AksiomatikMetodologi dan Statistika

Uji Homogenitas (Levene)

Uji kesamaan variansi antar kelompok menggunakan Levene (berbasis rata‑rata), dengan keputusan uji dan langkah perhitungan lengkap.

Apa yang dihitung?

Absolut deviasi dari mean tiap kelompok lalu dilakukan ANOVA pada deviasi tersebut untuk memperoleh statistik F (Levene), df1=k−1, df2=N−k, dan Sig.

Konfigurasi

Input Dataset (Group Y)

Baris pertama opsional sebagai header. Gunakan spasi/tab/koma sebagai pemisah.

Status data

Observasi: 15 • Kelompok: 3

Siap menghitung uji homogenitas.

Levene's Test for Equality of Variances

N = 15 • k = 3

Tabel Data Penelitian

KelompokNilai
Kontrol85.2000
Kontrol87.1000
Kontrol83.8000
Kontrol86.4000
Kontrol84.9000
Perlakuan_A92.3000
Perlakuan_A94.1000
Perlakuan_A90.8000
Perlakuan_A93.5000
Perlakuan_A91.7000
Perlakuan_B78.9000
Perlakuan_B81.2000
Perlakuan_B79.8000
Perlakuan_B80.6000
Perlakuan_B77.4000

Test of Homogeneity of Variances

Levene Statisticdf1df2Sig.
0.0469212–

Keputusan (α=0.05): p‑value tidak tersedia.

Langkah Perhitungan Levene (berdasar mean)

Sumber Nilai

k
3
[jumlah kelompok]
N
15
[jumlah observasi]
df1
2
[k − 1, antar‑kelompok]
df2
12
[N − k, dalam‑kelompok]
Zij=∣Yij−Yˉi.∣Z_{ij} = |Y_{ij} - \bar{Y}_{i.}|Zˉi.=1ni∑jZij ,Zˉ..=∑ini Zˉi.N\bar{Z}_{i.} = \frac{1}{n_i}\sum_j Z_{ij}\quad\ ,\quad \bar{Z}_{..} = \frac{\sum_i n_i\,\bar{Z}_{i.}}{N}

Rata‑rata per kelompok

Kelompok Kontrol: nKontrol = 5; Y = [85.2000, 87.1000, 83.8000, 86.4000, 84.9000]

YˉKontrol.=85.2000+87.1000+83.8000+86.4000+84.90005=85.4800\displaystyle \bar{Y}_{Kontrol.} = \frac{85.2000 + 87.1000 + 83.8000 + 86.4000 + 84.9000}{5} = 85.4800

Kelompok Perlakuan_A: nPerlakuan_A = 5; Y = [92.3000, 94.1000, 90.8000, 93.5000, 91.7000]

YˉPerlakuanA.=92.3000+94.1000+90.8000+93.5000+91.70005=92.4800\displaystyle \bar{Y}_{Perlakuan_A.} = \frac{92.3000 + 94.1000 + 90.8000 + 93.5000 + 91.7000}{5} = 92.4800

Kelompok Perlakuan_B: nPerlakuan_B = 5; Y = [78.9000, 81.2000, 79.8000, 80.6000, 77.4000]

YˉPerlakuanB.=78.9000+81.2000+79.8000+80.6000+77.40005=79.5800\displaystyle \bar{Y}_{Perlakuan_B.} = \frac{78.9000 + 81.2000 + 79.8000 + 80.6000 + 77.4000}{5} = 79.5800

Deviasi absolut & rata‑rata deviasi per kelompok

Kelompok Kontrol: Z = [ 0.2800, 1.6200, 1.6800, 0.9200, 0.5800 ]

ZˉKontrol.=0.2800+1.6200+1.6800+0.9200+0.58005=1.0160\displaystyle \bar{Z}_{Kontrol.} = \frac{0.2800 + 1.6200 + 1.6800 + 0.9200 + 0.5800}{5} = 1.0160

Kelompok Perlakuan_A: Z = [ 0.1800, 1.6200, 1.6800, 1.0200, 0.7800 ]

ZˉPerlakuanA.=0.1800+1.6200+1.6800+1.0200+0.78005=1.0560\displaystyle \bar{Z}_{Perlakuan_A.} = \frac{0.1800 + 1.6200 + 1.6800 + 1.0200 + 0.7800}{5} = 1.0560

Kelompok Perlakuan_B: Z = [ 0.6800, 1.6200, 0.2200, 1.0200, 2.1800 ]

ZˉPerlakuanB.=0.6800+1.6200+0.2200+1.0200+2.18005=1.1440\displaystyle \bar{Z}_{Perlakuan_B.} = \frac{0.6800 + 1.6200 + 0.2200 + 1.0200 + 2.1800}{5} = 1.1440
Zˉ..=1.0160imes5+1.0560imes5+1.1440imes515=1.0720\displaystyle \bar{Z}_{..} = \frac{1.0160 imes 5 + 1.0560 imes 5 + 1.1440 imes 5}{15} = 1.0720

ANOVA pada |deviasi|

SSB=∑i=1kni(Zˉi.−Zˉ..)2\displaystyle SS_B = \sum_{i=1}^{k} n_i (\bar{Z}_{i.} - \bar{Z}_{..})^2
SSB=5(1.0160−1.0720)2+5(1.0560−1.0720)2+5(1.1440−1.0720)2=0.0429\displaystyle SS_B = 5( 1.0160 - 1.0720 )^2 + 5( 1.0560 - 1.0720 )^2 + 5( 1.1440 - 1.0720 )^2 = 0.0429
SSW=∑i=1k∑j=1ni(Zij−Zˉi.)2\displaystyle SS_W = \sum_{i=1}^{k} \sum_{j=1}^{n_i} (Z_{ij} - \bar{Z}_{i.})^2
SSWKontrol=(0.2800−1.0160)2+(1.6200−1.0160)2+(1.6800−1.0160)2+(0.9200−1.0160)2+(0.5800−1.0160)2=1.5467\displaystyle SSW_{Kontrol} = (0.2800 - 1.0160)^2 + (1.6200 - 1.0160)^2 + (1.6800 - 1.0160)^2 + (0.9200 - 1.0160)^2 + (0.5800 - 1.0160)^2 = 1.5467
SSWPerlakuanA=(0.1800−1.0560)2+(1.6200−1.0560)2+(1.6800−1.0560)2+(1.0200−1.0560)2+(0.7800−1.0560)2=1.5523\displaystyle SSW_{Perlakuan_A} = (0.1800 - 1.0560)^2 + (1.6200 - 1.0560)^2 + (1.6800 - 1.0560)^2 + (1.0200 - 1.0560)^2 + (0.7800 - 1.0560)^2 = 1.5523
SSWPerlakuanB=(0.6800−1.1440)2+(1.6200−1.1440)2+(0.2200−1.1440)2+(1.0200−1.1440)2+(2.1800−1.1440)2=2.3843\displaystyle SSW_{Perlakuan_B} = (0.6800 - 1.1440)^2 + (1.6200 - 1.1440)^2 + (0.2200 - 1.1440)^2 + (1.0200 - 1.1440)^2 + (2.1800 - 1.1440)^2 = 2.3843
SSW=SSWKontrol+SSWPerlakuanA+SSWPerlakuanB=5.4834\displaystyle SS_W = SSW_{Kontrol} + SSW_{Perlakuan_A} + SSW_{Perlakuan_B} = 5.4834
MSB=SSBdf1=0.04292=0.0214\displaystyle MS_B = \frac{SS_B}{df_1} = \frac{0.0429}{2} = 0.0214
MSW=SSWdf2=5.483412=0.4569\displaystyle MS_W = \frac{SS_W}{df_2} = \frac{5.4834}{12} = 0.4569
W=MSBMSW=0.02140.4569=0.0469\displaystyle W = \frac{MS_B}{MS_W} = \frac{0.0214}{0.4569} = 0.0469

p‑value dihitung dari F(df1=2, df2=12). Keputusan (α=0.05): p‑value tidak tersedia.