Chapter 15 Textbook Study Guide

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Department
Operations Management and Information System
Course
OMIS 2010
Professor
Alan Marshall
Semester
Fall

Description
Chapter 15: Analysis of Variance 15.1 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. 2. 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 Tukeys multiple comparison procedure. 5. How to conduct Bartlett's test. 15.2 Single-Factor (One-Way) Analysis of Variance: Independent Samples When the k 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 k nj 2 SS(Total) = (xij x) j 1i1 k SST = nj(xj x) 2 n 1x 1 x) n (x2 2) n (x k k x)2 1 k nj SSE = (xij x j2 j 1i1 n n n 1 2 k = (xi1 x1) (xi2 x2) (x ik xk)2 1 i 1 i 1 = (n 1)s2 (n 1)s 2 (n 1)s2 1 1 2 2 k k MST = SST k 1 SSE MSE = n k 179 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 k 1 SST MST F Error n k SSE MSE Total n 1 SS(Total) Example 15.1 A major computer manufacturer has received numerous complaints concerning the short life of its disk drives. Most need repair within two years. Since the cost of repairs often exceeds the cost of a new disk drive, the manufacturer is concerned. In his search for a better disk drive, he finds three new prod- ucts. He decides to test these three plus his current disk drive to determine if differences in lifetimes exist among the products. He takes a random sample of five disk drives of each type and links it with a computer. The number of weeks until the drive breaks 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 disk drives? Type 1 Type 2 Type 3 Current Product 78 125 143 101 92 110 125 96 101 116 133 88 105 88 108 125 98 128 121 128 x1 94.8 x2 113.4 x3 126.0 x4 107.6 s2 110.7 s2 252.8 s2 172.0 s 2 320.3 1 2 3 4 Solution The problem objective is to compare the populations of lifetimes of the four disk 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 0:1 2 3 4 180 H 1 At least two means differ. The rejection region is F > , k 1, n kF .05, 3, 163.24 The test statistic is computed as follows: n 1 = n2= n 3 n 4 5 x = 5(94.80) 5(113.4) 5(126.0) 5(107.6) 2,209 110.45 20 20 SST = 5(94.8 110.45) 2+ 5(113.4 110.45) + 5(126.0 110.45) + 5(107.6 110.45)2 = 2,517.75 SSE = 4(110.7) + 4(252.8) + 4(172.0) + 4(320.3) = 3,423.2 MST = 2,517.75 839.25 3 3,423.2 MSE = 16 213.95 839.25 F = 3.92 213.95 The complete ANOVA table follows. Source d.f. Sums of Squares Mean Squares F-ratio Treatments 3 2,517.75 839.25 3.92 Error 16 3,423.2 213.95 Total 19 5,940.95 Since the F-ratio (3.92) exceedsF, k 1, n k24), we reject 0 and conclude that at least two means differ. 181 EXERCISES 15.1 Complete the ANOVA table and F-test with = .05. Source d.f. Sums of Squares Mean Squares F-ratio Treatments 5 Error 3,000 Total 20 3,500 15.2 Develop the ANOVA table from the following information and perform the F-test with = .01. SS(Total) = 150 SST= 90 SSE= 60 k = 5 n1= 7 n2= 5 n3= 8 n4= 9 n5= 4 ANOVA Table Source d.f. Sums of Squares Mean Squares F-ratio 182
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