3] - t *s *sqrt(tX.X.inv[3,3]), "\n",
"Batas Atas :", b[3] + t *s *sqrt(tX.X.inv[3,3])
)## Selang kepercayaan 95% bagi beta_2
## Batas Bawah : 1.016254
## Batas Atas : 1.020913Didapatkan hasil :
\[\mathbf{\beta_2} \pm t_{\left(n-p;\frac{\alpha}{2} \right)} s \sqrt{ (\mathbf{X}^t\mathbf{X})^-1}_{(2,2)} = [1.016254 ; 1.020913] \]
2.2.4 Selang kepercayaan \(95\%\) bagi \(\beta_3\)
cat(" Selang kepercayaan 95% bagi beta_3\n",
"Batas Bawah :", b[4] - t *s *sqrt(tX.X.inv[4,4]), "\n",
"Batas Atas :", b[4] + t *s *sqrt(tX.X.inv[4,4])
)## Selang kepercayaan 95% bagi beta_3
## Batas Bawah : 0.4525098
## Batas Atas : 0.5001561Didapatkan hasil :
\[\mathbf{\beta_3} \pm t_{\left(n-p;\frac{\alpha}{2} \right)} s \sqrt{ (\mathbf{X}^t\mathbf{X})^-1}_{(3,3)} = [0.4525098 ; 0.5001561] \]
2.2.5 Selang kepercayaan \(95\%\) bagi \(\beta_4\)
cat(" Selang kepercayaan 95% bagi beta_4\n",
"Batas Bawah :", b[5] - t *s *sqrt(tX.X.inv[5,5]), "\n",
"Batas Atas :", b[5] + t *s *sqrt(tX.X.inv[5,5])
)## Selang kepercayaan 95% bagi beta_4
## Batas Bawah : 0.181106
## Batas Atas : 0.2092906Didapatkan hasil :
\[\mathbf{\beta_4} \pm t_{\left(n-p;\frac{\alpha}{2} \right)} s \sqrt{ (\mathbf{X}^t\mathbf{X})^-1}_{(4,4)} = [0.181106 ; 0.2092906] \]