Analysis of Variance(ANOVA)

Penjelasan Umum• Seringkali kita ingin menguji apakah tiga atau lebih populasi

memiliki rata-rata yg sama. • Contoh:

Apakah bahan bakar/km yg digunakan untuk beberapa merek mobil sama?

Apakah pendapatan pekerja pada beberapa lapangan pekerjaan sama?

Atau apakah biaya produksi yg menggunakan beberapa proses yg berbeda adalah sama?

Penjelasan Umum• Kita dapat menggunakan cara seperti yg lalu untuk menguji

kesamaan rata-rata dua populasi, tetapi hal tersebut akan memakan waktu dan perhitungan yg lebih lama.Contoh: jika ada 5 pop, maka ada 5C2 cara/perhitungan yg harus dilakukan.

• Untuk itu kita dapat melakukan uji secara simultan /keseluruhan populasi tersebut dengan menggunakan distribusi F dan metoda yg disebut ANOVA (Analysis of Variance)

• Assumptions– Populations are normally distributed– Populations have equal variances– Samples are randomly and independently drawn

One-Way Analysis of Variance

Hipotesis untukOne-Way ANOVA

• – Seluruh rata-rata populasi adalah sama – Artinya: Tidak ada efek treatment (tidak ada keragaman rata-

rata dalam kelompok)

• – Minimal salah satu rata-rata populasi ada yang tidak sama– Artinya: Terdapat efek treatment (terdapat keragaman rata-

rata dalam kelompok)– Tidak berarti bahwa semua rata-rata populasi tidak sama

(beberapa pasang mungkin sama)

k3210 μμμμ:H

sama populasi rata-rata semuaTidak :AH

One-Factor ANOVA

All Means are the same:The Null Hypothesis is True

(No Treatment Effect)

k3210 μμμμ:H

same the are μ all Not:H iA

321 μμμ

One-Factor ANOVA

At least one mean is different:The Null Hypothesis is NOT true

(Treatment Effect is present)

k3210 μμμμ:H

same the are μ all Not:H iA

321 μμμ 321 μμμ

or

(continued)

Partitioning the Variation

Partitioning the Variation

• Total variation can be split into two parts:

SST = Total Sum of SquaresSSB = Sum of Squares BetweenSSW = Sum of Squares Within

SST = SSB + SSW

Partitioning the Variation

Total Variation = jumlah kuadrat total (SST), yang mengukur keragaman total dalam data

Within-Sample Variation = kergaman di dalam masing-masing kelompok populasi (SSW)

Between-Sample Variation = keragaman antar kelompok populasi (SSB)

SST = SSB + SSW

(continued)

Partition of Total Variation

Variation Due to Factor (SSB)

Variation Due to Random Sampling (SSW)

Total Variation (SST)

Commonly referred to as: Sum of Squares Within Sum of Squares Error Sum of Squares Unexplained Within Groups Variation

Commonly referred to as: Sum of Squares Between Sum of Squares Among Sum of Squares Explained Among Groups Variation

= +

Total Sum of Squares

k

i

n

jij

i

)xx(SST1 1

2

Dimana:

SST = Total sum of squares (Jumlah kuadrat total)k = jumlah populasi (kelompok, level atau

treatment)

ni = sample size dari populasi ke-i

xij = pengamatan ke-jth dari populasi ke-i

x = rata-rata total (rata-rata dari seluruh data)

SST = SSB + SSW

Total Variation(continued)

G rou p 1 G rou p 2 G rou p 3

Resp on se , X

X

2212

211 )xx(...)xx()xx(SST

kkn

Sum of Squares Between

Dimana:

SSB = Sum of squares betweenk = jumlah populasi (kelompok, level, atau

treatment)

ni = sample size dari populasi ke-i

xi = rata-rata sample dari populasi ke-i

x = rata-rata total (rata-rata dari seluruh data)

2

1

)xx(nSSB i

k

ii

SST = SSB + SSW

Between-Group Variation

Variation Due to Differences Among Groups

i j

2

1

)xx(nSSB i

k

ii

1

kSSBMSB

Mean Square Between = SSB/degrees of freedom

Between-Group Variation(continued)

G rou p 1 G rou p 2 G rou p 3

Resp on se , X

X1X 2X

3X

2222

211 )xx(n...)xx(n)xx(nSSB kk

Sum of Squares Within

Dimana:

SSW = Sum of squares withink = jumlah populasi (kelompok, level, atau

treatment)

ni = sample size dari populasi ke-i

xi = rata-rata sample dari populasi ke-i

xij = pengamatan ke-jth dari populasi ke-i

2

11

)xx(SSW iij

n

j

k

i

j

SST = SSB + SSW

Within-Group Variation

Summing the variation within each group and then adding over all groups

i

kNSSWMSW

Mean Square Within = SSW/degrees of freedom

2

11

)xx(SSW iij

n

j

k

i

j

Within-Group Variation(continued)

G rou p 1 G rou p 2 G rou p 3

Resp on se , X

1X2X

3X

22221

2111 )(...)()( kkn xxxxxxSSW

k

One-Way ANOVA Table

Source of Variation

dfSS MS

Between Samples SSB MSB =

Within Samples N - kSSW MSW =

Total N - 1SST =SSB+SSW

k - 1 MSBMSW

F ratio

k = jumlah populasi (kelompok, level, atau treatment)N = jumlah seluruh pengamatandf = derajat bebas

SSBk - 1SSWN - k

F =

One-Factor ANOVAF Test Statistic

• Test statistic

MSB is mean squares between variancesMSW is mean squares within variances

• Degrees of freedom– df1 = k – 1 (k = number of populations)– df2 = N – k (N = sum of sample sizes from all populations)

MSWMSBF

H0: μ1= μ2 = … = μ k

HA: At least two population means are different

• The F statistic is the ratio of the between estimate of variance and the within estimate of variance– The ratio must always be positive– df1 = k -1 will typically be small– df2 = N - k will typically be large

The ratio should be close to 1 if H0: μ1= μ2 = … = μk is true

The ratio will be larger than 1 if H0: μ1= μ2 = … = μk is false

Interpreting One-Factor ANOVA F Statistic

One-Factor ANOVA F Test Example

You want to see if three different golf clubs yield different distances. You randomly select five measurements from trials on an automated driving machine for each club. At the .05 significance level, is there a difference in mean distance?

Club 1 Club 2 Club 3254 234 200263 218 222241 235 197237 227 206251 216 204

••••

•

One-Factor ANOVA Example: Scatter Diagram

270

260

250

240

230

220

210

200

190

•••••

•••••

Distance

1X

2X

3X

X

227.0 x

205.8 x 226.0x 249.2x 321

Club 1 Club 2 Club 3254 234 200263 218 222241 235 197237 227 206251 216 204

Club1 2 3

One-Factor ANOVA Example Computations

Club 1 Club 2 Club 3254 234 200263 218 222241 235 197237 227 206251 216 204

x1 = 249.2

x2 = 226.0

x3 = 205.8

x = 227.0

n1 = 5

n2 = 5

n3 = 5

N = 15

k = 3

SSB = 5 [ (249.2 – 227)2 + (226 – 227)2 + (205.8 – 227)2 ] = 4716.4

SSW = (254 – 249.2)2 + (263 – 249.2)2 +…+ (204 – 205.8)2 = 1119.6

MSB = 4716.4 / (3-1) = 2358.2

MSW = 1119.6 / (15-3) = 93.325.275

93.32358.2F

F = 25.275

One-Factor ANOVA Example Solution

H0: μ1 = μ2 = μ3

HA: μi not all equal = .05df1= 2 df2 = 12

Test Statistic:

Decision:

Conclusion:Reject H0 at = 0.05

There is evidence that at least one μi differs from the rest0

= .05

F.05 = 3.885Reject H0Do not

reject H0

25.27593.3

2358.2MSWMSBF

Critical Value:

F = 3.885

Uji Wilayah Berganda• Dari hasil pengujian kesamaan rata-rata populasi dgn ANOVA,

jika keputusan adalah menolak Ho. Maka kita dapati kesimpulan bahwa tidak semua µ sama (paling sedikit ada dua yang tidak sama). Namun kita tidak tahu mana yang berbeda.

• Untuk mencari µ mana yang berbeda nyata→ UJI WILAYAH BERGANDA DUNCAN DAN UJI TUKEY

xμ 1 = μ 2 μ 3

Uji Duncan

Prosedur:1. Urutkan rata-rata sampel untuk masing-

masing populasi (kelompok) dari yang terkecil hingga terbesar

2. Hitung wilayah nyata terpendek dari berbagai rata-rata

Uji Duncan

3. Kriteria pengujianBandingkan selisih kedua rata-rata yang ingin dilihat perbedaannya dengan kriteria sbb: xi – xj ≤ Rp (Tidak berbeda nyata) xi – xj > Rp (Berbeda nyata)

Contoh: Uji Duncan

1. Urutkan rata-rata sampel:Club 1 Club 2 Club 3

254 234 200263 218 222241 235 197237 227 206251 216 204

2. Hitung wilayah nyata terpendek dari berbagai rata-rata:

205.8 226.0 249.2

p 2 3rp 3.082 3.225Rp 13.313 13.931

α = 0.05, df = 12

Contoh: Uji Duncan

3. Bandingkan selisih rata-rata dengan Rp:

(Beda nyata)

(Beda nyata)

(Beda nyata)

Uji Tukey-Kramer

Dimana:q = Nilai dari standardized range table dengan df = k

dan N - k MSW = Mean Square Within ni dan nj = Sample sizes dari populasi (kelompok) ke-i & ke-j

ji n1

n1

2MSWqRange Critical

Contoh: Uji Tukey-Kramer

1. Compute absolute mean differences:Club 1 Club 2 Club 3

254 234 200263 218 222241 235 197237 227 206251 216 204 20.2205.8226.0xx

43.4205.8249.2xx

23.2226.0249.2xx

32

31

21

2. Find the q value from the table Tukey with k and N - k degrees of freedom for the desired level of

3.77qα

Contoh: Uji Tukey-Kramer

5. All of the absolute mean differences are greater than critical range. Therefore there is a significant difference between each pair of means at 5% level of significance.

16.28551

51

293.33.77

n1

n1

2MSWqRange Critical

jiα

3. Compute Critical Range:

20.2xx

43.4xx

23.2xx

32

31

21

4. Compare: