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12-1 Chapter Twelve McGraw-Hill/Irwin © 2005 The McGraw-Hill Companies, Inc., All Rights Reserved.

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12-2 Chapter Twelve Analysis of Variance GOALS When you have completed this chapter, you will be able to: ONE List the characteristics of the F distribution. TWO Conduct a test of hypothesis to determine whether the variances of two populations are equal. THREE Discuss the general idea of analysis of variance. FOUR Organize data into a one-way and a two-way ANOVA table. Goals

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12-3 Chapter Twelve continued Analysis of Variance GOALS When you have completed this chapter, you will be able to: FIVE Define and understand the terms treatments and blocks. SIX Conduct a test of hypothesis among three or more treatment means. SEVEN Develop confidence intervals for the difference between treatment means. EIGHT Conduct a test of hypothesis to determine if there is a difference among block means. Goals

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12-4 Characteristics of F- Distribution Its values range from 0 to . As F the curve approaches the X- axis but never touches it. Characteristics of the F-Distribution There is a “family” of F Distributions. Each member of the family is determined by two parameters: the numerator degrees of freedom and the denominator degrees of freedom. F cannot be negative, and it is a continuous distribution. The F distribution is positively skewed.

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12-5 Test for Equal Variances of Two Populations For the two tail test, the test statistic is given by Test for Equal Variances of Two Populations and are the sample variances for the two samples. The larger s is placed in the denominator. The degrees of freedom are n 1 -1 for the numerator and n 2 -1 for the denominator. The null hypothesis is rejected if the computed value of the test statistic is greater than the critical value.

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12-6 Example 1 The mean rate of return on a sample of 8 utility stocks was 10.9 percent with a standard deviation of 3.5 percent. At the.05 significance level, can Colin conclude that there is more variation in the software stocks? Colin, a stockbroker at Critical Securities, reported that the mean rate of return on a sample of 10 internet stocks was 12.6 percent with a standard deviation of 3.9 percent.

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12-7 Example 1 continued Step 3: The test statistic is the F distribution. Step 1: The hypotheses are Step 2: The significance level is.05.

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12-8 Example 1 continued Step 5: The value of F is computed as follows. The p(F>1.2416) is.3965. H 0 is not rejected. There is insufficient evidence to show more variation in the internet stocks. Step 4: H 0 is rejected if F>3.68 or if p <.05. The degrees of freedom are n 1 -1 or 9 in the numerator and n 1 -1 or 7 in the denominator.

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12-9 The ANOVA Test of Means The null and alternate hypotheses for four sample means is given as: H o : 1 = 2 = 3 = 4 H 1 : 1 = 2 = 3 = 4 The ANOVA Test of Means The F distribution is also used for testing whether two or more sample means came from the same or equal populations. This technique is called analysis of variance or ANOVA

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12-10 The populations have equal standard deviations. ANOVA requires the following conditions Underlying assumptions for ANOVA The sampled populations follow the normal distribution. The samples are independent

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12-11 F = Estimate of the population variance based on the differences among the sample means Estimate of the population variance based on the variation within the samples ANOVA Test of Means Degrees of freedom for the F statistic in ANOVA If there are k populations being sampled, the numerator degrees of freedom is k – 1 If there are a total of n observations the denominator degrees of freedom is n – k.

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12-12 In the following table, i stands for the i th observation x G is the overall or grand mean k is the number of treatment groups ANOVA Test of Means Total Variation Treatment Variation Random Variation ANOVA divides the Total Variation into the variation due to the treatment, Treatment Variation, and to the error component, Random Variation.

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12-13 ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square F Treatments (k) SST k n k (X k -X G ) 2 k-1SST/(k-1) =MSTMST MSE ErrorSSE i k (X i.k -X k ) 2 n-kSSE/(n-k) =MSE TotalTSS i (X i -X G ) 2 n-1 Anova Table Treatment variation Random variation Total variation

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12-14 Rosenbaum Restaurants specialize in meals for families. Katy Polsby, President, recently developed a new meat loaf dinner. Before making it a part of the regular menu she decides to test it in several of her restaurants. Example 2 She would like to know if there is a difference in the mean number of dinners sold per day at the Anyor, Loris, and Lander restaurants. Use the.05 significance level.

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12-15 Number of Dinners Sold by Restaurant Restaurant Day AynorLorisLander Day 1 Day 2 Day 3 Day 4 Day 5 13 12 14 12 10 12 13 11 18 16 17 Example 2 continued

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12-16 Step One: State the null hypothesis and the alternate hypothesis. H o : Aynor = Loris = Landis H 1 : Aynor = Loris = Landis Step Two: Select the level of significance. This is given in the problem statement as.05. Step Three: Determine the test statistic. The test statistic follows the F distribution. Example 2 continued

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12-17 Step Five: Select the sample, perform the calculations, and make a decision. Step Four: Formulate the decision rule. The numerator degrees of freedom, k-1, equal 3-1 or 2. The denominator degrees of freedom, n-k, equal 13-3 or 10. The value of F at 2 and 10 degrees of freedom is 4.10. Thus, H 0 is rejected if F>4.10 or p< of.05. Example 2 continued Using the data provided, the ANOVA calculations follow.

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12-18 Anyor #sold SS(Anyor)Loris #sold SS(Loris)Lander #sold SS(Lander) 13 12 14 12 (13-12.75) 2 (12-12.75) 2 (14-12.75) 2 (12-12.75) 2 2.75 10 12 13 11 (10-11.5) 2 (12-11.5) 2 (13-11.5) 2 (11-11.5) 2 5 18 16 17 (18-17) 2 (16-17) 2 (17-17) 2 2 XkXk 12.7511.517 SSE: 2.75 + 5 + 2 = 9.75 X G : 14.00 Computation of SSE i k (X i.k -X k ) 2

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12-19 Anyor #sold TSS(Anyor)Loris #sold TSS(Loris)Lander #sold TSS(Lander) 13 12 14 12 (13-14) 2 (12-14) 2 (14-14) 2 (12-14) 2 9.00 10 12 13 11 (10-14) 2 (12-14) 2 (13-14) 2 (11-14) 2 30 18 16 17 (18-14) 2 (16-14) 2 (17-14) 2 47 TSS: 9.00 + 30 + 47 = 86.00 SSE: 9.75 X G : 14.00 Computation of TSS i (X i -X G ) 2 Example 2 continued Computation of TSS

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12-20 Computation of SST k n k (X k -X G ) 2 RestaurantXTXT SST Anyor Loris Lander 12.75 11.50 17.00 4(12.75-14) 2 4(11.50-14) 2 5(17.00-14) 2 76.25 Shortcut: SST = TSS – SSE = 86 – 9.75 = 76.25 Example 2 continued Computation of SST

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12-21 ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square F Treatments76.253-1 =2 76.25/2 =38.125 38.125.975 = 39.103 Error 9.7513-3 =10 9.75/10 =.975 Total86.0013-1 =12 Example 2 continued

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12-22 Example 2 continued The ANOVA tables on the next two slides are from the Minitab and EXCEL systems. The p(F> 39.103) is.000018. The mean number of meals sold at the three locations is not the same. Since an F of 39.103 > the critical F of 4.10, the p of.000018 < a of.05, the decision is to reject the null hypothesis and conclude that At least two of the treatment means are not the same.

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12-23 Example 2 continued Analysis of Variance Source DF SS MS F P Factor 2 76.250 38.125 39.10 0.000 Error 10 9.750 0.975 Total 12 86.000 Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ---------+---------+---------+--- ---- Aynor 4 12.750 0.957 (---*---) Loris 4 11.500 1.291 (---*---) Lander 5 17.000 0.707 (---*---) ---------+---------+---------+--- ---- Pooled StDev = 0.987 12.5 15.0 17.5

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12-24 Example 2 continued

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12-25 Inferences About Treatment Means One of the simplest procedures is through the use of confidence intervals around the difference in treatment means. When I reject the null hypothesis that the means are equal, I want to know which treatment means differ.

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12-26 Confidence Interval for the Difference Between Two Means If the confidence interval around the difference in treatment means includes zero, there is not a difference between the treatment means. t is obtained from the t table with degrees of freedom (n - k). MSE = [SSE/(n - k)]

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12-27 EXAMPLE 3 95% confidence interval for the difference in the mean number of meat loaf dinners sold in Lander and Aynor Can Katy conclude that there is a difference between the two restaurants?

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12-28 Example 3continued The mean number of meals sold in Aynor is different from Lander. Because zero is not in the interval, we conclude that this pair of means differs.

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12-29 Two-Factor ANOVA SSB = r (X b – X G ) 2 where r is the number of blocks X b is the sample mean of block b X G is the overall or grand mean In the following ANOVA table, all sums of squares are computed as before, with the addition of the SSB. Sometimes there are other causes of variation. For the two- factor ANOVA we test whether there is a significant difference between the treatment effect and whether there is a difference in the blocking effect (a second treatment variable).

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12-30 ANOVA Table Source of Variation Sum of SquaresDegrees of Freedom Mean Square F Treatments (k) SST k-1SST/(k-1) =MSTMST MSE MSB MSE Blocks (b) SSBb-1SSB/(b-1) =MSB ErrorSSE (TSS – SST –SSB) (k-1)(b-1)SSE/(n-k) =MSE TotalTSSn-1 Two factor ANOVA table

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12-31 Example 4 At the.05 significance level, can we conclude there is a difference in the mean production by shift and in the mean production by employee? The Bieber Manufacturing Co. operates 24 hours a day, five days a week. The workers rotate shifts each week. Todd Bieber, the owner, 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.

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12-32 Example 4 continued

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12-33 Step 5: Perform the calculations and make a decision. Step 4: Formulate the decision rule. Ho is rejected if F > 4.46, the degrees of freedom are 2 and 8, or if p <.05. Example 4 continued Step 1: State the null hypothesis and the alternate hypothesis. H 1 : Not all means are equal. Treatment Effect Step 2: Select the level of significance. Given as.05. Step 3: Determine the test statistic. The test statistic follows the F distribution.

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12-34 Step 1: State the null hypothesis and the alternate hypothesis. H 1 : Not all means are equal. Step 2: Select the level of significance. Given as =.05. Step 3: Determine the test statistic. The test statistic follows the F distribution. Example 4 continued Block Effect Step 4: Formulate the decision rule. H 0 is rejected if F>3.84, df =(4,8) or if p <.05. Step 5: Perform the calculations and make a decision.

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12-35 Block Sums of Squares Effects of time of day and worker on productivity DayEveningNightEmployee xSSB McCartney 31253530.333(30.33-28.87) 2 = 6.42 Neary 33263330.673(30.67-28.87) 2 = 9.68 Schoen 28243027.333(27.33-28.87) 2 7.08 Thompson 30292829.003(29.00-28.87) 2.09 Wagner28262727.003(27.00-28.87) 2 10.49 SSB = 6.42 + 9.68 + 7.08 +.05 + 10.49= 33.73 Note: x G = 28.87

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12-36 Example 4 continued Compute the remaining sums of squares as before: TSS = 139.73 SST = 62.53 SSE = 43.47 (139.73-62.53-33.73) df(block) = 4 (b-1) df(treatment) = 2 (k-1) df(error)=8 (k-1)(b-1)

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12-37 Example 4 continued ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square F Treatments (k) 62.53 262.53/2 =31.275 31.27/5.43 = 5.75 Blocks (b) 33.73433.73/4 =8.43 8.43/5.43 =1.55 Error43.47843.47/8 =5.43 Total139.7314

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12-38 Example 4 continued Block Effect Since the computed F of 1.55 of.05, H 0 is not rejected since there is no significant difference in the average number of units produced for the different employees. Treatment Effect Since the computed F of 5.75 > the critical F of 4.10, the p of.03 < of.05, H 0 is rejected. There is a difference in the mean number of units produced for the different time periods.

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12-39 Example 4 continued Two-way ANOVA: Units versus Worker, Shift Analysis of Variance for Units Source DF SS MS F P Worker 4 33.73 8.43 1.55 0.276 Shift 2 62.53 31.27 5.75 0.028 Error 8 43.47 5.43 Total 14 139.73 Minitab output

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12-40 Output Using EXCEL Example 4 continued

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