Chapter 5: Analysis of Variance 5. Introduction In this chapter, we introduced the analysis of variance technique, which deals with problems whose objective is to compare two or more populations of quantitative data. You are expected to learn how to do the following: l. Recognize when the analysis of variance is to be employed.. Recognize which of the three models introduced in this chapter is to be used. 3. How to interpret the ANOVA table. 4. How to perform Fisher's LSD method, the Bonferonni adjustment and Tuey s multiple comparison procedure. 5. How to conduct Bartlett's test. 5. Single-Factor (One-Way) Analysis of Variance: Independent Samples When the samples are drawn independently of each other, we partition the total sum of squares into two sources of variability: sum of squares for treatment and sum of squares for error. The F-test is then used to complete the technique. The important formulas are n j SS(Total) = ( x ij x ) j = i= SST = n j ( x j x ) = n ( x x ) + n ( x x ) + K + n ( x x ) j = n j SSE = ( x ij x j ) j = i= n n n = ( x i x ) + ( x i x ) + K + ( x i x ) i = i = i = = ( n )s MST = SST MSE = SSE n + ( n )s + K + ( n ) s 79
F = MST MSE The ANOVA table for the completely randomized design is shown below. Source d.f. Sums of Squares Mean Squares F-ratio Treatments SST MST F Error n SSE MSE Total n SS(Total) Example 5. A major computer manufacturer has received numerous complaints concerning the short life of its dis drives. Most need repair within two years. Since the cost of repairs often exceeds the cost of a new dis drive, the manufacturer is concerned. In his search for a better dis drive, he finds three new products. He decides to test these three plus his current dis drive to determine if differences in lifetimes exist among the products. He taes a random sample of five dis drives of each type and lins it with a computer. The number of wees until the drive breas down is recorded and is shown below. Do these data allow us to conclude at the 5% significance level that there are differences among the dis drives? Type Type Type 3 Current Product 78 5 43 0 9 0 5 96 0 6 33 88 05 88 08 5 98 8 8 x = 94.8 x = 3.4 x 3 = 6.0 x 4 = 07.6 s = 0.7 s = 5.8 s 3 = 7.0 s 4 = 30.3 Solution The problem objective is to compare the populations of lifetimes of the four dis drives, and the data are quantitative. Because the samples are independent, the appropriate technique is the completely randomized design of the analysis of variance. The null and alternative hypotheses automatically follow. H 0 : µ = µ = µ 3 = µ 4 80
H : At least two means differ. The rejection region is F > F α,, n = F.05, 3, 6 = 3.4 The test statistic is computed as follows: n = n = n 3 = n 4 = 5 x = 5(94.80 ) + 5(3.4) + 5(6.0 ) + 5(07.6 ) 0 =,09 0 = 0.45 SST = 5(94.8 0.45) + 5(3.4 0.45) + 5(6.0 0.45) + 5(07.6 0.45) =,57.75 SSE = 4(0.7) + 4(5.8) + 4(7.0) + 4(30.3) = 3,43. MST = MSE = F =,57.75 3 3,43. 6 839.5 3.95 = 839.5 = 3.95 = 3.9 The complete ANOVA table follows. Source d.f. Sums of Squares Mean Squares F-ratio Treatments 3,57.75 839.5 3.9 Error 6 3,43. 3.95 Total 9 5,940.95 Since the F-ratio (3.9) exceeds F α,, n (3.4), we reject H 0 and conclude that at least two means differ. 8
EXERCISES 5. Complete the ANOVA table and F-test with α =.05. Source d.f. Sums of Squares Mean Squares F-ratio Treatments 5 Error 3,000 Total 0 3,500 5. Develop the ANOVA table from the following information and perform the F-test with α =.0. SS(Total) = 50 SST = 90 SSE = 60 = 5 n = 7 n = 5 n 3 = 8 n 4 = 9 n 5 = 4 ANOVA Table Source d.f. Sums of Squares Mean Squares F-ratio 8
5.3 Test the hypotheses (with α =.0) H 0 : µ = µ = µ 3 H : At least two means differ. given the following statistics: x = 5 x = x 3 = 3 s = 6 s = s 3 = 9 n = 5 n = 8 n 3 = 0 ANOVA Table Source d.f. Sums of Squares Mean Squares F-ratio 5.4 Perform the analysis of variance test with α =.05 using the following information: Treatment 3 4 x ij 30 60 540 40 x ij 9,650 8,50 0,00 4,800 n j 0 8 5 H 0 : H : Rejection region: Value of the test statistic: 83
ANOVA Table Source d.f. Sums of Squares Mean Squares F-ratio Conclusion: 5.5 A nationwide real estate chain is in the process of examining condominium prices across the country. The company has hired a statistician who taes a random sample of six sales in each of four cities. The results are recorded to the nearest thousand dollars and are shown below. Can we conclude from these data that there are differences in the selling prices of condominiums among the four cities? (Use α =.05.) New Yor Chicago Dallas Los Angeles 53 99 73 305 78 56 85 88 85 76 66 35 3 03 38 48 77 9 75 96 89 4 44 x i =,50 x i =,438 x i =,80 x i =,706 x i = 388,98 x i = 35,636 x i = 79,076 x i = 489,470 H 0 : H : Rejection region: Value of the test statistic: 84
ANOVA Table Source d.f. Sum of Squares Mean Squares F-ratio Conclusion: 5.3 Analysis of Variance Models In this section we described some of the models that are available to statisticians. However, we covered only three of these. 5.4 Single-Factor Analysis of Variance: Randomized Blocs If the experimental design is randomized bloc, the total variation as measured by SS(Total) must be partitioned into three sources of variation: treatments (measured by SST), blocs (measured by SSB), and error (measured by SSE). The general form of the ANOVA table is shown below. ANOVA Table for Randomized Bloc Design Source d.f. Sums of Squares Mean Squares F-ratio Treatments SST MST F = MST/MSE Blocs b l SSB MSB F = MSB/MSE Error n b + SSE MSE Total n SS(Total) The rejection region for testing the treatment means is F > F α,, n b + 85
The rejection region for testing the bloc means is F > F α, b, n b + Example 5. The statistician in Exercise 5.5 decides to redo the experiment to eliminate the variation among condominium prices. In each city, the selling prices of a,000-square-foot, a,500-square-foot, a,000- square-foot, a,500-square-foot and a 3,000-square-foot condominium are randomly selected. The results are shown below. Condominium Size New Yor Chicago Dallas Los Angeles,000 square feet 65 85 73 00,500 square feet 98 93 8 96,000 square feet 5 5 97 78,500 square feet 3 68 9 33 3,000 square feet 405 38 94 446 The following statistics were computed: SST = 5,90.9 SSB = 07,57.7 SSE = 8,360.3 Complete the ANOVA table to determine if we can conclude at the 5% significance level that there are differences in condominium prices among the four cities. Solution We test the hypotheses H 0 : µ = µ = µ 3 = µ4 H : At least two treatment means differ. Rejection region: F > F α,, n b + = F.05, 3, = 3.49 Value of the test statistic: From the table below, we find F = 7.3. 86
Source d.f. Sums of Squares Mean Squares F-ratio Treatments 3 5,90.9 5,063.6 7.3 Blocs 4 07,57.7 6,789.4 38.5 Error 8,360.3 696.7 Total 9 30,708.9 Conclusion: Reject H 0. There is sufficient evidence to conclude that there are differences in condominium prices among the four cities. EXERCISES 5.6 Complete the following ANOVA table (randomized bloc design). Source d.f. Sums of Squares Mean Squares F-ratio Treatments 5 300 Blocs 8 Error 400 Total 53,000 5.7 Refer to Exercise 5.6. Test to determine if there are differences among the treatment means with α =.0. 5.8 Refer to Exercise 5.6. Test to determine if there are differences among the bloc means with α =.0. 87
5.9 A large catalogue chain store has been experimenting with several methods of advertising its extensive variety of bicycles. Three inds of catalogues have been prepared. In one, a side view of each bicycle is shown. In another, each bicycle s excellent record of longevity is extolled. In the third, pictures of the bicycles with their riders are shown. The company s management would lie to now if there are differences in sales among the stores that use the different catalogues. The monthly sales of bicycles of three randomly selected stores, each using a different catalogue, are shown below. Do these data allow us to conclude that there are differences in bicycle sales among the stores using the three catalogues? Use α =.0. Monthly Bicycle Sales Catalogue Month 3 March 7 8 4 April 5 9 0 May 0 7 88 June 7 95 4 July 83 6 303 August 85 5 6 September 90 38 85 October 5 4 SST(Catalogues) = 53 SSB(Months) = 05,35 SS(Total) = 4,08 H 0 : H : Rejection region: Value of the test statistic: 88
ANOVA Table Source d.f. Sums of Squares Mean Squares F-ratio Conclusion: 5.5 Two-Factor Analysis of Variance: Independent Samples When the treatments are defined by two factors, we are often interested in determining whether treatment differences (if they exist) are due to factor A, factor B, or interaction between the two factors. The answer is provided by applying the two-factor analysis of variance. The format of the ANOVA table appears below. Source d.f. Sums of Squares Mean Squares F-Ratios Factor A a SS(A) MS(A) = SS(A) (a ) Factor B b SS(B) SS( B) MS(B) = (b ) Interaction (a )(b ) SS(AB) SS(AB) MS(AB) = (a )(b ) Error n ab SSE SSE MSE = (n ab) F = MS(A) MSE F = MS(B) MSE F = MS(AB) MSE Total n SS(Total) Example 5.3 The dean of a business school was examining the factors that lead to success in the MBA program. She felt that the type of degree and whether the student had previous wor experience were liely to be critical factors. To test her beliefs she too a random sample of 00 students. For each she recorded their MBA grade point average (response variable), the type of undergraduate degree (B.A., B.Sc., B.Eng, B.B.A.) (factor A), and whether the student had wor experience (factor B). The sums of squares are listed below. What conclusions can be reached from these results? 89
SS(Total )= 68 SS(A) = 7 SS(B) = 6 SS(AB) = 68 Solution The number of levels of factor A (type of degree) is 4; the number of factors of level B (previous wor experience yes or no) is. The ANOVA table is shown below. Source d.f. Sums of Squares Mean Squares F-Ratios Factor A 3 7 9.0.60 Factor B 6 6.0.07 Interaction 3 68.67 4.03 Error 9 57 5.6 Total 99 68 We first test for interaction. At the 5% significance level the rejection region is F > F α, (a )(b ), n ab = F.05, 3, 9.68 Since F = 4.03 we conclude that there is evidence of an interaction effect. We do not conduct tests to determine if factors A and B are significant. EXERCISES 5.0 Complete the following ANOVA table. Source d.f. Sums of Squares Mean Squares F-Ratios Factor A 4 80 Factor B 3 450 Interaction 40 Error 330 Total 30 00 5. Refer to Exercise 5.0. Test to determine whether there is evidence of an interaction effect. Use α =.05. 90
5. Refer to Exercises 5.0 and 5.. Test to determine whether Factors A and B are significant. Use α =.05. 5.6 Operations Management Application: Finding and Reducing Variation In this section we presented a common application of the analysis of variance. We also discussed the Taguchi method, which involves designing experiments to produce operations methods that lead to improved quality. 5.7 Multiple Comparisons If we wish to determine which treatment means differ, we can use one of several multiple comparison procedures. In this section we introduced Fisher's least significant difference (LSD) method, the Bonferroni adjustment to LSD, and Tuey s multiple comparison method. These techniques are applied when the samples are independent, the sample sizes are equal, and all conditions required to perform the F-test of the analysis of variance are satisfied. Fisher's Least Significant Difference Method If the analysis of variance reveals that there is evidence that at least two means differ, we can determine which ones differ by computing the difference between each pair of means and comparing the difference to LSD. For the means of sample i and sample j we define the least significant difference as LSD = t α / MSE ni + n j The degrees of freedom are n -. The value of αis the same as used in the analysis of variance, often 5%. The problem with this approach is that we increase the probability of concluding that some differences 9
exit when in fact they do not. We can lower that probability by using the Bonferonni adjustment which sets the significance level as α/c where C = (-)/. Tuey's Multiple Comparison Method where The ey statistic is ω = q α (,v ) MSE / n g = number of samples v = number of degrees of freedom associated with MSE (v = n ) n g = number of observations in each sample ( n g = n = n =... = n ) q α (, v) = critical value of the studentized range (from Table 7 in Appendix B) Example 5.4 Refer to Example 5.. Apply the LSD method, LSD method with the Bonferroni adjustment, and Tuey's multiple comparison method to determine which dis drives are different at the 5% significance level? Solution The sample means are x = 94.8 x = 3.4 x 3 = 6.0 x 4 = 07.6 The sample sizes are equal to 5 and MSE = 3.95 The least significant difference is defined as LSD = t α / MSE ni + n j 9
The number of degrees of freedom are n - = 0-4 = 6. With a 5% significance level we have t Thus, α /, n =.05, 6 t =.0 LSD =. 0 3. 95 + = 9.6 5 5 The least significant difference using the Bonferroni adjustment is calculated by altering the significance level. Because there are 4(3)/ = 6 possible pairs of means we divide 5% by 6. We used Excel to find t Thus, t = 3.00 α /, n =.0047, 6 LSD = 3. 00 3. 95 + = 7.85 5 5 The critical value of Tuey s multiple comparison method is ω = q α (, v) MSE / n g = q.05 ( 4, 6 ) 3.95 / 5 = (4.05)(6.54) = 6.49 Putting the sample means in ascending order, we find x x 4 x x 3 94.8 07.6 3.4 6.0 The only difference that exceeds LSD = 9.6, LSD (Bonferroni) = 7.85, and ω = 6.49 is x 3 x = 3. Therefore, there is sufficient evidence at the 5% significance level to conclude that µ and µ 3 differ no matter which method is used. There is not enough evidence to indicate that there are any other differences among the population means. 93
EXERCISES 5.3 Use the least significant difference method to determine which means differ. Use α =.05. = 4 n g = 7 MSE = 5 x = 0 x = 9 x 3 = 53 x 4 = 54 5.4 Refer to Exercise 5.3 Use the Bonferroni adjustment to determine which means differ. 5.5 Refer to Exercise 5.3 Use Tuey s multiple comparison method to determine which means differ. 5.8 Bartlett's Test Bartlett's test is a diagnostic tool that is used to test one of the requirements of the analysis of variance. It tests the quality of variance. The test statistic is where B = ( n ) ln(mse ) ( n )ln(s C i= C = + 3( ) n n i i = i i ) 94
The statistic is chi-squared distributed with - degrees of freedom. Example 5.5 Refer to Example 5.. Is there enough evidence at the 5% significance level to infer that the population variances are unequal? Solution We test the following hypotheses. H 0 : σ = σ = σ 3 = σ 4 H : At least one variance differs C = + 3( ) n n i i = 4 = + 3(4 ) 5 0 i = 4 =.04 B = ( n ) ln( MSE) ( n i ) ln( s i C i=.04 ) = [(0 4) ln(3.95) {(5 ) ln(0.7) + (5 ) ln(5.8) + (5 ) ln(7.0) + (5 ) ln(30.3)}] =.04 [6(5.366)-{4(4.707)+4(5.533)+4(5.47)+4(5.769)} =.6 The rejection region is >, χ.05, 3 B χ α = = 7.8473 Conclusion: There is not enough evidence to infer that any variances differ. The equality of variances requirement appears to be satisfied. EXERCISES 5.6 Refer to Example 5.. Apply Bartlett's test at the 5% significance level to determine whether the population variances differ. H 0 : H : 95
Test statistic: Rejection region: Value of the test statistic: 5.7 Refer to Exercise 5.5. Apply Bartlett's test at the 5% significance level to determine whether the population variances differ. H 0 : H : Test statis tic: Rejection region: Value of the test statistic: 96