Prakt PML 10 - Hipotesis linier One-Way Classification

Model Linier Umum dalam Klasifikasi Satu Arah

Model Linier

\[ y_{ij} = \mu + \tau_i + \epsilon_{ij}, \quad i = 1, \dots, k \quad j = 1, \dots, n_i \]

\(y_{ij}\): pengamatan ke-\(j\) pada perlakuan ke-\(i\),
\(\mu\): rataan umum,
\(\tau_i\): efek dari perlakuan ke-\(i\),
\(\epsilon_{ij}\): kesalahan acak (\(\epsilon_{ij} \sim N(0, \sigma^2)\)).

Asumsi Utama

Kesalahan \(\epsilon_{ij}\) berdistribusi normal dengan rata-rata nol dan variansi homogen.
Perlakuan \(\tau_i\) adalah pengaruh tetap dari perlakuan ke-\(i\) terhadap respons \(y_{ij}\).

Model dalam Bentuk Matriks

\[ \mathbf{y} = X \boldsymbol{\beta} + \boldsymbol{\epsilon} \]

\(\mathbf{y}\) adalah vektor pengamatan,
\(X\) adalah matriks desain yang mengkode perlakuan (sering disebut sebagai matriks model atau desain),
\(\boldsymbol{\beta}\) adalah vektor parameter yang akan diestimasi (terdiri dari \(\mu\) dan efek perlakuan \(\tau_i\)),
\(\boldsymbol{\epsilon}\) adalah vektor kesalahan acak.

Uji Hipotesis

\[ H_0: \tau_1 = \tau_2 = \dots = \tau_k \]

Artinya, tidak ada perbedaan antara efek dari masing-masing perlakuan. Hipotesis ini juga bisa dituliskan dalam bentuk matriks sebagai:

\[ H_0: C \beta = 0 \]

di mana:

\(C\) adalah matriks kontras yang mendefinisikan hubungan antar parameter (dalam hal ini perbedaan antar perlakuan),
\(\beta\) adalah vektor parameter.

Proses Uji Hipotesis

MKT

\[ \hat{\beta} = (X'X)^{-1} X'y \]

di mana:

\(X'\) adalah transpos dari matriks desain,
\(y\) adalah vektor pengamatan.

Statistik Uji F

\[ F = \frac{C \hat{\beta}' (C (X'X)^{-1} C')^{-1} C \hat{\beta}}{s^2} \]

di mana:

\(C \hat{\beta}\) adalah hasil estimasi dari kontras-kontras parameter,
\((X'X)^{-1}\) adalah invers dari matriks informasi \(X'X\),
\(s^2\) adalah varians residual yang dihitung sebagai:

\[ s^2 = \frac{\sum \text{residual}^2}{n - r} \]

Keputusan Uji Hipotesis

Bandingkan dengan nilai kritis dari distribusi F (\(F_{\text{tabel}}\)) untuk tingkat signifikansi yang diberikan (\(\alpha = 0.05\)):

Jika \(F_{\text{hitung}} > F_{\text{tabel}}\), kita tolak hipotesis nol dan menyimpulkan bahwa ada perbedaan signifikan antar perlakuan.
Jika \(F_{\text{hitung}} \leq F_{\text{tabel}}\), kita gagal menolak hipotesis nol, yang berarti tidak ada bukti yang cukup untuk menyimpulkan adanya perbedaan antar perlakuan.

Cara Pengerjaan

Membuat Data dan Model Linier

# Data Observasi
y <- c(3, 2, 4, 7, 6, 5)  # Observasi
treatment <- factor(rep(1:3, each = 2))  # Faktor Perlakuan (1, 2, 3 dengan 2 ulangan)

# Membuat model linier
model <- lm(y ~ treatment)

# Ringkasan dari model
summary(model)

## 
## Call:
## lm(formula = y ~ treatment)
## 
## Residuals:
##    1    2    3    4    5    6 
##  0.5 -0.5 -1.5  1.5  0.5 -0.5 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)  
## (Intercept)   2.5000     0.9574   2.611   0.0796 .
## treatment2    3.0000     1.3540   2.216   0.1135  
## treatment3    3.0000     1.3540   2.216   0.1135  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.354 on 3 degrees of freedom
## Multiple R-squared:  0.6857, Adjusted R-squared:  0.4762 
## F-statistic: 3.273 on 2 and 3 DF,  p-value: 0.1762

ANOVA untuk Pengujian Hipotesis

Untuk menguji hipotesis nol menggunakan ANOVA, kita bisa menjalankan fungsi anova():

# Menghitung ANOVA untuk model
anova_result <- anova(model)

# Menampilkan hasil ANOVA
anova_result

## Analysis of Variance Table
## 
## Response: y
##           Df Sum Sq Mean Sq F value Pr(>F)
## treatment  2   12.0  6.0000  3.2727 0.1762
## Residuals  3    5.5  1.8333

Hasil ANOVA akan memberikan statistik F, nilai p, dan apakah kita bisa menolak atau menerima hipotesis nol.

Pengujian Kontras dengan `linearHypothesis()`

# Install dan load paket 'car' jika belum terpasang
# install.packages("car")
library(car)

## Loading required package: carData

# Matriks kontras untuk menguji H0: tau1 = tau2 dan tau2 = tau3
C <- rbind(c(1, -1, 0), c(0, 1, -1))

# Menguji hipotesis kontras
linearHypothesis(model, C)

## Linear hypothesis test
## 
## Hypothesis:
## (Intercept) - treatment2 = 0
## treatment2 - treatment3 = 0
## 
## Model 1: restricted model
## Model 2: y ~ treatment
## 
##   Res.Df    RSS Df Sum of Sq      F Pr(>F)
## 1      5 5.6111                           
## 2      3 5.5000  2   0.11111 0.0303 0.9704

Fungsi ini akan menghitung statistik uji untuk hipotesis kontras dan memberikan hasil apakah hipotesis bisa ditolak atau tidak.

Menghitung F-Hitung Secara Manual

# Menghitung matriks XtX dan XtY
XtX <- t(model.matrix(model)) %*% model.matrix(model)
XtY <- t(model.matrix(model)) %*% y

# Estimasi parameter (beta)
beta_hat <- solve(XtX) %*% XtY

# Matriks kontras
C <- matrix(c(1, -1, 0, 0, 1, -1), nrow = 2, byrow = TRUE)

# Menghitung F-hitung
s2 <- sum(resid(model)^2) / model$df.residual
F_value <- t(C %*% beta_hat) %*% solve(C %*% solve(XtX) %*% t(C)) %*% (C %*% beta_hat) / s2

F_value

##            [,1]
## [1,] 0.06060606

Contoh Soal

Pertimbangkan model klasifikasi satu arah dengan efek tetap dan \(k = 3\). Periksa setiap hipotesis nol berikut untuk testability.

\(H_0: \tau_1 = \tau_3 \text{ dan } \tau_1 - 2\tau_2 + \tau_3 = 0\)
\(H_0: \tau_1 = \tau_2 \text{ dan } \tau_1 = \tau_3 \text{ dan } 2\tau_1 - \tau_2 - \tau_3 = 0\)

\(H_0: \tau_1 = \tau_3\) dan \(H_0: \tau_1 - 2\tau_2 + \tau_3 = 0\) apakah testable?

Jawaban:

Model statistik:

\[ y_{ij} = \mu + \tau_i + \varepsilon_{ij} \quad i = 1,2,3 \quad j = 1,2,...,n_i \]

\[ \tau_1 - 2\tau_2 + \tau_3 = 0 \]

\[ C_2 = \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} \]

\[ \tau_1 = \tau_3 \]

\[ C_1 = \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ C = \begin{bmatrix} 0 & 1 & -2 & 1 \\ 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 = 0 \] \[ \Rightarrow -2\alpha_1 + 0 = 0 \] \[ \Rightarrow \alpha_1 - \alpha_2 = 0 \]

Maka, kesimpulannya:

\[ \alpha_1 = \alpha_2 = 0 \]

Sehingga \(C\) independen dan \(H_0\) testable.

b). apakah \[H_0: \tau_1 = \tau_2 \quad \text{dan} \quad \tau_1 = \tau_3 \quad \text{dan} \quad 2\tau_1 - \tau_2 - \tau_3 = 0\] testable?

Jawaban

\[ C = \begin{bmatrix} 0 & 1 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 2 & -1 & -1 \end{bmatrix} \quad \beta = \begin{bmatrix} \mu \\ \tau_1 \\ \tau_2 \\ \tau_3 \end{bmatrix} \]

\[ c_1\beta = \begin{bmatrix} 0 & 1 & -1 & 0 \end{bmatrix} \beta = \tau_1 - \tau_2 \] \[ c_2\beta = \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} \beta = \tau_1 - \tau_3 \] \[ c_3\beta = \begin{bmatrix} 0 & 2 & -1 & -1 \end{bmatrix} \beta = 2\tau_1 - \tau_2 - \tau_3 \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -1 & 0 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} + \alpha_3 \begin{bmatrix} 0 & 2 & -1 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 + 2\alpha_3 = 0 \] \[ \Rightarrow - \alpha_1 + 0 - \alpha_3 = 0 \] \[ \Rightarrow 0 - \alpha_2 - \alpha_3 = 0 \]

Maka:

\[ \alpha_1 = 0 \] \[ \alpha_1 = \alpha_2 = -\alpha_3 = 0 \] \[ \alpha_1 = \alpha_2 = \alpha_3 = 0 \]

Sehingga \(C\) independen dan \(H_0\) testable ✓.

Latihan Soal Praktikum

Soal 1: Pertimbangkan model klasifikasi satu arah dengan efek tetap

\(k = 3\). Periksa setiap hipotesis nol berikut untuk testability.

\(H_0: \tau_1 + \tau_2 = 2\tau_3 \text{ dan } \tau_1 - \tau_2 = 0\)
\(H_0: \tau_1 = \tau_2 \text{ dan } \tau_3 = \tau_2 - \tau_1 \text{ dan } \tau_1 + \tau_3 = 0\)

Jawaban:

\(H_0: \tau_1 + \tau_2 = 2\tau_3 \text{ dan } \tau_1 - \tau_2 = 0\)

Model statistik:

\[ y_{ij} = \mu + \tau_i + \varepsilon_{ij} \quad i = 1,2,3 \quad j = 1,2,...,n_i \]

\[ \tau_1 - 2\tau_2 + \tau_3 = 0 \]

\[ C_2 = \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} \]

\[ \tau_1 = \tau_3 \]

\[ C_1 = \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ C = \begin{bmatrix} 0 & 1 & -2 & 1 \\ 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 = 0 \] \[ \Rightarrow -2\alpha_1 + 0 = 0 \] \[ \Rightarrow \alpha_1 - \alpha_2 = 0 \]

Maka, kesimpulannya:

\[ \alpha_1 = \alpha_2 = 0 \]

Sehingga \(C\) independen dan \(H_0\) testable.

\(H_0: \tau_1 = \tau_2 \text{ dan } \tau_3 = \tau_2 - \tau_1 \text{ dan } \tau_1 + \tau_3 = 0\)

\[ C = \begin{bmatrix} 0 & 1 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 2 & -1 & -1 \end{bmatrix} \quad \beta = \begin{bmatrix} \mu \\ \tau_1 \\ \tau_2 \\ \tau_3 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -1 & 0 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} + \alpha_3 \begin{bmatrix} 0 & 2 & -1 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 + 2\alpha_3 = 0 \] \[ \Rightarrow - \alpha_1 + 0 - \alpha_3 = 0 \] \[ \Rightarrow 0 - \alpha_2 - \alpha_3 = 0 \]

Maka:

\[ \alpha_1 = 0 \] \[ \alpha_1 = \alpha_2 = -\alpha_3 = 0 \] \[ \alpha_1 = \alpha_2 = \alpha_3 = 0 \]Sehingga \(C\) independen dan \(H_0\) testable ✓.

Soal 2: Pertimbangkan model klasifikasi satu arah dengan efek tetap

\(k = 3\). Periksa setiap hipotesis nol berikut untuk testability.

\(H_0: \tau_1 = \tau_3 \text{ dan } 3\tau_2 - \tau_1 = \tau_3\)

Model statistik:

\[ y_{ij} = \mu + \tau_i + \varepsilon_{ij} \quad i = 1,2,3 \quad j = 1,2,...,n_i \]

\[ \tau_1 - 2\tau_2 + \tau_3 = 0 \]

\[ C_2 = \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} \]

\[ \tau_1 = \tau_3 \]

\[ C_1 = \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ C = \begin{bmatrix} 0 & 1 & -2 & 1 \\ 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 = 0 \] \[ \Rightarrow -2\alpha_1 + 0 = 0 \] \[ \Rightarrow \alpha_1 - \alpha_2 = 0 \]

Maka, kesimpulannya:

\[ \alpha_1 = \alpha_2 = 0 \]

Sehingga \(C\) independen dan \(H_0\) testable.

\(H_0: \tau_1 = 2\tau_2 - \tau_3 \text{ dan } \tau_1 + \tau_2 = \tau_3\)

\[ C = \begin{bmatrix} 0 & 1 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 2 & -1 & -1 \end{bmatrix} \quad \beta = \begin{bmatrix} \mu \\ \tau_1 \\ \tau_2 \\ \tau_3 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -1 & 0 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} + \alpha_3 \begin{bmatrix} 0 & 2 & -1 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 + 2\alpha_3 = 0 \] \[ \Rightarrow - \alpha_1 + 0 - \alpha_3 = 0 \] \[ \Rightarrow 0 - \alpha_2 - \alpha_3 = 0 \]

Maka:

\[ \alpha_1 = 0 \] \[ \alpha_1 = \alpha_2 = -\alpha_3 = 0 \] \[ \alpha_1 = \alpha_2 = \alpha_3 = 0 \]

Sehingga \(C\) independen dan \(H_0\) testable ✓.

Soal 3: Pertimbangkan model klasifikasi satu arah dengan efek tetap

\(k = 3\). Periksa setiap hipotesis nol berikut untuk testability.

\(H_0: \tau_1 + \tau_2 = 3\tau_3 \text{ dan } \tau_1 - \tau_3 = \tau_2\)

Model statistik:

\[ y_{ij} = \mu + \tau_i + \varepsilon_{ij} \quad i = 1,2,3 \quad j = 1,2,...,n_i \]

\[ \tau_1 - 2\tau_2 + \tau_3 = 0 \]

\[ C_2 = \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} \]

\[ \tau_1 = \tau_3 \]

\[ C_1 = \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ C = \begin{bmatrix} 0 & 1 & -2 & 1 \\ 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 = 0 \] \[ \Rightarrow -2\alpha_1 + 0 = 0 \] \[ \Rightarrow \alpha_1 - \alpha_2 = 0 \]

Maka, kesimpulannya:

\[ \alpha_1 = \alpha_2 = 0 \]

Sehingga \(C\) independen dan \(H_0\) testable.

\(H_0: \tau_2 = \tau_1 - \tau_3 \text{ dan } \tau_1 + \tau_2 = 0\)

\[ C = \begin{bmatrix} 0 & 1 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 2 & -1 & -1 \end{bmatrix} \quad \beta = \begin{bmatrix} \mu \\ \tau_1 \\ \tau_2 \\ \tau_3 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -1 & 0 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} + \alpha_3 \begin{bmatrix} 0 & 2 & -1 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 + 2\alpha_3 = 0 \] \[ \Rightarrow - \alpha_1 + 0 - \alpha_3 = 0 \] \[ \Rightarrow 0 - \alpha_2 - \alpha_3 = 0 \]

Maka:

\[ \alpha_1 = 0 \] \[ \alpha_1 = \alpha_2 = -\alpha_3 = 0 \] \[ \alpha_1 = \alpha_2 = \alpha_3 = 0 \]Sehingga \(C\) independen dan \(H_0\) testable ✓.

Soal 4: Pertimbangkan model klasifikasi satu arah dengan efek tetap

\(k = 3\). Periksa setiap hipotesis nol berikut untuk testability.

\(H_0: \tau_1 = 2\tau_3 \text{ dan } \tau_1 - \tau_2 = \tau_3\)

Model statistik:

\[ y_{ij} = \mu + \tau_i + \varepsilon_{ij} \quad i = 1,2,3 \quad j = 1,2,...,n_i \]

\[ \tau_1 - 2\tau_2 + \tau_3 = 0 \]

\[ C_2 = \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} \]

\[ \tau_1 = \tau_3 \]

\[ C_1 = \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ C = \begin{bmatrix} 0 & 1 & -2 & 1 \\ 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 = 0 \] \[ \Rightarrow -2\alpha_1 + 0 = 0 \] \[ \Rightarrow \alpha_1 - \alpha_2 = 0 \]

Maka, kesimpulannya:

\[ \alpha_1 = \alpha_2 = 0 \]

Sehingga \(C\) independen dan \(H_0\) testable.

\(H_0: 3\tau_1 = \tau_2 \text{ dan } \tau_1 + \tau_3 = 2\tau_2\)

\[ C = \begin{bmatrix} 0 & 1 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 2 & -1 & -1 \end{bmatrix} \quad \beta = \begin{bmatrix} \mu \\ \tau_1 \\ \tau_2 \\ \tau_3 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -1 & 0 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} + \alpha_3 \begin{bmatrix} 0 & 2 & -1 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 + 2\alpha_3 = 0 \] \[ \Rightarrow - \alpha_1 + 0 - \alpha_3 = 0 \] \[ \Rightarrow 0 - \alpha_2 - \alpha_3 = 0 \]

Maka:

\[ \alpha_1 = 0 \] \[ \alpha_1 = \alpha_2 = -\alpha_3 = 0 \] \[ \alpha_1 = \alpha_2 = \alpha_3 = 0 \]Sehingga \(C\) independen dan \(H_0\) testable ✓.

Soal 5: Pertimbangkan model klasifikasi satu arah dengan efek tetap

\(k = 3\). Periksa setiap hipotesis nol berikut untuk testability.

\(H_0: 2\tau_1 - \tau_2 = \tau_3 \text{ dan } \tau_1 = \tau_2\)

Model statistik:

\[ y_{ij} = \mu + \tau_i + \varepsilon_{ij} \quad i = 1,2,3 \quad j = 1,2,...,n_i \]

\[ \tau_1 - 2\tau_2 + \tau_3 = 0 \]

\[ C_2 = \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} \]

\[ \tau_1 = \tau_3 \]

\[ C_1 = \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ C = \begin{bmatrix} 0 & 1 & -2 & 1 \\ 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 = 0 \] \[ \Rightarrow -2\alpha_1 + 0 = 0 \] \[ \Rightarrow \alpha_1 - \alpha_2 = 0 \]

Maka, kesimpulannya:

\[ \alpha_1 = \alpha_2 = 0 \]

Sehingga \(C\) independen dan \(H_0\) testable.

\(H_0: \tau_1 = \tau_3 \text{ dan } \tau_2 - \tau_3 = \tau_1 - \tau_2\)

\[ C = \begin{bmatrix} 0 & 1 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 2 & -1 & -1 \end{bmatrix} \quad \beta = \begin{bmatrix} \mu \\ \tau_1 \\ \tau_2 \\ \tau_3 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -1 & 0 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} + \alpha_3 \begin{bmatrix} 0 & 2 & -1 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 + 2\alpha_3 = 0 \] \[ \Rightarrow - \alpha_1 + 0 - \alpha_3 = 0 \] \[ \Rightarrow 0 - \alpha_2 - \alpha_3 = 0 \]

Maka:

\[ \alpha_1 = 0 \] \[ \alpha_1 = \alpha_2 = -\alpha_3 = 0 \] \[ \alpha_1 = \alpha_2 = \alpha_3 = 0 \]Sehingga \(C\) independen dan \(H_0\) testable ✓.

Soal 6: Pertimbangkan model klasifikasi satu arah dengan efek tetap

\(k = 3\). Periksa setiap hipotesis nol berikut untuk testability.

\(H_0: \tau_1 + 2\tau_2 = \tau_3 \text{ dan } \tau_2 = \tau_1 + \tau_3\)

Model statistik:

\[ y_{ij} = \mu + \tau_i + \varepsilon_{ij} \quad i = 1,2,3 \quad j = 1,2,...,n_i \]

\[ \tau_1 - 2\tau_2 + \tau_3 = 0 \]

\[ C_2 = \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} \]

\[ \tau_1 = \tau_3 \]

\[ C_1 = \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ C = \begin{bmatrix} 0 & 1 & -2 & 1 \\ 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 = 0 \] \[ \Rightarrow -2\alpha_1 + 0 = 0 \] \[ \Rightarrow \alpha_1 - \alpha_2 = 0 \]

Maka, kesimpulannya:

\[ \alpha_1 = \alpha_2 = 0 \]

Sehingga \(C\) independen dan \(H_0\) testable.

\(H_0: \tau_1 - \tau_3 = \tau_2 \text{ dan } 2\tau_1 = \tau_3\)

\[ C = \begin{bmatrix} 0 & 1 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 2 & -1 & -1 \end{bmatrix} \quad \beta = \begin{bmatrix} \mu \\ \tau_1 \\ \tau_2 \\ \tau_3 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -1 & 0 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} + \alpha_3 \begin{bmatrix} 0 & 2 & -1 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 + 2\alpha_3 = 0 \] \[ \Rightarrow - \alpha_1 + 0 - \alpha_3 = 0 \] \[ \Rightarrow 0 - \alpha_2 - \alpha_3 = 0 \]

Maka:

\[ \alpha_1 = 0 \] \[ \alpha_1 = \alpha_2 = -\alpha_3 = 0 \] \[ \alpha_1 = \alpha_2 = \alpha_3 = 0 \]Sehingga \(C\) independen dan \(H_0\) testable ✓.

Soal 7: Pertimbangkan model klasifikasi satu arah dengan efek tetap

\(k = 3\). Periksa setiap hipotesis nol berikut untuk testability.

\(H_0: \tau_1 = \tau_2 = \tau_3\)

Model statistik:

\[ y_{ij} = \mu + \tau_i + \varepsilon_{ij} \quad i = 1,2,3 \quad j = 1,2,...,n_i \]

\[ \tau_1 - 2\tau_2 + \tau_3 = 0 \]

\[ C_2 = \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} \]

\[ \tau_1 = \tau_3 \]

\[ C_1 = \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ C = \begin{bmatrix} 0 & 1 & -2 & 1 \\ 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 = 0 \] \[ \Rightarrow -2\alpha_1 + 0 = 0 \] \[ \Rightarrow \alpha_1 - \alpha_2 = 0 \]

Maka, kesimpulannya:

\[ \alpha_1 = \alpha_2 = 0 \]

Sehingga \(C\) independen dan \(H_0\) testable.

\(H_0: 2\tau_1 + \tau_3 = \tau_2 \text{ dan } \tau_1 = \tau_2 - \tau_3\)

\[ C = \begin{bmatrix} 0 & 1 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 2 & -1 & -1 \end{bmatrix} \quad \beta = \begin{bmatrix} \mu \\ \tau_1 \\ \tau_2 \\ \tau_3 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -1 & 0 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} + \alpha_3 \begin{bmatrix} 0 & 2 & -1 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 + 2\alpha_3 = 0 \] \[ \Rightarrow - \alpha_1 + 0 - \alpha_3 = 0 \] \[ \Rightarrow 0 - \alpha_2 - \alpha_3 = 0 \]

Maka:

\[ \alpha_1 = 0 \] \[ \alpha_1 = \alpha_2 = -\alpha_3 = 0 \] \[ \alpha_1 = \alpha_2 = \alpha_3 = 0 \]Sehingga \(C\) independen dan \(H_0\) testable ✓.

Soal 8: Pertimbangkan model klasifikasi satu arah dengan efek tetap

\(k = 3\). Periksa setiap hipotesis nol berikut untuk testability.

\(H_0: \tau_1 + \tau_3 = 2\tau_2 \text{ dan } \tau_1 = 2\tau_3\)

Model statistik:

\[ y_{ij} = \mu + \tau_i + \varepsilon_{ij} \quad i = 1,2,3 \quad j = 1,2,...,n_i \]

\[ \tau_1 - 2\tau_2 + \tau_3 = 0 \]

\[ C_2 = \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} \]

\[ \tau_1 = \tau_3 \]

\[ C_1 = \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ C = \begin{bmatrix} 0 & 1 & -2 & 1 \\ 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 = 0 \] \[ \Rightarrow -2\alpha_1 + 0 = 0 \] \[ \Rightarrow \alpha_1 - \alpha_2 = 0 \]

Maka, kesimpulannya:

\[ \alpha_1 = \alpha_2 = 0 \]

Sehingga \(C\) independen dan \(H_0\) testable.

\(H_0: \tau_2 = 2\tau_1 - \tau_3 \text{ dan } \tau_3 = \tau_2 + \tau_1\)

\[ C = \begin{bmatrix} 0 & 1 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 2 & -1 & -1 \end{bmatrix} \quad \beta = \begin{bmatrix} \mu \\ \tau_1 \\ \tau_2 \\ \tau_3 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -1 & 0 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} + \alpha_3 \begin{bmatrix} 0 & 2 & -1 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 + 2\alpha_3 = 0 \] \[ \Rightarrow - \alpha_1 + 0 - \alpha_3 = 0 \] \[ \Rightarrow 0 - \alpha_2 - \alpha_3 = 0 \]

Maka:

\[ \alpha_1 = 0 \] \[ \alpha_1 = \alpha_2 = -\alpha_3 = 0 \] \[ \alpha_1 = \alpha_2 = \alpha_3 = 0 \]Sehingga \(C\) independen dan \(H_0\) testable ✓.

Soal 9: Pertimbangkan model klasifikasi satu arah dengan efek tetap

\(k = 3\). Periksa setiap hipotesis nol berikut untuk testability.

\(H_0: \tau_1 + \tau_3 = \tau_2 \text{ dan } 3\tau_1 - 2\tau_2 = \tau_3\)

Model statistik:

\[ y_{ij} = \mu + \tau_i + \varepsilon_{ij} \quad i = 1,2,3 \quad j = 1,2,...,n_i \]

\[ \tau_1 - 2\tau_2 + \tau_3 = 0 \]

\[ C_2 = \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} \]

\[ \tau_1 = \tau_3 \]

\[ C_1 = \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ C = \begin{bmatrix} 0 & 1 & -2 & 1 \\ 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 = 0 \] \[ \Rightarrow -2\alpha_1 + 0 = 0 \] \[ \Rightarrow \alpha_1 - \alpha_2 = 0 \]

Maka, kesimpulannya:

\[ \alpha_1 = \alpha_2 = 0 \]

Sehingga \(C\) independen dan \(H_0\) testable.

\(H_0: \tau_2 = \tau_3 \text{ dan } \tau_1 + \tau_2 = 3\tau_3\)

\[ C = \begin{bmatrix} 0 & 1 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 2 & -1 & -1 \end{bmatrix} \quad \beta = \begin{bmatrix} \mu \\ \tau_1 \\ \tau_2 \\ \tau_3 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -1 & 0 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} + \alpha_3 \begin{bmatrix} 0 & 2 & -1 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 + 2\alpha_3 = 0 \] \[ \Rightarrow - \alpha_1 + 0 - \alpha_3 = 0 \] \[ \Rightarrow 0 - \alpha_2 - \alpha_3 = 0 \]

Maka:

\[ \alpha_1 = 0 \] \[ \alpha_1 = \alpha_2 = -\alpha_3 = 0 \] \[ \alpha_1 = \alpha_2 = \alpha_3 = 0 \]Sehingga \(C\) independen dan \(H_0\) testable ✓.

Soal 10: Pertimbangkan model klasifikasi satu arah dengan efek tetap

\(k = 3\). Periksa setiap hipotesis nol berikut untuk testability.

\(H_0: 2\tau_1 = \tau_3 \text{ dan } \tau_2 + \tau_3 = \tau_1\)

Model statistik:

\[ y_{ij} = \mu + \tau_i + \varepsilon_{ij} \quad i = 1,2,3 \quad j = 1,2,...,n_i \]

\[ \tau_1 - 2\tau_2 + \tau_3 = 0 \]

\[ C_2 = \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} \]

\[ \tau_1 = \tau_3 \]

\[ C_1 = \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ C = \begin{bmatrix} 0 & 1 & -2 & 1 \\ 0 & 1 & 0 & -1 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -2 & 1 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 = 0 \] \[ \Rightarrow -2\alpha_1 + 0 = 0 \] \[ \Rightarrow \alpha_1 - \alpha_2 = 0 \]

Maka, kesimpulannya:

\[ \alpha_1 = \alpha_2 = 0 \]

Sehingga \(C\) independen dan \(H_0\) testable.

\(H_0: \tau_1 = \tau_3 \text{ dan } \tau_1 + \tau_2 = 3\tau_3\)

\[ C = \begin{bmatrix} 0 & 1 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 2 & -1 & -1 \end{bmatrix} \quad \beta = \begin{bmatrix} \mu \\ \tau_1 \\ \tau_2 \\ \tau_3 \end{bmatrix} \]

\[ \alpha_1 \begin{bmatrix} 0 & 1 & -1 & 0 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix} + \alpha_3 \begin{bmatrix} 0 & 2 & -1 & -1 \end{bmatrix} = 0 \]

\[ \Rightarrow \alpha_1 + \alpha_2 + 2\alpha_3 = 0 \] \[ \Rightarrow - \alpha_1 + 0 - \alpha_3 = 0 \] \[ \Rightarrow 0 - \alpha_2 - \alpha_3 = 0 \]

Maka:

\[ \alpha_1 = 0 \] \[ \alpha_1 = \alpha_2 = -\alpha_3 = 0 \] \[ \alpha_1 = \alpha_2 = \alpha_3 = 0 \]

Sehingga \(C\) independen dan \(H_0\) testable ✓.

Prakt PML 10 - Hipotesis linier One-Way Classification