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Analysis of Variance : the simultaneous comparizon of several population means F-distribution : used to test whether two samples are from population having.

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Presentation on theme: "Analysis of Variance : the simultaneous comparizon of several population means F-distribution : used to test whether two samples are from population having."— Presentation transcript:

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2 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.

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

4 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 routeInterstate-75 route

5 US-Route-25Interstate-75 Rata-rata = Std deviasi = Rata-rata = Std deviasi = 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.

6 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 Hasil ini menunjukkan bahwa Ho di tolak Berikan kesimpulan anda.

7 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

8 SS-total = Σ(X-X G ) 2 SSE = Σ(X-X C ) 2 SST = Σ(X c -X G ) 2 SST = SS-total - SSE

9 ANOVA Table Source of VariationSum of Squares Degrees of Freedom Mean SquareF Treatments Error Total SST SSE SS- Tolal k-1 n-k n-1 SST/(k-1) = MST SSE/(n-k) = MSE MST/ MSE

10 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 !

11 Course Grades ExcellentGoodFairPoorTotal Column Total n Mean

12 Ho : μ 1 = μ 2 = μ 3 = μ 4 H1 : The mean scores are not all equal

13 ANOVA Table Source of VariationSum of Squares Degrees of Freedom Mean SquareF Treatments Error Total SSE SS- Tolal Nilai F tabel untuk α = 0.01, df numerator = 3 dan df denominator =18, adalah 5.09 Berikan kesimpulan anda !

14 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

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

16 ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean SquareF 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

17 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 ?

18 Employee Units Produced DayAfternoonNight Skaff Lum Clark Treece Morgan282627

19 ANOVA Table Source of VariationSum of Squares Degrees of Freedom Mean SquareF Treatment (rotate shift) Blocks (employee) Error Total For treatment : Ho :μ 1 = μ 1 = μ 3 H1 : Not all means equal Reject if F > 4.46 For Block : Ho :μ 1 = μ 2 = μ 3 = μ 4 = μ 5 H1 : Not all means equal Reject if F > 3.84 There is a difference in shifts but not by employees.


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