Analysis of Variance

Analysis of Variance : the simultaneous comparizon of several population means F-distribution : used to test whether two samples are from population having equal variances and it is also applied when we want to compare several population means simultaneously.

Comparing two population variance 2 Ho : σ = σ H1 : σ = σ Test statistic for comparing two variances : F = ~ F- distribution 2 2 1 2 2 1 2 2 s 1 2 s 2

Contoh : Sean Lammers, president of the Lammers Limos Service Company, membandingkan variasi waktu tempuh (menit) dari Cityhall di Toledo, Ohio ke Metro Airport di Detroit melalui 2 rute ( via US -25 dan Interstate-75), dengan taraf α = 10 % US-25 route Interstate-75 route 52 67 56 45 70 54 64 59 60 61 51 63 57 65

US-Route-25 Interstate-75 Rata-rata = 28.29 Std deviasi = 8.9947 Rata-rata = 59.00 Std deviasi = 4.3753 US-Route25 is more variation then Interstate-75 route,This is somewhat consistent with his knowledge of two route; US-25 contain more stoplights, I-75 is limited-access interstate highway, but I-75 is several miles longer.

Nilai F = 4.23, Nilai tabel F distribution pada Appendix G pada buku Douglas A. Lind,dkk,Statistical Techniques in Business of Economics,International Edition,yaitu : nilai F dengan α/2 = 5 %, dan df numerator 6 dan df denominator 7 adalah sebesar 3.87. Hasil ini menunjukkan bahwa Ho di tolak Berikan kesimpulan anda.

The Anova Test Total Variation : the sum of the squared differences between each observation and the overall mean (grand mean); SS-total Treatment Variation : the sum of the squared differences between each treatment mean and the overall mean (grand mean); SST Random Variation : the sum of the squared differences between each observation and its treatment mean; SSE

SS-total = Σ(X-XG)2 SSE = Σ(X-XC)2 SST = Σ(Xc -XG)2 SST = SS-total - SSE

ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square F Treatments Error Total SST SSE SS- Tolal k-1 n-k n-1 SST/(k-1) = MST SSE/(n-k) = MSE MST/ MSE

Example : Prof. James had students in his marketing class rate his performance as Excellent, Good, Fair or Poor. A graduate student collected the rate and from records office, Prof. James was matched with his or her course grade. The sample information is reported below. Is there a difference in the mean score of the students in each of the four rating categories ? Use the 0.01 significance level. (note : The rating is the treatment variable) Formulasikan hipotesis, buat tabel anova, tentukan nilai F tabel dan buat keputusan !

Course Grades Excellent Good Fair Poor Total 94 90 85 80 75 68 77 83 88 70 73 76 78 65 72 74 Column 349 391 510 414 1664 n 4 5 7 6 22 Mean 87.25 78.20 72.86 69.00 75.64

Ho : μ1= μ2 = μ3 = μ4 H1 : The mean scores are not all equal

Berikan kesimpulan anda ! ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square F Treatments Error Total 890.68 594.41SSE SS- Tolal 3 18 21 296.89 33.02 8.99 Nilai F tabel untuk α = 0.01, df numerator = 3 dan df denominator =18, adalah 5.09 Berikan kesimpulan anda !

Two-way Anova We have the second treatment variable, that is Blocking variable Blocking variable : A second treatment variable that when included in the ANOVA analysis will have the effect of reducing the SSE term

SSB = k Σ(Xb-XG)2 k is the number of treatment b is the number of blocks Xb is the sample mean of block b XG is the overall or grand mean SSE = SS total – SST - SSB

ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square F Treatments Blocks Error Total SST SSB SSE SS- Tolal k-1 b-1 (k-1)(b-1) n-1 SST/(k-1) = MST SSB/(b-1) =MSB SSE/(k-1)(b-1) = MSE MST/ MSE MSB/MSE

Example The Chapin Manufacturing Company operates 24 hours a day, five days a week. The workers rotate shifts each week. Management is interested in whether there is a difference in the number of units produced when the employees work on various shifts. A sample of five workers is selected and their output recorded on each shift. At the 0.05 significance level, can we conclude there is a difference in the mean production rate by shift or by employee ?

Employee Units Produced Day Afternoon Night Skaff 31 25 35 Lum 33 26 Clark 28 24 30 Treece 29 Morgan 27

ANOVA Table For Block : For treatment : Ho :μ1 = μ2 = μ3 = μ4 = μ5 H1 : Not all means equal Reject if F > 4.46 For Block : Ho :μ1 = μ2 = μ3 = μ4 = μ5 H1 : Not all means equal Reject if F > 3.84 ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square F Treatment (rotate shift) Blocks (employee) Error Total 62.53 33.73 43.47 139.73 2 4 8 14 31.2650 8.4325 5.4338 5.75 1.55 There is a difference in shifts but not by employees.