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Chapter15

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  • Transcript

    • 1. 1 Chapter 15 The Analysis of Variance
    • 2. 2 A study was done on the survival time of patients with advanced cancer of the stomach, bronchus, colon, ovary or breast when treated with ascorbate1 . In this study, the authors wanted to determine if the survival times differ based on the affected organ. 1 Cameron, E. and Pauling, L. (1978) Supplemental ascorbate in the supportive treatment of cancer: re-evaluation of prolongation of survival time in terminal human cancer. Proceedings of the National Academy of Science, USA, 75, 4538-4542. A Problem
    • 3. 3 A comparative dotplot of the survival times is shown below. A Problem 3000200010000 Survival Time (in days) Dotplot for Survival Time Cancer Type Breast Bronchus Colon Ovary Stomach
    • 4. 4 H0: µstomach = µbronchus = µcolon = µovary= µbreast Ha: At least two of the µ’s are different A Problem The hypotheses used to answer the question of interest are The question is similar to ones encountered in chapter 11 where we looked at tests for the difference of means of two different variables. In this case we are interested in looking a more than two variable.
    • 5. 5 A single-factor analysis of variance (ANOVA) problems involves a comparison of k population or treatment means µ1, µ2, … , µk. The objective is to test the hypotheses: H0: µ1 = µ2 = µ3 = … = µk Ha: At least two of the µ’s are different Single-factor Analysis of Variance (ANOVA)
    • 6. 6 The analysis is based on k independently selected samples, one from each population or for each treatment. In the case of populations, a random sample from each population is selected independently of that from any other population. When comparing treatments, the experimental units (subjects or objects) that receive any particular treatment are chosen at random from those available for the experiment. Single-factor Analysis of Variance (ANOVA)
    • 7. 7 A comparison of treatments based on independently selected experimental units is often referred to as a completely randomized design. Single-factor Analysis of Variance (ANOVA)
    • 8. 8 70 60 50 40 Fertilizer Yield Dotplots of Yield by Fertilizer (group means are indicated by lines) Type 1 Type 2 Type 3 Notice that in the above comparative dotplot, the differences in the treatment means is large relative to the variability within the samples. Single-factor Analysis of Variance (ANOVA)
    • 9. 9 Statistics Psychology Economics Business 85 75 65 SubjectPrice Dotplots of Price by Subject (group means are indicated by lines) Notice that in the above comparative dotplot, the differences in the treatment means is not easily understood relative to the sample variability. ANOVA techniques will allow us to determined if those differences are significant. Single-factor Analysis of Variance (ANOVA)
    • 10. 10 ANOVA Notation k = number of populations or treatments being compared Population or treatment 1 2 … k Population or treatment mean µ1 µ2 … µk Sample mean …1x 2x kx Population or treatment variance …2 1σ 2 2σ 2 kσ Sample variance … 2 1s 2 2s 2 ks Sample size n1 n2 … nk
    • 11. 11 N = n1 + n2 + … + nk (Total number of observations in the data set) ANOVA Notation T x grand mean N = = T = grand total = sum of all N observations 1 1 2 2 k kn x n x n x= + + +L
    • 12. 12 Assumptions for ANOVA 1. Each of the k populations or treatments, the response distribution is normal. 2. σ1 = σ2 = … = σk (The k normal distributions have identical standard deviations. 3. The observations in the sample from any particular one of the k populations or treatments are independent of one another. 4. When comparing population means, k random samples are selected independently of one another. When comparing treatment means, treatments are assigned at random to subjects or objects.
    • 13. 13 Definitions ( ) ( ) ( ) 2 2 2 1 1 2 2 k kSSTr n x x n x x n x x= − + − + + −L A measure of disparity among the sample means is the treatment sum of squares, denoted by SSTr is given by A measure of variation within the k samples, called error sum of squares and denoted by SSE is given by ( ) ( ) ( )2 2 2 1 1 2 2 k kSSE n 1 s n 1 s n 1 s= − + − + + −L
    • 14. 14 Definitions A mean square is a sum of squares divided by its df. In particular, The error df comes from adding the df’s associated with each of the sample variances: (n1 - 1) + (n2 - 1) + …+ (nk - 1) = n1 + n2 … + nk - 1 - 1 - … - 1 = N - k mean square for treatments = MSTr = SSTr k 1− mean square for error = MSE = SSE N k−
    • 15. 15 Example Three filling machines are used by a bottler to fill 12 oz cans of soda. In an attempt to determine if the three machines are filling the cans to the same (mean) level, independent samples of cans filled by each were selected and the amounts of soda in the cans measured. The samples are given below. Machine 1 12.033 11.985 12.009 12.009 12.033 12.025 12.054 12.050 Machine 2 12.031 11.985 11.998 11.992 11.985 12.027 11.987 Machine 3 12.034 12.021 12.038 12.058 12.001 12.020 12.029 12.011 12.021
    • 16. 16 Example ( ) ( ) ( ) 2 2 2 1 1 2 2 k k 2 2 2 SSTr n x x n x x n x x 8(0.0065833) 7(-0.0174524) 9(0.0077222) 0.000334672+0.00213210+0.00053669 0.00301552 = − + − + + − = + + = = L 1 1 1n 8, x 12.0248, s 0.02301= = = 3 3 3n 9, x 12.0259, s 0.01650= = = 2 2 2n 7, x 12.0007, s 0.01989= = = x 12.018167=
    • 17. 17 Example ( ) ( ) ( )2 2 2 1 1 2 2 k k 2 2 2 SSE n 1 s n 1 s n 1 s 7(0.0230078) 6(0.0198890) 8(0.01649579) 0.0037055 0.0023734 0.0021769 0.00825582 = − + − + + − = + + = + + = L 1 1 1n 8, x 12.0248, s 0.02301= = = 3 3 3n 9, x 12.0259, s 0.01650= = = 2 2 2n 7, x 12.0007, s 0.01989= = = x 12.018167=
    • 18. 18 Example SSTr k 1− mean square for treatments = MSTr = SSTr 0.00301552 MSTr 0.0015078 k 1 3 1 = = = − − mean square for error = MSE = SSE N k− SSE 0.0082579 MSE 0.00039313 N k 24 3 = = = − − 1 1 1n 8, x 12.0248, s 0.02301= = = 3 3 3n 9, x 12.0259, s 0.01650= = = 2 2 2n 7, x 12.0007, s 0.01989= = = x 12.018167=
    • 19. 19 Comments Both MSTr and MSE are quantities that are calculated from sample data. As such, both MSTr and MSE are statistics and have sampling distributions. More specifically, when H0 is true, µMSTr = µMSE. However, when H0 is false, µMSTr = µMSE and the greater the differences among the µ’s, the larger µMSTr will be relative to µMSE.
    • 20. 20 The Single-Factor ANOVA F Test Null hypothesis: H0: µ1 = µ2 = µ3 = … = µk Alternate hypothesis: At least two of the µ’s are different Test Statistic: MSTr F MSE =
    • 21. 21 The Single-Factor ANOVA F Test When H0 is true and the ANOVA assumptions are reasonable, F has an F distribution with df1 = k - 1 and df2 = N - k. Values of F more contradictory to H0 than what was calculated are values even farther out in the upper tail, so the P-value is the area captured in the upper tail of the corresponding F curve.
    • 22. 22 Example Consider the earlier example involving the three filling machines. Machine 1 12.033 11.985 12.009 12.009 12.033 12.025 12.054 12.050 Machine 2 12.031 11.985 11.998 11.992 11.985 12.027 11.987 Machine 3 12.034 12.021 12.038 12.058 12.001 12.020 12.029 12.011 12.021
    • 23. 23 Example SSTr 0.00301552= SSE 0.00825582= MSTr 0.0015078= MSE 0.00039313= 1 1 1n 8, x 12.0248, s 0.02301= = = 3 3 3n 9, x 12.0259, s 0.01650= = = 2 2 2n 7, x 12.0007, s 0.01989= = = x 12.018167=
    • 24. 24 Example 1. Let µ1, µ2 and µ3 denote the true mean amount of soda in the cans filled by machines 1, 2 and 3, respectively. 2. H0: µ1 = µ2 = µ3 3. Ha: At least two among are µ1, µ2 and µ3 different 4. Significance level: α = 0.01 5. Test statistic: MSTr F MSE =
    • 25. 25 Example 6. Looking at the comparative dotplot, it seems reasonable to assume that the distributions have the same σ’s. We shall look at the normality assumption on the next slide.* 12.0612.0512.0412.0312.0212.0112.0011.99 Fill Dotplot for Fill Machine Machine 1 Machine 2 Machine 3 *When the sample sizes are large, we can make judgments about both the equality of the standard deviations and the normality of the underlying populations with a comparative boxplot.
    • 26. 26 Example 6. Looking at normal plots for the samples, it certainly appears reasonable to assume that the samples from Machine’s 1 and 2 are samples from normal distributions. Unfortunately, the normal plot for the sample from Machine 2 does not appear to be a sample from a normal population. So as to have a computational example, we shall continue and finish the test, treating the result with a “grain of salt.” P-Value: 0.692 A-Squared: 0.235 Anderson-Darling Normality Test N: 8 StDev: 0.0230078 Average: 12.0248 12.05512.04512.03512.02512.01512.00511.99511.985 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Machine 1 Normal Probability Plot P-Value: 0.031 A-Squared: 0.729 Anderson-Darling Normality Test N: 7 StDev: 0.0198890 Average: 12.0007 12.0312.0212.0112.0011.99 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Machine 2 Normal Probability Plot P-Value: 0.702 A-Squared: 0.237 Anderson-Darling Normality Te N: 9 StDev: 0.0164958 Average: 12.0259 12.0612.0512.0412.0312.0212.0112.00 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Machine 3 Normal Probability Plot
    • 27. 27 Example 7. Computation: SSTr 0.00301552= SSE 0.00825582= MSTr 0.0015078= MSE 0.00039313= 1 1 1n 8, x 12.0248, s 0.02301= = = 3 3 3n 9, x 12.0259, s 0.01650= = = 2 2 2n 7, x 12.0007, s 0.01989= = = x 12.018167= 1 2 3N n n n 8 7 9 24, k 3= + + = + + = = 1 2 MSTr 0.0015078 F 3.835 MSE 0.00039313 df treatment df k 1 3 1 2 df error df N k 24 3 21 = = = = = − = − = = = − = − =
    • 28. 28 Example 8. P-value: 3.835 dfden / dfnum α 2 21 0.100 2.57 0.050 3.47 0.025 4.42 0.010 5.78 0.001 9.77 From the F table with numerator df1 = 2 and denominator df2 = 21 we can see that 0.025 < P-value < 0.05 (Minitab reports this value to be 0.038 1 2 MSTr 0.0015078 F 3.835 MSE 0.00039313 df treatment df k 1 3 1 2 df error df N k 24 3 21 = = = = = − = − = = = − = − = Recall 1 2 MSTr 0.0015078 F 3.835 MSE 0.00039313 df treatment df k 1 3 1 2 df error df N k 24 3 21 = = = = = − = − = = = − = − = Recall
    • 29. 29 Example 9. Conclusion: Since P-value > α = 0.01, we fail to reject H0. We are unable to show that the mean fills are different and conclude that the differences in the mean fills of the machines show no statistically significant differences.
    • 30. 30 Total Sum of Squares The relationship between the three sums of squares is SSTo = SSTr + SSE which is often called the fundamental identity for single-factor ANOVA. Informally this relation is expressed as Total variation = Explained variation + Unexplained variation Total sum of squares, denoted by SSTo, is given by with associated df = N - 1. all N obs. 2 SSTo (x x)= −∑
    • 31. 31 Single-factor ANOVA Table The following is a fairly standard way of presenting the important calculations from an single-factor ANOVA. The output from most statistical packages will contain an additional column giving the P-value.
    • 32. 32 Single-factor ANOVA Table The ANOVA table supplied by Minitab One-way ANOVA: Fills versus Machine Analysis of Variance for Fills Source DF SS MS F P Machine 2 0.003016 0.001508 3.84 0.038 Error 21 0.008256 0.000393 Total 23 0.011271
    • 33. 33 Another Example A food company produces 4 different brands of salsa. In order to determine if the four brands had the same sodium levels, 10 bottles of each Brand were randomly (and independently) obtained and the sodium content in milligrams (mg) per tablespoon serving was measured. The sample data are given on the next slide. Use the data to perform an appropriate hypothesis test at the 0.05 level of significance.
    • 34. 34 Another Example Brand A 43.85 44.30 45.69 47.13 43.35 45.59 45.92 44.89 43.69 44.59 Brand B 42.50 45.63 44.98 43.74 44.95 42.99 44.95 45.93 45.54 44.70 Brand C 45.84 48.74 49.25 47.30 46.41 46.35 46.31 46.93 48.30 45.13 Brand D 43.81 44.77 43.52 44.63 44.84 46.30 46.68 47.55 44.24 45.46
    • 35. 35 Another Example 1. Let µ1, µ2 , µ3 and µ4 denote the true mean sodium content per tablespoon in each of the brands respectively. 2. H0: µ1 = µ2 = µ3 = µ4 3. Ha: At least two among are µ1, µ2, µ3 and µ4 are different 4. Significance level: α = 0.05 5. Test statistic: MSTr F MSE =
    • 36. 36 6. Looking at the following comparative boxplot, it seems reasonable to assume that the distributions have the equal σ’s as well as the samples being samples from normal distributions. Another Example BrandD BrandC BrandB BrandA 49 48 47 46 45 44 43 42 Boxplots of Brand A - Brand D (means are indicated by solid circles)
    • 37. 37 Example Treatment df = k - 1 = 4 - 1 = 3 7. Computation: Brand k si Brand A 10 44.900 1.180 Brand B 10 44.591 1.148 Brand C 10 47.056 1.331 Brand D 10 45.180 1.304 xi = − + − + − + − = − + − + − + − = 2 2 2 2 1 1 2 2 3 3 4 4 2 2 2 2 SSTr n (x x) n (x x) n (x x) n (x x) 10(44.900 45.432) 10(44.591 45.432) 10(47.056 45.432) 10(45.180 45.432) 36.912 =x 45.432
    • 38. 38 Example 7. Computation (continued): Error df = N - k = 40 - 4 = 36 ( ) ( ) ( ) ( )= − + − + − + − = + + + = 2 2 2 2 1 1 2 2 3 3 4 4 2 2 2 2 SSE n 1 s n 1 s n 1 s n 1 s 9(1.180) 9(1.148) 9(1.331) 9(1.304) 55.627 = = = = =SSTr SSE SSTr 36.912 MSTr 12.304df 3F 7.963 SSE 55.627MSE 1.5452 df 36
    • 39. 39 Example 8.P-value: F = 7.96 with dfnumerator= 3 and dfdenominator= 36 Using df = 30 we find P-value < 0.001 7.96
    • 40. 40 Example 9. Conclusion: Since P-value < α = 0.001, we reject H0. We can conclude that the mean sodium content is different for at least two of the Brands. We need to learn how to interpret the results and will spend some time on developing techniques to describe the differences among the µ’s.
    • 41. 41 Multiple Comparisons A multiple comparison procedure is a method for identifying differences among the µ’s once the hypothesis of overall equality (H0) has been rejected. The technique we will present is based on computing confidence intervals for difference of means for the pairs. Specifically, if k populations or treatments are studied, we would create k(k-1)/2 differences. (i.e., with 3 treatments one would generate confidence intervals for µ1 - µ2, µ1 - µ3 and µ2 - µ3.) Notice that it is only necessary to look at a confidence interval for µ1 - µ2 to see if µ1 and µ2 differ.
    • 42. 42 The Tukey-Kramer Multiple Comparison Procedure When there are k populations or treatments being compared, k(k-1)/2 confidence intervals must be computed. If we denote the relevant Studentized range critical value by q, the intervals are as follows: For µi - µj: Two means are judged to differ significantly if the corresponding interval does not include zero. i j i j MSE 1 1 ( ) q 2 n n   µ − µ ± +   
    • 43. 43 The Tukey-Kramer Multiple Comparison Procedure When all of the sample sizes are the same, we denote n by n = n1 = n2 = n3 = … = nk, and the confidence intervals (for µi - µj) simplify to i j MSE ( ) q n µ − µ ±
    • 44. 44 Example (continued) Continuing with example dealing with the sodium content for the four Brands of salsa we shall compute the Tukey-Kramer 95% Tukey- Kramer confidence intervals for µA - µB, µA - µC, µA - µD, µB - µC, µB - µD and µC - µD. = = = = = = =   =  ÷   = = A B C D 55.627 MSE 1.5452, n n n n n 10 36 Interpolating from the table q 3.81 i.e. 60% of the way from 3.85 to 3.79 MSE 1.5452 q 3.81 1.498 n 10
    • 45. 45 Example (continued) Difference 95% Confidence Limits 95% Confidence Interval µA - µB 0.309 ± 1.498 (-1.189, 1.807) µA - µC -2.156 ± 1.498 (-3.654, -0.658) µA - µD -0.280 ± 1.498 (-1.778, 1.218) µB - µC -2.465 ± 1.498 (-3.963, -0.967) µB - µD -0.589 ± 1.498 (-2.087, 0.909) µC - µD 1.876 ± 1.498 (0.378, 3.374) Notice that the confidence intervals for µA – µB, µA – µC and µC – µD do not contain 0 so we can infer that the mean sodium content for Brands C is different from Brands A, B and D.
    • 46. 46 Example (continued) We also illustrate the differences with the following listing of the sample means in increasing order with lines underneath those blocks of means that are indistinguishable. Brand B Brand A Brand D Brand C 44.591 44.900 45.180 47.056 Notice that the confidence interval for µA – µC, µB – µC, and µC – µD do not contain 0 so we can infer that the mean sodium content for Brand C and all others differ.
    • 47. 47 Minitab Output for Example One-way ANOVA: Sodium versus Brand Analysis of Variance for Sodium Source DF SS MS F P Brand 3 36.91 12.30 7.96 0.000 Error 36 55.63 1.55 Total 39 92.54 Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ------+---------+---------+---------+ Brand A 10 44.900 1.180 (-----*------) Brand B 10 44.591 1.148 (------*-----) Brand C 10 47.056 1.331 (------*------) Brand D 10 45.180 1.304 (------*-----) ------+---------+---------+---------+ Pooled StDev = 1.243 44.4 45.6 46.8 48.0
    • 48. 48 Minitab Output for Example Tukey's pairwise comparisons Family error rate = 0.0500 Individual error rate = 0.0107 Critical value = 3.81 Intervals for (column level mean) - (row level mean) Brand A Brand B Brand C Brand B -1.189 1.807 Brand C -3.654 -3.963 -0.658 -0.967 Brand D -1.778 -2.087 0.378 1.218 0.909 3.374
    • 49. 49 Simultaneous Confidence Level The Tukey-Kramer intervals are created in a manner that controls the simultaneous confidence level. For example at the 95% level, if the procedure is used repeatedly on many different data sets, in the long run only about 5% of the time would at least one of the intervals not include that value of what it is estimating. We then talk about the family error rate being 5% which is the maximum probability of one or more of the confidence intervals of the differences of mean not containing the true difference of mean.
    • 50. 50 Randomized Block Experiment Suppose that experimental units (individuals or objects to which the treatments are applied) are first separated into groups consisting of k units in such a way that the units within each group are as similar as possible. Within any particular group, the treatments are then randomly allocated so that each unit in a group receives a different treatment. The groups are often called blocks and the experimental design is referred to as a randomized block design. This slide as well as all remaining slides in this show refer to materials that are available on the Testbook website.
    • 51. 51 Example When choosing a variety of melon to plant, one thing that a farmer might be interested in is the length of time (in days) for the variety to bear harvestable fruit. Since the growing conditions (soil, temperature, humidity) also affect this, a farmer might experiment with three hybrid melons (denoted hybrid A, hybrid B and hybrid C) by taking each of the four fields that he wants to use for growing melons and subdividing each field into 3 subplots (1, 2 and 3) and then planting each hybrid in one subplot of each field. The blocks are the fields and the treatments are the hybrid that is planted. The question of interest would be “Are the mean times to bring harvestable fruit the same for all three hybrids?”
    • 52. 52 Assumptions and Hypotheses The single observation made on any particular treatment in a given block is assumed to be selected from a normal distribution. The variance of this distribution is σ2 , the same for each block-treatment combinations. However, the mean value may depend separately both on the treatment applied and on the block. The hypotheses of interest are as follows: H0: The mean value does not depend on which treatment is applied Ha: The mean value does depend on which treatment is applied
    • 53. 53 Summary of the Randomized Block F Test Notation: Let k = number of treatments l = number of blocks = average of all observations in block I ib ix = average if all observations for treatment i = average of all kl observations in the experiment (the grand mean) x
    • 54. 54 Summary of the Randomized Block F Test Sums of squares and associated df’s are as follows. Sum of Squares Symbol df Formula Treatments SSTr k - 1 Blocks SSBl l - 1 Error SSE (k - 1)(l - 1) Total SSTo kl - 1 2 2 2 1 2 kSSTr l[(x x) (x x) ... (x x) ]= −+−++− SSE SSTo SSTr SSBl= − − 2 2 2 1 2 lSSBl k[(b x) (b x) ... (b x) ]= −+−++− ( ) 2 all x x x−∑
    • 55. 55 Summary of the Randomized Block F Test SSE is obtained by subtraction through the use of the fundamental identity SSTo = SSTr + SSBl + SSE The test is based on df1 = k - 1 and df2 = (k - 1)(l - 1) Test statistic: MSTr F MSE = where SSTr SSE MSTr and MSE k 1 (k 1)(l 1) = = − − −
    • 56. 56 The ANOVA Table for a Randomized Block Experiment Source of Variation df Sum of Squares Mean Square F Treatments k –1 SSTr SSTr MSTr k 1 = − MSTr F MSE = Blocks l -1 SSBl SSBl MSBl l 1 = − Error (k–1)(l–1) SSE SSE MSE (k 1)(l 1) = − − Total kl - 1 SSTo
    • 57. 57 Multiple Comparisons As before, in single-factor ANOVA, once H0 has been rejected, declare that treatments I and j differ significantly if the interval does not include zero, where q is based on a comparison of k treatments and error df = (k - 1)(l - 1). i j MSE ( ) q l µ − µ ±
    • 58. 58 Example (Food Prices) In an attempt to measure which of 3 grocery chains has the best overall prices, it was felt that there would be a great deal of variability of prices if items were randomly selected from each of the chains, so a randomized block experiment was devised to answer the question. A list of standard items was developed (typically a fairly large representative list would be used, but do to a problem with insufficient planning, only 7 items were left “in the shopping cart.” and the price recorded for each of these items in each of the stores.
    • 59. 59 Example (Food Prices) Because of the problem that the Blocking variable (the item) wasn’t set up with a well designed, representative sample of the items in a typical shopping basket, the results should be taken with a “grain of salt.” For the purposes of showing the calculations, we shall treat this as being the contents of a “representative” shopping basket. The data appear in the next slide along with the hypotheses.
    • 60. 60 Example (Food Prices) H0: µA = µB = µC Ha: At least two among are µA, µB and µC are different Product Store A Store B Store C Tide (100 oz liquid detergent) 6.39 5.59 5.24 1 lb Land O'Lakes Butter 3.99 3.49 2.98 1 dozen Large Grade AA eggs 1.49 1.49 0.72 Tropicana (no pulp, non-conc) OJ (64 oz) 3.99 2.99 2.50 2 Liter Diet Coke 1.39 1.50 1.04 1 loaf Wonderbread 2.09 2.09 1.43 18 oz jar Skippy Peanut Butter 2.49 2.49 1.77
    • 61. 61 Calculations Treatments: k 3= Blocks: l 7= 57.15 x 2.7214 21 = = 2 2 2 1 2 3 2 2 2 SSTr l[(x x) (x x) (x x) ] 7[(3.1186 2.7214) (2.8057 2.7214) (2.2400 2.7214) ] 7[0.15772 0.00710 0.23177] 7[0.39660] 2.7762 = − + − + − = − + − + − = + + = = SSTr 2.7762 MSTr 1.3881 k 1 3 1 = = = − −
    • 62. 62 Calculations 2 2 2 1 2 7 2 2 2 2 2 2 2 SSBl k[(b x) (b x) ... (b x) ] 3[(5.7400 2.7214) (3.4867 2.7214) (1.2333 2.7214) (3.1600 2.7214) (1.3100 2.7214) (1.8700 2.7214) (2.2500 2.7214) ] 3[9.1118+0.58559+2.21443+0.19234+1.9921+0. = − + − + + − = − + − + − + − + − + − + − = 72493+0.22224] 3[15.04344] 45.1303= = SSTr 2.7762 MSTr 1.3881 k 1 3 1 = = = − −
    • 63. 63 Calculations SSE SSTo SSTr SSBl 48.6356 2.7762 45.1303 0.72893 = − − = − − = SSE 0.72893 MSE (k 1)(l 1) (3 1)(7 1) 0.06074 = = − − − − = MSTr 1.3881 F 22.85 MSE 0.060744 = = = den num df (k 1)(l 1) (3 1)(7 1) 12 df k 1 3 1 2 = − − = − − = = − = − =
    • 64. 64 den num df (k 1)(l 1) (3 1)(7 1) 12 df k 1 3 1 2 = − − = − − = = − = − = Conclusions We can reject the hypothesis that the mean prices are the same in all three stores. The actual differences can be estimated with confidence intervals. MSTr 1.3881 F 22.85 MSE 0.060744 = = =
    • 65. 65 Conclusions We find q = 4.34 for the 95% Tukey confidence intervals. The confidence intervals are Difference 95% Confidence Limits 95% Confidence Interval µA - µB 0.313 ± 0.404 (-0.091, 0.717) µA - µC 0.879 ± 0.404 (0.474, 1.283) µB - µC 0.566 ± 0.404 (0.161, 0.970) Store C Store B Store A $2.24 $2.81 $3.20 We therefore conclude that Store A is cheaper on the average than Store B and Store C.
    • 66. 66 Two-Factor ANOVA Notation: k = number of levels of factor A l = number of levels of factor B kl = number of treatments (each one a combination of a factor A level and a factor B level) m = number of observations on each treatment
    • 67. 67 Two-Factor ANOVA Example A grocery store has two stocking supervisors, Fred & Wilma. The store is open 24 hours a day and would like to schedule these two individuals in a manner that is most effective. To help determine how to schedule them, a sample of their work was obtained by scheduling each of them for 5 times in each of the three shifts and then tracked the number of cases of groceries that were emptied and stacked during the shift. The data follows on the next slide.
    • 68. 68 Two-Factor ANOVA Example Supervisor Day Swing Night 495 547 481 457 500 578 504 496 485 607 517 515 428 518 497 481 520 498 508 471 560 572 550 583 533 507 518 578 625 598 Shift Fred Wilma
    • 69. 69 Interactions There is said to be an interaction between the factors, if the change in true average response when the level of one factor changes depend on the level of the other factor. One can look at the possible interaction between two factors by drawing an interactions plot, which is a graph of the means of the response for one factor plotted against the values of the other factor.
    • 70. 70 Two-Factor ANOVA Example Supervisor Day Swing Night Fred 529.40 495.60 500.00 508.33 Wilma 507.80 527.00 585.60 540.13 Mean Output for Each Shift 518.60 511.30 542.80 524.23 Mean Output for Each Supervisor Shift A table of the sample means for the 30 observations.
    • 71. 71 Two-Factor ANOVA Example Typically, only one of these interactions plots will be constructed. As you can see from these diagrams, there is a suggestion that Fred does better during the day and Wilma is better at night or during the swing shift. The question to ask is “Are these differences significant?” Specifically is there an interaction between the supervisor and the shift. Fred Wilma SwingNightDay 590 580 570 560 550 540 530 520 510 500 Shift Supervisor Mean Interaction Plot - Data Means for Cases Day Night Swing WilmaFred 590 580 570 560 550 540 530 520 510 500 Supervisor Shift Mean Interaction Plot - Data Means for Cases
    • 72. 72 Interactions If the graphs of true average responses are connected line segments that are parallel, there is no interaction between the factors. In this case, the change in true average response when the level of one factor is changed is the same for each level of the other factor. Special cases of no interaction are as follows: 1.The true average response is the same for each level of factor A (no factor A main effects). 2.The true average response is the same for each level of factor B (no factor B main effects).
    • 73. 73 Basic Assumptions for Two- Factor ANOVA The observations on any particular treatment are independently selected from a normal distribution with variance σ2 (the same variance for each treatment), and samples from different treatments are independent of one another.
    • 74. 74 Two-Factor ANOVA Table The following is a fairly standard way of presenting the important calculations for an two- factor ANOVA. The fundamental identity is SSTo = SSA + SSB + SSAB +SSE
    • 75. 75 Two-Factor ANOVA Example Source df Sum of Squares Mean Square F Shift 2 5437 2719 1.82 Supervisor 1 7584 7584 5.07 Interaction 2 14365 7183 4.80 Error 24 35878 1495 Total 29 63265
    • 76. 76 Two-Factor ANOVA Example Minitab output for the Two-Factor ANOVA Two-way ANOVA: Cases versus Shift, Supervisor Analysis of Variance for Cases Source DF SS MS F P Shift 2 5437 2719 1.82 0.184 Supervis 1 7584 7584 5.07 0.034 Interaction 2 14365 7183 4.80 0.018 Error 24 35878 1495 Total 29 63265 1. Test of H0: no interaction between supervisor and Shift There is evidence of an interaction.

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