Tugas Uji Indrawi 2

TUGAS UJI INDERAWI

ANALISIS DATA METODE ANOVA, KRUSKAL WALLIS DAN FRIEDMAN

SOAL

Hasil uji kesukaan rasa yoghurt sinbiotik pada 9 kombinasi perlakuan, dengan 20

panelis adalah sebagai berikut :

Tabel 1. Data Hasil Uji Kesukaan Rasa Yoghurt Sinbiotik

PanelisUlangan 1 Ulangan 2 Ulangan 3

K1G1 K1G2 K1G3 K2G1 K2G2 K2G3 K3G1 K3G2 K3G31 4 2 2 4 4 4 4 3 32 3 3 3 3 2 2 2 2 23 5 5 5 5 5 5 5 5 54 2 2 2 2 2 2 2 2 25 3 3 3 2 3 3 3 3 46 4 3 3 4 3 3 4 4 37 3 4 2 3 2 4 2 3 38 4 4 4 4 3 5 3 4 49 3 3 3 4 3 3 2 4 4

10 4 3 2 4 4 4 4 3 211 4 5 5 3 4 5 3 3 312 2 4 4 4 2 4 2 2 213 3 3 3 3 3 3 3 3 314 3 4 2 3 2 3 2 3 315 3 3 2 2 3 4 3 4 516 4 3 3 3 4 3 3 3 317 4 3 3 3 2 3 3 3 318 4 3 4 4 4 4 4 4 419 3 2 4 3 2 5 4 5 220 3 4 2 3 2 3 3 4 3

Berilah kesimpulan saudara jika data tersebut dianalisis menggunakan :

a. ANOVA dan Uji lanjut

b. Kruskal Wallis dan Uji lanjut

c. Friedman dan Uji lanjut

Nama : Nur Diana SeptiNPM : 1233010021

JAWAB

Tabel 2. Data Hasil Uji Kesukaan Rasa Yoghurt Sinbiotik Ulangan 1

Sampel

Panelis Ulangan 1

K1G1 1 4K1G2 1 2K1G3 1 2K1G1 2 3K1G2 2 3K1G3 2 3K1G1 3 5K1G2 3 5K1G3 3 5K1G1 4 2K1G2 4 2K1G3 4 2K1G1 5 3K1G2 5 3K1G3 5 3K1G1 6 4K1G2 6 3K1G3 6 3K1G1 7 3K1G2 7 4K1G3 7 2K1G1 8 4K1G2 8 4K1G3 8 4K1G1 9 3K1G2 9 3K1G3 9 3K1G1 10 4K1G2 10 3K1G3 10 2K1G1 11 4K1G2 11 5K1G3 11 5K1G1 12 2K1G2 12 4K1G3 12 4K1G1 13 3K1G2 13 3K1G3 13 3K1G1 14 3K1G2 14 4

K1G3 14 2K1G1 15 3K1G2 15 3K1G3 15 2K1G1 16 4K1G2 16 3K1G3 16 3K1G1 17 4K1G2 17 3K1G3 17 3K1G1 18 4K1G2 18 3K1G3 18 4K1G1 19 3K1G2 19 2K1G3 19 4K1G1 20 3K1G2 20 4K1G3 20 2


Sampel

Panelis Ulangan 2

K2G1 1 4K2G2 1 4K2G3 1 4K2G1 2 3K2G2 2 2K2G3 2 2K2G1 3 5K2G2 3 5K2G3 3 5K2G1 4 2K2G2 4 2K2G3 4 2K2G1 5 2K2G2 5 3K2G3 5 3K2G1 6 4K2G2 6 3K2G3 6 3K2G1 7 3K2G2 7 2K2G3 7 4K2G1 8 4K2G2 8 3K2G3 8 5K2G1 9 4K2G2 9 3K2G3 9 3K2G1 10 4K2G2 10 4K2G3 10 4K2G1 11 3K2G2 11 4K2G3 11 5K2G1 12 4K2G2 12 2K2G3 12 4K2G1 13 3K2G2 13 3K2G3 13 3K2G1 14 3K2G2 14 2K2G3 14 3K2G1 15 2K2G2 15 3K2G3 15 4K2G1 16 3

K2G2 16 4K2G3 16 3K2G1 17 3K2G2 17 2K2G3 17 3K2G1 18 4K2G2 18 4K2G3 18 4K2G1 19 3K2G2 19 2K2G3 19 5K2G1 20 3K2G2 20 2K2G3 20 3


Sampel

Panelis Ulangan 3

K3G1 1 4K3G2 1 3

K3G3 1 3K3G1 2 2K3G2 2 2K3G3 2 2K3G1 3 5K3G2 3 5K3G3 3 5K3G1 4 2K3G2 4 2K3G3 4 2K3G1 5 3K3G2 5 3K3G3 5 4K3G1 6 4K3G2 6 4K3G3 6 3K3G1 7 2K3G2 7 3K3G3 7 3K3G1 8 3K3G2 8 4K3G3 8 4K3G1 9 2K3G2 9 4K3G3 9 4K3G1 10 4K3G2 10 3K3G3 10 2K3G1 11 3K3G2 11 3K3G3 11 3K3G1 12 2K3G2 12 2K3G3 12 2K3G1 13 3K3G2 13 3K3G3 13 3K3G1 14 2K3G2 14 3K3G3 14 3K3G1 15 3K3G2 15 4K3G3 15 5K3G1 16 3K3G2 16 3K3G3 16 3

K3G1 17 3K3G2 17 3K3G3 17 3K3G1 18 4K3G2 18 4K3G3 18 4K3G1 19 4K3G2 19 5K3G3 19 2K3G1 20 3K3G2 20 4K3G3 20 3

a. Anova dan Uji Lanjut1. Ulangan 1

One-way ANOVA: Ulangan 1 versus Sampel

Source DF SS MS F PSampel 2 1,300 0,650 0,84 0,436Error 57 43,950 0,771Total 59 45,250

S = 0,8781 R-Sq = 2,87% R-Sq(adj) = 0,00%

Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev -+---------+---------+---------+--------K1G1 20 3,4000 0,7539 (------------*------------)K1G2 20 3,3000 0,8645 (------------*------------)K1G3 20 3,0500 0,9987 (------------*------------) -+---------+---------+---------+-------- 2,70 3,00 3,30 3,60

Pooled StDev = 0,8781

Grouping Information Using Tukey Method

Sampel N Mean GroupingK1G1 20 3,4000 AK1G2 20 3,3000 AK1G3 20 3,0500 A

Means that do not share a letter are significantly different.

Tukey 95% Simultaneous Confidence IntervalsAll Pairwise Comparisons among Levels of Sampel

Individual confidence level = 98,05%

Sampel = K1G1 subtracted from:

Sampel Lower Center Upper +---------+---------+---------+-----K1G2 -0,7676 -0,1000 0,5676 (------------*------------)K1G3 -1,0176 -0,3500 0,3176 (------------*------------) +---------+---------+---------+----- -1,00 -0,50 0,00 0,50


Sampel Lower Center Upper +---------+---------+---------+-----K1G3 -0,9176 -0,2500 0,4176 (------------*------------) +---------+---------+---------+----- -1,00 -0,50 0,00 0,50

Kesimpulan

Pada ulangan 1 tingkat kesukaan panelis terhadap rasa yoghurt sinbiotik

sampel K1G1, K1G2, dan K1G3 tidak berbeda nyata (P>0,05)

2. Ulangan 2



S = 0,8979 R-Sq = 8,44% R-Sq(adj) = 5,22%

Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev ------+---------+---------+---------+---K2G1 20 3,3000 0,8013 (---------*----------)K2G2 20 2,9500 0,9445 (---------*---------)K2G3 20 3,6000 0,9403 (---------*---------) ------+---------+---------+---------+--- 2,80 3,20 3,60 4,00








Sampel Lower Center Upper ---------+---------+---------+--------+K2G2 -1,0326 -0,3500 0,3326 (---------*---------)K2G3 -0,3826 0,3000 0,9826 (--------*---------) ---------+---------+---------+--------+ -0,70 0,00 0,70 1,40


Sampel Lower Center Upper ---------+---------+---------+---------+K2G3 -0,0326 0,6500 1,3326 (--------*---------) ---------+---------+---------+---------+ -0,70 0,00 0,70 1,40

Kesimpulan



3. Ulangan 3



S = 0,8988 R-Sq = 1,99% R-Sq(adj) = 0,00%

Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev --+---------+---------+---------+-------K3G1 20 3,0500 0,8870 (-------------*------------)K3G2 20 3,3500 0,8751 (-------------*------------)K3G3 20 3,1500 0,9333 (------------*------------) --+---------+---------+---------+------- 2,70 3,00 3,30 3,60








Sampel Lower Center Upper --------+---------+---------+---------+-K3G2 -0,3833 0,3000 0,9833 (-------------*-------------)K3G3 -0,5833 0,1000 0,7833 (-------------*-------------) --------+---------+---------+---------+- -0,50 0,00 0,50 1,00


Sampel Lower Center Upper --------+---------+---------+---------+K3G3 -0,8833 -0,2000 0,4833 (-------------*-------------) --------+---------+---------+---------+ -0,50 0,00 0,50 1,00

Kesimpulan



b. Kruskal – Wallis dan Uji Lanjut

1. Ulangan 1

Kruskal-Wallis Test: Ulangan 1 versus Sampel

Kruskal-Wallis Test on Ulangan 1

Sampel N Median Ave Rank ZK1G1 20 3,000 33,8 1,04K1G2 20 3,000 31,3 0,24K1G3 20 3,000 26,4 -1,29Overall 60 30,5

H = 1,87 DF = 2 P = 0,393H = 2,10 DF = 2 P = 0,349 (adjusted for ties)

Kesimpulan :

Hasilnya adalah p > 0,05, artinya H0 tidak ditolak sehingga disimpulkan

sampel K1G1, K1G2, dan K1G3 itu tidak memberikan efek terhadap hasil.

2. Ulangan 2



Sampel N Median Ave Rank ZK2G1 20 3,000 31,1 0,19K2G2 20 3,000 24,6 -1,87K2G3 20 3,500 35,9 1,68Overall 60 30,5


Kesimpulan :



3. Ulangan 3



Sampel N Median Ave Rank ZK3G1 20 3,000 28,1 -0,74K3G2 20 3,000 33,7 1,00K3G3 20 3,000 29,6 -0,27Overall 60 30,5


Kesimpulan :



c. Friedman dan Uji lanjut

1. Ulangan 1

Friedman Test: Ulangan 1 versus Sampel blocked by Panelis

S = 1,83 DF = 2 P = 0,402S = 3,32 DF = 2 P = 0,190 (adjusted for ties)

Sum ofSampel N Est Median RanksK1G1 20 3,0000 44,0K1G2 20 3,0000 40,5K1G3 20 3,0000 35,5

Grand median = 3,0000

Kesimpulan:

Hasilnya adalah P > 0,05, artinya H0 tidak ditolak sehingga dapat

disimpulkan sampel K1G1, K1G2, dan K1G3 tidak menunjukan perbedaan.

2. Ulangan 2



Sum ofSampel N Est Median RanksK2G1 20 3,2500 41,5K2G2 20 2,9167 33,0K2G3 20 3,0833 45,5


Kesimpulan :

Hasilnya adalah P < 0,05, artinya H0 ditolak sehingga dapat disimpulkan

sampel K3G1, K3G2, dan K3G3 menunjukan perbedaan.

3. Ulangan 3



Sum ofSampel N Est Median Ranks

K3G1 20 3,0000 36,5K3G2 20 3,0000 43,5K3G3 20 3,0000 40,0


Kesimpulan:

Hasilnya adalah P > 0,05, artinya H0 tidak ditolak sehingga dapat

disimpulkan sampel K3G1, K3G2, dan K3G3 tidak menunjukan perbedaan.

Tugas Uji Indrawi 2

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