Anova (1)
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This document provides an overview of one-way analysis of variance (ANOVA) and randomized block ANOVA. It introduces ANOVA as a technique to compare two or more population means by analyzing sample variances. For one-way ANOVA, it demonstrates how to calculate sums of squares for treatments (SST), errors (SSE), and test statistics to determine if there are differences in population means. For randomized block ANOVA, it shows how total variability is partitioned into sums of squares for treatments, blocks, and errors, and how mean squares are used to calculate test statistics to analyze differences between treatments and blocks.
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Randomized Blocks](https://image.slidesharecdn.com/anova1-150723050133-lva1-app6892/85/Anova-1-36-320.jpg)
![Calculating the sums of squares
• Formulai for the calculation of the sums of squares
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![Calculating the sums of squares
• Formulai for the calculation of the sums of squares
Treatment
Block 1 2 k Block mean
1 X11 X12 . . . X1k
2 X21 X22 X2k
.
.
.
b Xb1 Xb2 Xbk
Treatment mean
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2]B[x
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![Sums of squares
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![• The procedure:
– Compute the number of pairwise comparisons (C)
[C=k(k-1)/2], where k is the number of populations.
– Set α = αE/C, where αE is the true probability of making at
least one Type I error (called experimentwise Type I error).
– We can conclude that µi and µj differ (at α/C% significance
level if
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Bonferroni Adjustment](https://image.slidesharecdn.com/anova1-150723050133-lva1-app6892/85/Anova-1-76-320.jpg)
Anova (1)
- 2. 15.1 Introduction • Analysis of variance compares two or more populations of interval data. • Specifically, we are interested in determining whether differences exist between the population means. • The procedure works by analyzing the sample variance.
- 3. • The analysis of variance is a procedure that tests to determine whether differences exits between two or more population means. • To do this, the technique analyzes the sample variances 15.2 One Way Analysis of Variance
- 4. • Example 15.1 – An apple juice manufacturer is planning to develop a new product -a liquid concentrate. – The marketing manager has to decide how to market the new product. – Three strategies are considered • Emphasize convenience of using the product. • Emphasize the quality of the product. • Emphasize the product’s low price. One Way Analysis of Variance
- 5. • Example 15.1 - continued – An experiment was conducted as follows: • In three cities an advertisement campaign was launched . • In each city only one of the three characteristics (convenience, quality, and price) was emphasized. • The weekly sales were recorded for twenty weeks following the beginning of the campaigns. One Way Analysis of Variance
- 6. One Way Analysis of Variance Convnce Quality Price 529 804 672 658 630 531 793 774 443 514 717 596 663 679 602 719 604 502 711 620 659 606 697 689 461 706 675 529 615 512 498 492 691 663 719 733 604 787 698 495 699 776 485 572 561 557 523 572 353 584 469 557 634 581 542 580 679 614 624 532 See file Xm15 -01 Weekly sales Weekly sales Weekly sales
- 7. • Solution – The data are interval – The problem objective is to compare sales in three cities. – We hypothesize that the three population means are equal One Way Analysis of Variance
- 8. H0: µ1 = µ2= µ3 H1: At least two means differ To build the statistic needed to test the hypotheses use the following notation: • Solution Defining the Hypotheses
- 9. Independent samples are drawn from k populations (treatments). 1 2 k X11 x21 . . . Xn1,1 1 1 x n X12 x22 . . . Xn2,2 2 2 x n X1k x2k . . . Xnk,k k k x n Sample size Sample mean First observation, first sample Second observation, second sample X is the “response variable”. The variables’ value are called “responses”. Notation
- 10. Terminology • In the context of this problem… Response variable – weekly sales Responses – actual sale values Experimental unit – weeks in the three cities when we record sales figures. Factor – the criterion by which we classify the populations (the treatments). In this problems the factor is the marketing strategy. Factor levels – the population (treatment) names. In this problem factor levels are the marketing trategies.
- 11. Two types of variability are employed when testing for the equality of the population means The rationale of the test statistic
- 12. Graphical demonstration: Employing two types of variability
- 13. 20 25 30 1 7 Treatment 1 Treatment 2 Treatment 3 10 12 19 9 Treatment 1Treatment 2Treatment 3 20 16 15 14 11 10 9 10x1 = 15x2 = 20x3 = 10x1 = 15x2 = 20x3 = The sample means are the same as before, but the larger within-sample variability makes it harder to draw a conclusion about the population means. A small variability within the samples makes it easier to draw a conclusion about the population means.
- 14. The rationale behind the test statistic – I • If the null hypothesis is true, we would expect all the sample means to be close to one another (and as a result, close to the grand mean). • If the alternative hypothesis is true, at least some of the sample means would differ. • Thus, we measure variability between sample means.
- 15. • The variability between the sample means is measured as the sum of squared distances between each mean and the grand mean. This sum is called the Sum of Squares for Treatments SST In our example treatments are represented by the different advertising strategies. Variability between sample means
- 16. 2 k 1j jj )xx(nSST ∑= −= There are k treatments The size of sample j The mean of sample j Sum of squares for treatments (SST) Note: When the sample means are close to one another, their distance from the grand mean is small, leading to a small SST. Thus, large SST indicates large variation between sample means, which supports H1.
- 17. • Solution – continued Calculate SST 2 k 1j jj 321 )xx(nSST 65.608x00.653x577.55x ∑= −= === = 20(577.55 - 613.07)2 + + 20(653.00 - 613.07)2 + + 20(608.65 - 613.07)2 = = 57,512.23 The grand mean is calculated by k21 kk2211 n...nn xn...xnxn X +++ +++ = Sum of squares for treatments (SST)
- 18. Is SST = 57,512.23 large enough to reject H0 in favor of H1? See next. Sum of squares for treatments (SST)
- 19. • Large variability within the samples weakens the “ability” of the sample means to represent their corresponding population means. • Therefore, even though sample means may markedly differ from one another, SST must be judged relative to the “within samples variability”. The rationale behind test statistic – II
- 20. • The variability within samples is measured by adding all the squared distances between observations and their sample means. This sum is called the Sum of Squares for Error SSEIn our example this is the sum of all squared differences between sales in city j and the sample mean of city j (over all the three cities). Within samples variability
- 21. • Solution – continued Calculate SSE Sum of squares for errors (SSE) ∑∑= = −= === k j jij n i xxSSE sss j 1 2 1 2 3 2 2 2 1 )( 24.670,811,238,700.775,10 = (n1 - 1)s1 2 + (n2 -1)s2 2 + (n3 -1)s3 2 = (20 -1)10,774.44 + (20 -1)7,238.61+ (20-1)8,670.24 = 506,983.50
- 22. Is SST = 57,512.23 large enough relative to SSE = 506,983.50 to reject the null hypothesis that specifies that all the means are equal? Sum of squares for errors (SSE)
- 23. To perform the test we need to calculate the mean squaresmean squares as follows: The mean sum of squares Calculation of MST - Mean Square for Treatments 12.756,28 13 23.512,57 1 = − = − = k SST MST Calculation of MSE Mean Square for Error 45.894,8 360 50.983,509 = − = − = kn SSE MSE
- 24. Calculation of the test statistic 23.3 45.894,8 12.756,28 = = = MSE MST F with the following degrees of freedom: v1=k -1 and v2=n-k Required Conditions: 1. The populations tested are normally distributed. 2. The variances of all the populations tested are equal.
- 25. And finally the hypothesis test: H0: µ1 = µ2 = …=µk H1: At least two means differ Test statistic: R.R: F>Fα,k-1,n-k MSE MST F= The F test rejection region
- 26. The F test Ho: µ1 = µ2= µ3 H1: At least two means differ Test statistic F= MST/ MSE= 3.23 15.3FFF:.R.R 360,13,05.0knk ≈=> −−−,−,α 1 Since 3.23 > 3.15, there is sufficient evidence to reject Ho in favor of H1,and argue that at least one of the mean sales is different than the others. 23.3 17.894,8 12.756,28 MSE MST F = = =
- 27. -0.02 0 0.02 0.04 0.06 0.08 0.1 0 1 2 3 4 • Use Excel to find the p-value – fx Statistical FDIST(3.23,2,57) = .0467 The F test p- value p Value = P(F>3.23) = .0467
- 28. Excel single factor ANOVA SS(Total) = SST + SSE Anova: Single Factor SUMMARY Groups Count Sum Average Variance Convenience 20 11551 577.55 10775.00 Quality 20 13060 653.00 7238.11 Price 20 12173 608.65 8670.24 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 57512 2 28756 3.23 0.0468 3.16 Within Groups 506984 57 8894 Total 564496 59 Xm15-01.xls
- 29. 15.3 Analysis of Variance Experimental Designs • Several elements may distinguish between one experimental design and others. – The number of factors. • Each characteristic investigated is called a factor. • Each factor has several levels.
- 30. Factor A Level 1Level2 Level 1 Factor B Level 3 Two - way ANOVA Two factors Level2 One - way ANOVA Single factor Treatment 3 (level 1) Response Response Treatment 1 (level 3) Treatment 2 (level 2)
- 31. • Groups of matched observations are formed into blocks, in order to remove the effects of “unwanted” variability. • By doing so we improve the chances of detecting the variability of interest. Independent samples or blocks
- 32. • Fixed effects – If all possible levels of a factor are included in our analysis we have a fixed effect ANOVA. – The conclusion of a fixed effect ANOVA applies only to the levels studied. • Random effects – If the levels included in our analysis represent a random sample of all the possible levels, we have a random-effect ANOVA. – The conclusion of the random-effect ANOVA applies to all the levels (not only those studied). Models of Fixed and Random Effects
- 33. • In some ANOVA models the test statistic of the fixed effects case may differ from the test statistic of the random effect case. • Fixed and random effects - examples – Fixed effects - The advertisement Example (15.1): All the levels of the marketing strategies were included – Random effects - To determine if there is a difference in the production rate of 50 machines, four machines are randomly selected and there production recorded. Models of Fixed and Random Effects.
- 34. 15.4 Randomized Blocks (Two-way) Analysis of Variance • The purpose of designing a randomized block experiment is to reduce the within-treatments variation thus increasing the relative amount of between treatment variation. • This helps in detecting differences between the treatment means more easily.
- 35. Treatment 4 Treatment 3 Treatment 2 Treatment 1 Block 1Block3 Block2 Block all the observations with some commonality across treatments Randomized Blocks
- 36. Treatment Block 1 2 k Block mean 1 X11 X12 . . . X1k 2 X21 X22 X2k . . . b Xb1 Xb2 Xbk Treatment mean 1]B[x 2]B[x b]B[x 1]T[x 2]T[x k]T[x Block all the observations with some commonality across treatments Randomized Blocks
- 37. • The sum of square total is partitioned into three sources of variation – Treatments – Blocks – Within samples (Error) SS(Total) = SST + SSB + SSESS(Total) = SST + SSB + SSE Sum of square for treatments Sum of square for blocks Sum of square for error Recall. For the independent samples design we have: SS(Total) = SST + SSE Partitioning the total variability
- 38. Calculating the sums of squares • Formulai for the calculation of the sums of squares Treatment Block 1 2 k Block mean 1 X11 X12 . . . X1k 2 X21 X22 X2k . . . b Xb1 Xb2 Xbk Treatment mean 1]B[x 2]B[x 1]T[x 2]T[x k]T[x x 2 1 X)]T[x(b − ...X)]T[x(b 2 2 + −+ 2 k X)]T[x(b −+SST = + − 2 1 X)]B[x(k + − 2 2 X)]B[x(k 2 k X)]B[x(k − SSB= =+−+−+ +−+−++−+−= ...)()(... )()(...)()()( 2 2 2 1 2 22 2 12 2 21 2 11 XxXX XxXxXxXxTotalSS kk
- 39. Calculating the sums of squares • Formulai for the calculation of the sums of squares Treatment Block 1 2 k Block mean 1 X11 X12 . . . X1k 2 X21 X22 X2k . . . b Xb1 Xb2 Xbk Treatment mean 1]B[x 2]B[x 1]T[x 2]T[x k]T[x x 2 1 X)]T[x(b − ...X)]T[x(b 2 2 + −+ 2 k X)]T[x(b −+SST = + − 2 1 X)]B[x(k + − 2 2 X)]B[x(k 2 k X)]B[x(k − SSB= ...)X]B[x]T[xx()X]B[x]T[xx( ...)X]B[x]T[xx()X]B[x]T[xx( ...)X]B[x]T[xx()X]B[x]T[xx(SSE 2 2kk2 2 1kk1 2 2222 2 1212 2 2121 2 1111 ++−−++−− ++−−++−− ++−−++−−=
- 40. To perform hypothesis tests for treatments and blocks we need • Mean square for treatments • Mean square for blocks • Mean square for error Mean Squares 1k SST MST − = 1b SSB MSB − = 1bkn SSE MSE +−− =
- 41. Test statistics for the randomized block design ANOVA MSE MST F = MSE MSB F = Test statistic for treatments Test statistic for blocks
- 42. • Testing the mean responses for treatments F > Fα,k-1,n-k-b+1 • Testing the mean response for blocks F> Fα,b-1,n-k-b+1 The F test rejection regions
- 43. • Example 15.2 – Are there differences in the effectiveness of cholesterol reduction drugs? – To answer this question the following experiment was organized: • 25 groups of men with high cholesterol were matched by age and weight. Each group consisted of 4 men. • Each person in a group received a different drug. • The cholesterol level reduction in two months was recorded. – Can we infer from the data in Xm15-02 that there are differences in mean cholesterol reduction among the four drugs? Randomized Blocks ANOVA - Example
- 44. • Solution – Each drug can be considered a treatment. – Each 4 records (per group) can be blocked, because they are matched by age and weight. – This procedure eliminates the variability in cholesterol reduction related to different combinations of age and weight. – This helps detect differences in the mean cholesterol reduction attributed to the different drugs. Randomized Blocks ANOVA - Example
- 45. BlocksTreatments b-1 MST / MSE MSB / MSE Conclusion: At 5% significance level there is sufficient evidence to infer that the mean “cholesterol reduction” gained by at least two drugs are different. K-1 Randomized Blocks ANOVA - Example ANOVA Source of Variation SS df MS F P-value F crit Rows 3848.7 24 160.36 10.11 0.0000 1.67 Columns 196.0 3 65.32 4.12 0.0094 2.73 Error 1142.6 72 15.87 Total 5187.2 99
- 46. Analysis of Variance Chapter 15 - continued
- 47. 15.5 Two-Factor Analysis of Variance - • Example 15.3 – Suppose in Example 15.1, two factors are to be examined: • The effects of the marketing strategy on sales. – Emphasis on convenience – Emphasis on quality – Emphasis on price • The effects of the selected media on sales. – Advertise on TV – Advertise in newspapers
- 48. • Solution – We may attempt to analyze combinations of levels, one from each factor using one-way ANOVA. – The treatments will be: • Treatment 1: Emphasize convenience and advertise in TV • Treatment 2: Emphasize convenience and advertise in newspapers • ……………………………………………………………………. • Treatment 6: Emphasize price and advertise in newspapers Attempting one-way ANOVA
- 49. • Solution – The hypotheses tested are: H0: µ1= µ2= µ3= µ4= µ5= µ6 H1: At least two means differ. Attempting one-way ANOVA
- 50. City1 City2 City3 City4 City5 City6 Convnce Convnce Quality Quality Price Price TV Paper TV Paper TV Paper – In each one of six cities sales are recorded for ten weeks. – In each city a different combination of marketing emphasis and media usage is employed. • Solution Attempting one-way ANOVA
- 51. • The p-value =.0452. • We conclude that there is evidence that differences exist in the mean weekly sales among the six cities. City1 City2 City3 City4 City5 City6 Convnce Convnce Quality Quality Price Price TV Paper TV Paper TV Paper • Solution Xm15-03 Attempting one-way ANOVA
- 52. • These result raises some questions: – Are the differences in sales caused by the different marketing strategies? – Are the differences in sales caused by the different media used for advertising? – Are there combinations of marketing strategy and media that interact to affect the weekly sales? Interesting questions – no answers
- 53. • The current experimental design cannot provide answers to these questions. • A new experimental design is needed. Two-way ANOVA (two factors)
- 54. Two-way ANOVA (two factors) City 1 sales City3 sales City 5 sales City 2 sales City 4 sales City 6 sales TV Newspapers Convenience Quality Price Are there differences in the mean sales caused by different marketing strategies? Factor A: Marketing strategy FactorB: Advertisingmedia
- 55. Test whether mean sales of “Convenience”, “Quality”, and “Price” significantly differ from one another. H0: µConv.= µQuality = µPrice H1: At least two means differ Calculations are based on the sum of square for factor A SS(A) Two-way ANOVA (two factors)
- 56. Two-way ANOVA (two factors) City 1 sales City 3 sales City 5 sales City 2 sales City 4 sales City 6 sales Factor A: Marketing strategy FactorB: Advertisingmedia Are there differences in the mean sales caused by different advertising media? TV Newspapers Convenience Quality Price
- 57. Test whether mean sales of the “TV”, and “Newspapers” significantly differ from one another. H0: µTV = µNewspapers H1: The means differ Calculations are based on the sum of square for factor B SS(B) Two-way ANOVA (two factors)
- 58. Two-way ANOVA (two factors) City 1 sales City 5 sales City 2 sales City 4 sales City 6 sales TV Newspapers Convenience Quality Price Factor A: Marketing strategy FactorB: Advertisingmedia Are there differences in the mean sales caused by interaction between marketing strategy and advertising medium? City 3 sales TV Quality
- 59. Test whether mean sales of certain cells are different than the level expected. Calculation are based on the sum of square for interaction SS(AB) Two-way ANOVA (two factors)
- 60. Graphical description of the possible relationships between factors A and B. Graphical description of the possible relationships between factors A and B.
- 61. Levels of factor A 1 2 3 Level 1 of factor B Level 2 of factor B 1 2 3 1 2 31 2 3 Level 1and 2 of factor B Difference between the levels of factor A No difference between the levels of factor B Difference between the levels of factor A, and difference between the levels of factor B; no interaction Levels of factor A Levels of factor A Levels of factor A No difference between the levels of factor A. Difference between the levels of factor B Interaction M R e e s a p n o n s e M R e e s a p n o n s e M R e e s a p n o n s e M R e e s a p n o n s e
- 62. Sums of squares ∑= −= a 1i 2 i )x]A[x(rb)A(SS })()()){(2(10( 222 . xxxxxx pricequalityconv −+−+− ∑= −= b 1j 2 j )x]B[x(ra)B(SS })()){(3)(10( 22 xxxx NewspaperTV −+− ∑∑ == +−−= b 1j 2 jiij a 1i )x]B[x]A[x]AB[x(r)AB(SS ∑∑∑ === −= r k ijijk b j a i ABxxSSE 1 2 11 )][(
- 63. F tests for the Two-way ANOVA • Test for the difference between the levels of the main factors A and B F= MS(A) MSE F= MS(B) MSE Rejection region: F > Fα,a-1 ,n-ab F > Fα, b-1, n-ab • Test for interaction between factors A and B F= MS(AB) MSE Rejection region: F > Fα,(a-1)(b-1),n-ab SS(A)/(a-1) SS(B)/(b-1) SS(AB)/(a-1)(b-1) SSE/(n-ab)
- 64. Required conditions: 1. The response distributions is normal 2. The treatment variances are equal. 3. The samples are independent.
- 65. • Example 15.3 – continued( Xm15-03) F tests for the Two-way ANOVA Convenience Quality Price TV 491 677 575 TV 712 627 614 TV 558 590 706 TV 447 632 484 TV 479 683 478 TV 624 760 650 TV 546 690 583 TV 444 548 536 TV 582 579 579 TV 672 644 795 Newspaper 464 689 803 Newspaper 559 650 584 Newspaper 759 704 525 Newspaper 557 652 498 Newspaper 528 576 812 Newspaper 670 836 565 Newspaper 534 628 708 Newspaper 657 798 546 Newspaper 557 497 616 Newspaper 474 841 587
- 66. • Example 15.3 – continued – Test of the difference in mean sales between the three marketing strategies H0: µconv. = µquality = µprice H1: At least two mean sales are different F tests for the Two-way ANOVA ANOVA Source of Variation SS df MS F P-value F crit Sample 13172.0 1 13172.0 1.42 0.2387 4.02 Columns 98838.6 2 49419.3 5.33 0.0077 3.17 Interaction 1609.6 2 804.8 0.09 0.9171 3.17 Within 501136.7 54 9280.3 Total 614757.0 59 Factor A Marketing strategies
- 67. • Example 15.3 – continued – Test of the difference in mean sales between the three marketing strategies H0: µconv. = µquality = µprice H1: At least two mean sales are different F = MS(Marketing strategy)/MSE = 5.33 Fcritical = Fα,a-1,n-ab = F.05,3-1,60-(3)(2)= 3.17; (p-value = .0077) – At 5% significance level there is evidence to infer that differences in weekly sales exist among the marketing strategies. F tests for the Two-way ANOVA MS(A)/MSE
- 68. • Example 15.3 - continued – Test of the difference in mean sales between the two advertising media H0: µTV. = µNespaper H1: The two mean sales differ F tests for the Two-way ANOVA Factor B = Advertising media ANOVA Source of Variation SS df MS F P-value F crit Sample 13172.0 1 13172.0 1.42 0.2387 4.02 Columns 98838.6 2 49419.3 5.33 0.0077 3.17 Interaction 1609.6 2 804.8 0.09 0.9171 3.17 Within 501136.7 54 9280.3 Total 614757.0 59
- 69. • Example 15.3 - continued – Test of the difference in mean sales between the two advertising media H0: µTV. = µNespaper H1: The two mean sales differ F = MS(Media)/MSE = 1.42 Fcritical = Fα,a-1,n-ab = F.05,2-1,60-(3)(2)= 4.02 (p-value = .2387) – At 5% significance level there is insufficient evidence to infer that differences in weekly sales exist between the two advertising media. F tests for the Two-way ANOVA MS(B)/MSE
- 70. • Example 15.3 - continued – Test for interaction between factors A and B H0: µTV*conv. = µTV*quality =…=µnewsp.*price H1: At least two means differ F tests for the Two-way ANOVA Interaction AB = Marketing*Media ANOVA Source of Variation SS df MS F P-value F crit Sample 13172.0 1 13172.0 1.42 0.2387 4.02 Columns 98838.6 2 49419.3 5.33 0.0077 3.17 Interaction 1609.6 2 804.8 0.09 0.9171 3.17 Within 501136.7 54 9280.3 Total 614757.0 59
- 71. • Example 15.3 - continued – Test for interaction between factor A and B H0: µTV*conv. = µTV*quality =…=µnewsp.*price H1: At least two means differ F = MS(Marketing*Media)/MSE = .09 Fcritical = Fα,(a-1)(b-1),n-ab = F.05,(3-1)(2-1),60-(3)(2)= 3.17 (p-value= .9171) – At 5% significance level there is insufficient evidence to infer that the two factors interact to affect the mean weekly sales. MS(AB)/MSE F tests for the Two-way ANOVA
- 72. 15.7 Multiple Comparisons • When the null hypothesis is rejected, it may be desirable to find which mean(s) is (are) different, and at what ranking order. • Three statistical inference procedures, geared at doing this, are presented: – Fisher’s least significant difference (LSD) method – Bonferroni adjustment – Tukey’s multiple comparison method
- 73. • Two means are considered different if the difference between the corresponding sample means is larger than a critical number. Then, the larger sample mean is believed to be associated with a larger population mean. • Conditions common to all the methods here: – The ANOVA model is the one way analysis of variance – The conditions required to perform the ANOVA are satisfied. – The experiment is fixed-effect 15.7 Multiple Comparisons
- 74. Fisher Least Significant Different (LSD) Method • This method builds on the equal variances t-test of the difference between two means. • The test statistic is improved by using MSE rather than sp 2 . • We can conclude that µi and µj differ (at α% significance level if |µi - µj| > LSD, where kn.f.d ) n 1 n 1 (MSEtLSD ji 2 −= += α
- 75. Experimentwise Type I error rate (αE) (the effective Type I error) • The Fisher’s method may result in an increased probability of committing a type I error. • The experimentwise Type I error rate is the probability of committing at least one Type I error at significance level of α. It is calculated by αE = 1-(1 – α)C where C is the number of pairwise comparisons (I.e. C = k(k-1)/2 • The Bonferroni adjustment determines the required Type I error probability per pairwise comparison (α) ,to secure a pre- determined overall αE.
- 76. • The procedure: – Compute the number of pairwise comparisons (C) [C=k(k-1)/2], where k is the number of populations. – Set α = αE/C, where αE is the true probability of making at least one Type I error (called experimentwise Type I error). – We can conclude that µi and µj differ (at α/C% significance level if kn.f.d ) n 1 n 1 (MSEt ji )C2(ji −= +>µ−µ α Bonferroni Adjustment
- 77. 35.4465.6080.653xx 10.3165.60855.577xx 45.750.65355.577xx 32 31 21 =−=− =−=− =−=− • Example 15.1 - continued – Rank the effectiveness of the marketing strategies (based on mean weekly sales). – Use the Fisher’s method, and the Bonferroni adjustment method • Solution (the Fisher’s method) – The sample mean sales were 577.55, 653.0, 608.65. – Then, 71.59)20/1()20/1(8894t ) n 1 n 1 (MSEt 2/05. ji 2 ≈+ =+α Fisher and Bonferroni Methods
- 78. • Solution (the Bonferroni adjustment) – We calculate C=k(k-1)/2 to be 3(2)/2 = 3. – We set α = .05/3 = .0167, thus t.0167/2,60-3 = 2.467 (Excel). 54.73)20/1()20/1(8894467.2 ) n 1 n 1 (MSEt ji 2 =+ =+α Again, the significant difference is between µ1 and µ2. 35.4465.6080.653xx 10.3165.60855.577xx 45.750.65355.577xx 32 31 21 =−=− =−=− =−=− Fisher and Bonferroni Methods
- 79. • The test procedure: – Find a critical number ω as follows: g n MSE ),k(q ν=ω α k = the number of samples ν =degrees of freedom = n - k ng = number of observations per sample (recall, all the sample sizes are the same) α = significance level qα(k,ν) = a critical value obtained from the studentized range table Tukey Multiple Comparisons
- 80. If the sample sizes are not extremely different, we can use the above procedure with ng calculated as the harmonic mean of the sample sizes. k21 n1...n1n1 k gn +++ = • Repeat this procedure for each pair of samples. Rank the means if possible. • Select a pair of means. Calculate the difference between the larger and the smaller mean. • If there is sufficient evidence to conclude that µmax > µmin . minmax xx − ω>− minmax xx Tukey Multiple Comparisons
- 81. City 1 vs. City 2: 653 - 577.55 = 75.45 City 1 vs. City 3: 608.65 - 577.55 = 31.1 City 2 vs. City 3: 653 - 608.65 = 44.35 • Example 15.1 - continued We had three populations (three marketing strategies). K = 3, Sample sizes were equal. n1 = n2 = n3 = 20, ν = n-k = 60-3 = 57, MSE = 8894. minmax xx − 70.71 20 8894 )57,3(.q n MSE ),k(q 05 g ==ν=ω α Take q.05(3,60) from the table. Population Sales - City 1 Sales - City 2 Sales - City 3 Mean 577.55 653 698.65 ω>− minmax xx Tukey Multiple Comparisons
- 82. Excel – Tukey and Fisher LSD method Xm15-01 Fisher’s LDS Bonferroni adjustments α = .05 α = .05/3 = .0167 Multiple Comparisons LSD Omega Treatment Treatment Difference Alpha = 0.05 Alpha = 0.05 Convenience Quality -75.45 59.72 71.70 Price -31.1 59.72 71.70 Quality Price 44.35 59.72 71.70 Multiple Comparisons LSD Omega Treatment Treatment Difference Alpha = 0.0167 Alpha = 0.05 Convenience Quality -75.45 73.54 71.70 Price -31.1 73.54 71.70 Quality Price 44.35 73.54 71.70