⫺0.2 Jay L. Devore Probability and Statistics

Example 11.4 Example 11.3 continued If H 0A is true, MSA is an unbiased estimator of s 2 , so F is a ratio of two unbiased estimators of s 2 . When H 0A is false, MSA tends to overestimate s 2 . Thus H 0A should be rejected when the ratio F A is too large. Similar comments apply to MSB and H 0B . Multiple Comparisons After rejecting either H 0A or H 0B , Tukey’s procedure can be used to identify signifi- cant differences between the levels of the factor under investigation. 1. For comparing levels of factor A, obtain . For comparing levels of factor B, obtain . 2. Compute Q a ,I,I21J21 2MSEJ for factor A comparisons means being compared w 5 Q estimated standard deviation of the sample Q a ,J,I21J21 Q a ,I,I21J21 Identification of significant differences among the four washing treatments requires and . The four factor B sample means column averages are now listed in increasing order, and any pair differing by less than .340 is underscored by a line segment: Washing treatment 1 appears to differ significantly from the other three treatments, but no other significant differences are identified. In particular, it is not apparent which among treatments 2, 3, and 4 is best at removing marks. ■ Randomized Block Experiments In using single-factor ANOVA to test for the presence of effects due to the I dif- ferent treatments under study, once the IJ subjects or experimental units have been chosen, treatments should be allocated in a completely random fashion. That is, J subjects should be chosen at random for the first treatment, then another sample of J chosen at random from the remaining subjects for the second treat- ment, and so on. It frequently happens, though, that subjects or experimental units exhibit het- erogeneity with respect to other characteristics that may affect the observed responses. Then, the presence or absence of a significant F value may be due to this extraneous variation rather than to the presence or absence of factor effects. This is why paired experiments were introduced in Chapter 9. The analogy to a paired exper- iment when is called a randomized block experiment. An extraneous factor, “blocks,” is constructed by dividing the IJ units into J groups with I units in each I . 2 IJ 2 J x 4 x 2 .300 337 x 3 x 1 .423 .803 w 5 4.902.014473 5 .340 Q .05,4,6 5 4.90 because, e.g., the standard deviation of is . 3. Arrange the sample means in increasing order, underscore those pairs differing by less than w, and identify pairs not underscored by the same line as correspon- ding to significantly different levels of the given factor. s 1J X i Q a ,J,I21J21 MSEI for factor B comparisons e 5 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook andor eChapters. Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Example 11.5 group. This grouping or blocking should be done so that within each block, the I units are homogeneous with respect to other factors thought to affect the responses. Then within each homogeneous block, the I treatments are randomly assigned to the I units or subjects. A consumer product-testing organization wished to compare the annual power con- sumption for five different brands of dehumidifier. Because power consumption depends on the prevailing humidity level, it was decided to monitor each brand at four different levels ranging from moderate to heavy humidity thus blocking on humidity level. Within each level, brands were randomly assigned to the five selected locations. The resulting observations annual kWh appear in Table 11.2, and the ANOVA calculations are summarized in Table 11.3. Table 11.2 Power Consumption Data for Example 11.5 Treatments Blocks humidity level brands 1 2 3 4 1 685 792 838 875 3190 797.50 2 722 806 893 953 3374 843.50 3 733 802 880 941 3356 839.00 4 811 888 952 1005 3656 914.00 5 828 920 978 1023 3749 937.25 3779 4208 4541 4797 17,325 755.80 841.60 908.20 959.40 866.25 x j x j x i x i Table 11.3 ANOVA Table for Example 11.5 Source of Variation df Sum of Squares Mean Square f Treatments brands 4 53,231.00 13,307.75 Blocks 3 116,217.75 38,739.25 Error 12 1671.00 139.25 Total 19 171,119.75 f B 5 278.20 f A 5 95.57 Since and , H is rejected in favor of H a . Power consumption appears to depend on the brand of humidifier. To identify significantly different brands, we use Tukey’s procedure. and . The underscoring indicates that the brands can be divided into three groups with respect to power consumption. Because the block factor is of secondary interest, is not needed, though the computed value of F B is clearly highly significant. Figure 11.4 shows SAS output for this data. At the top of the ANOVA table, the sums of squares SSs for treatments brands and blocks humidity levels are combined into a single “model” SS. F .05,3,12 x 1 x 3 x 2 x 4 x 5 797.50 839.00 843.50 914.00 937.25 w 5 4.512139.254 5 26.6 Q .05,5,12 5 4.51 f A 5 95.57 3.26 F .05,4,12 5 3.26 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook andor eChapters. Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Example 11.6 Analysis of Variance Procedure Dependent Variable: POWERUSE Sum of Mean Source DF Squares Square F Value Pr . F Model 7 169448.750 24206.964 173.84 0.0001 Error 12 1671.000 139.250 Corrected Total 19 171119.750 R-Square C.V. Root MSE POWERUSE Mean 0.990235 1.362242 11.8004 866.25000 Source DF Anova SS Mean Square F Value PR . F BRAND 4 53231.000 13307.750 95.57 0.0001 HUMIDITY 3 116217.750 38739.250 278.20 0.0001 Alpha ⫽ 0.05 df ⫽ 12 MSE ⫽ 139.25 Critical Value of Studentized Range ⫽ 4.508 Minimum Significant Difference ⫽ 26.597 Means with the same letter are not significantly different. Tukey Grouping Mean N BRAND A 937.250 4 5 A A 914.000 4 4 B 843.500 4 2 B B 839.000 4 3 C 797.500 4 1 How does string tension in tennis rackets affect the speed of the ball coming off the racket? The article “Elite Tennis Player Sensitivity to Changes in String Tension and the Effect on Resulting Ball Dynamics” Sports Engr., 2008: 31–36 described an experiment in which four different string tensions N were used, and balls projected from a machine were hit by 18 different players. The rebound speed kmh was then determined for each tension-player combination. Consider the following data in Table 11.4 from a similar experiment involving just six play- ers the resulting ANOVA is in good agreement with what was reported in the article. The ANOVA calculations are summarized in Table 11.5. The P-value for testing to see whether true average rebound speed depends on string tension is .049. Thus is barely rejected at significance level .05 in favor of the conclusion that true average speed does vary with tension . Application of Tukey’s procedure to identify significant differences among tensions requires . Then . The difference between the largest and smallest sample mean tensions is 6.87. So although the F test is significant, Tukey’s w 5 7.464 Q .05,4,15 5 4.08 F .05,3,15 5 3.29 H : a 1 5 a 2 5 a 3 5 a 4 5 Figure 11.4 SAS output for power consumption data ■ In many experimental situations in which treatments are to be applied to sub- jects, a single subject can receive all I of the treatments. Blocking is then often done on the subjects themselves to control for variability between subjects; each subject is then said to act as its own control. Social scientists sometimes refer to such exper- iments as repeated-measures designs. The “units” within a block are then the differ- ent “instances” of treatment application. Similarly, blocks are often taken as different time periods, locations, or observers. Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook andor eChapters. Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Table 11.4 Rebound Speed Data for Example 11.6 Player Tension 1 2 3 4 5 6 210 105.7 116.6 106.6 113.9 119.4 123.5 114.28 235 113.3 119.9 120.5 119.3 122.5 124.0 119.92 260 117.2 124.4 122.3 120.0 115.1 127.9 121.15 285 110.0 106.8 110.0 115.3 122.6 128.3 115.50 111.55 116.93 114.85 117.13 119.90 125.93 x .j x i . Table 11.5 ANOVA Table for Example 11.6 Source df SS MS f P Tension 3 199.975 66.6582 3.32 0.049 Player 5 477.464 95.4928 4.76 0.008 Error 15 301.188 20.0792 Total 23 978.626 method does not identify any significant differences. This occasionally happens when the null hypothesis is just barely rejected. The configuration of sample means in the cited article is similar to ours. The authors commented that the results were contrary to previous laboratory-based tests, where higher rebound speeds are typically associated with low string tension. ■ In most randomized block experiments in which subjects serve as blocks, the subjects actually participating in the experiment are selected from a large population. The subjects then contribute random rather than fixed effects. This does not affect the procedure for comparing treatments when one observation per “cell,” as in this section, but the procedure is altered if . We will shortly consider two-factor models in which effects are random. More on Blocking When , either the F test or the paired differences t test can be used to analyze the data. The resulting conclusion will not depend on which procedure is used, since and . Just as with pairing, blocking entails both a potential gain and a potential loss in precision. If there is a great deal of heterogeneity in experimental units, the value of the variance parameter s 2 in the one-way model will be large. The effect of block- ing is to filter out the variation represented by s 2 in the two-way model appropriate for a randomized block experiment. Other things being equal, a smaller value of s 2 results in a test that is more likely to detect departures from H i.e., a test with greater power. However, other things are not equal here, since the single-factor F test is based on degrees of freedom df for error, whereas the two-factor F test is based on df for error. Fewer error df results in a decrease in power, essentially because the denominator estimator of s 2 is not as precise. This loss in df can be especially serious if the experimenter can afford only a small number of observations. Nevertheless, if it appears that blocking will significantly reduce variability, the sacrifice of error df is sensible. I 2 1J 2 1 I J 2 1 t a 2,n 2 5 F a ,1,n T 2 5 F I 5 2 K ij 5 K . 1 K ij 5 1 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook andor eChapters. Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Models with Random and Mixed Effects In many experiments, the actual levels of a factor used in the experiment, rather than being the only ones of interest to the experimenter, have been selected from a much larger population of possible levels of the factor. If this is true for both factors in a two-factor experiment, a random effects model is appropriate. The case in which the levels of one factor are the only ones of interest and the levels of the other fac- tor are selected from a population of levels leads to a mixed effects model. The two- factor random effects model when is The , and ’s are all independent, normally distributed rv’s with mean 0 and variances , and s 2 , respectively. The hypotheses of interest are then level of factor A does not contribute to variation in the response versus and versus . Whereas as before, the expected mean squares for factors A and B are now Thus when is true, is still a ratio of two unbiased estimators of s 2 . It can be shown that a level a test for H 0A versus H aA still rejects if , and, similarly, the same procedure as before is used to decide between H 0B and H aB . If factor A is fixed and factor B is random, the mixed model is where and the B j ’s and are normally distributed with mean 0 and vari- ances and s 2 , respectively. Now the two null hypotheses are with expected mean squares The test procedures for H 0A versus H aA and H 0B versus H aB are exactly as before. For example, in the analysis of the color-change data in Example 11.1, if the four wash treatments were randomly selected, then because and is rejected in favor of . An estimate of the “variance component” is then given by . Summarizing, when , although the hypotheses and expected mean squares differ from the case of both effects fixed, the test procedures are identical. K ij 5 1 MSB 2 MSEI 5 .0485 s B 2 H aB : s B 2 . F .05,3,6 5 4.76, H 0B : s B 2 5 f B 5 11.05 E MSE 5 s 2 EMSA 5 s 2 1 J I 2 1 g a i 2 EMSB 5 s 2 1 Is B 2 H 0A : a 1 5 c5 a I 5 0 and H 0B : s B 2 5 s B 2 P ij ’s ga i 5 X ij 5 m 1 a i 1 B j 1 P ij i 5 1, c , I, j 5 1, c , J f A F a ,I21,I21J21 H 0A F A F B H 0A H 0B E MSA 5 s 2 1 Js A 2 E MSB 5 s 2 1 Is B 2 E MSE 5 s 2 H aB : s B 2 . H 0B : s B 2 5 H aA : s A 2 . H 0A : s A 2 5 s A 2 , s B 2 P ij A i ’s, B j ’s X ij 5 m 1 A i 1 B j 1 P ij i 5 1, c , I, j 5 1, c , J K ij 5 1 EXERCISES Section 11.1 1–15 1. The number of miles of useful tread wear in 1000s was determined for tires of each of five different makes of sub- compact car factor A, with in combination with each of four different brands of radial tires factor B, with , resulting in observations. The values , and were then computed. Assume that an additive model is appropriate. SSE 5 59.2 SSB 5 44.1 SSA 5 30.6, IJ 5 20 J 5 4 I 5 5 a. Test no differences in true average tire lifetime due to makes of cars versus H a : at least one using a level .05 test. b. no differences in true aver- age tire lifetime due to brands of tires versus H a : at least one using a level .05 test. b j

2 H

: b 1 5 b 2 5 b 3 5 b 4 5 a i

2 H

: a 1 5 a 2 5 a 3 5 a 4 5 a 5 5 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook andor eChapters. Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 2. Four different coatings are being considered for corrosion protection of metal pipe. The pipe will be buried in three dif- ferent types of soil. To investigate whether the amount of cor- rosion depends either on the coating or on the type of soil, 12 pieces of pipe are selected. Each piece is coated with one of the four coatings and buried in one of the three types of soil for a fixed time, after which the amount of corrosion depth of maximum pits, in .0001 in. is determined. The data appears in the table. Soil Type B 1 2 3 1 64 49 50 2 53 51 48 Coating A 3 47 45 50 4 51 43 52 a. Assuming the validity of the additive model, carry out the ANOVA analysis using an ANOVA table to see whether the amount of corrosion depends on either the type of coating used or the type of soil. Use . b. Compute , and . 3. The article “Adiabatic Humidification of Air with Water in a Packed Tower” Chem. Eng. Prog., 1952: 362–370 reports data on gas film heat transfer coefficient Btuhr ft 2 on °F as a function of gas rate factor A and liquid rate factor B. B 1190 2250 3300 4400 1200 200 226 240 261 2400 278 312 330 381 A 3700 369 416 462 517 41100 500 575 645 733 a. After constructing an ANOVA table, test at level .01 both the hypothesis of no gas-rate effect against the appropri- ate alternative and the hypothesis of no liquid-rate effect against the appropriate alternative. b. Use Tukey’s procedure to investigate differences in expected heat transfer coefficient due to different gas rates. c. Repeat part b for liquid rates. 4. In an experiment to see whether the amount of coverage of light-blue interior latex paint depends either on the brand of paint or on the brand of roller used, one gallon of each of four brands of paint was applied using each of three brands of roller, resulting in the following data number of square feet covered. Roller Brand 1 2 3 1 454 446 451 Paint 2 446 444 447 Brand 3 439 442 444 4 444 437 443 b ˆ 3 m ˆ , aˆ 1 , aˆ 2 , aˆ 3 , aˆ 4 , bˆ 1 , bˆ 2 a 5 .05 a. Construct the ANOVA table. [Hint: The computations can be expedited by subtracting 400 or any other convenient number from each observation. This does not affect the final results.] b. State and test hypotheses appropriate for deciding whether paint brand has any effect on coverage. Use . c. Repeat part b for brand of roller. d. Use Tukey’s method to identify significant differences among brands. Is there one brand that seems clearly preferable to the others? 5. In an experiment to assess the effect of the angle of pull on the force required to cause separation in electrical connec- tors, four different angles factor A were used, and each of a sample of five connectors factor B was pulled once at each angle “A Mixed Model Factorial Experiment in Testing Electrical Connectors,” Industrial Quality Control, 1960: 12–16. The data appears in the accompanying table. B 1 2 3 4 5 ° 45.3 42.2 39.6 36.8 45.8 2 ° 44.1 44.1 38.4 38.0 47.2 A 4 ° 42.7 42.7 42.6 42.2 48.9 6 ° 43.5 45.8 47.9 37.9 56.4 Does the data suggest that true average separation force is affected by the angle of pull? State and test the appropriate hypotheses at level .01 by first constructing an ANOVA table , and . 6. A particular county employs three assessors who are respon- sible for determining the value of residential property in the county. To see whether these assessors differ systematically in their assessments, 5 houses are selected, and each assessor is asked to determine the market value of each house. With factor A denoting assessors and factor B denoting houses , suppose and . a. Test at level .05. H states that there are no systematic differences among assessors. b. Explain why a randomized block experiment with only 5 houses was used rather than a one-way ANOVA experi- ment involving a total of 15 different houses, with each assessor asked to assess 5 different houses a different group of 5 for each assessor. 7. The article “Rate of Stuttering Adaptation Under Two Electro-Shock Conditions” Behavior Research Therapy, 1967: 49–54 gives adaptation scores for three different treat- ments: 1 no shock, 2 shock following each stuttered word, and 3 shock during each moment of stuttering. These treat- ments were used on each of 18 stutterers, resulting in , and . a. Construct the ANOVA table and test at level .05 to see whether the true average adaptation score depends on the treatment given. SSBl 5 2977.67 SST 5 3476.00, SSTr 5 28.78 H : a 1 5 a 2 5 a 3 5 SSE 5 25.6 SSB 5 113.5, SSA 5 11.7, J 5 5 I 5 3 SSB 5 246.97 SST 5 396.13, SSA 5 58.16 a 5 .05 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook andor eChapters. Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. b. Judging from the F ratio for subjects factor B, do you think that blocking on subjects was effective in this experiment? Explain. 8. The paper “Exercise Thermoregulation and Hyperprolac- tinaemia” Ergonomics, 2005: 1547–1557 discussed how various aspects of exercise capacity might depend on the temperature of the environment. The accompanying data on body mass loss kg after exercising on a semi-recumbent cycle ergometer in three different ambient temperatures 6°C, 18°C, and 30°C was provided by the paper’s authors. Cold Neutral Hot 1 .4 1.2 1.6 2 .4 1.5 1.9 3 1.4 .8 1.0 4 .2 .4 .7 Subject 5 1.1 1.8 2.4 6 1.2 1.0 1.6 7 .7 1.0 1.4 8 .7 1.5 1.3 9 .8 .8 1.1 a. Does temperature affect true average body mass loss? Carry out a test using a significance level of .01 as did the authors of the cited paper. b. Investigate significant differences among the temperatures. c. The residuals are .20, .30, , , .30, .00, .03, , , .13, .23, , , .03, , , .33, , , , .67, .11, , .27, .01, , .24. Use these as a basis for investigating the plausibility of the assump- tions that underlie your analysis in a. 9. The article “The Effects of a Pneumatic Stool and a One- Legged Stool on Lower Limb Joint Load and Muscular Activity During Sitting and Rising” Ergonomics, 1993: 519–535 gives the accompanying data on the effort required of a subject to arise from four different types of stools Borg scale. Perform an analysis of variance using , and fol- low this with a multiple comparisons analysis if appropriate. Subject 1 2 3 4 5 6 7 8 9 1 12 10 7 7 8 9 8 7 9 8.56 Type 2 15 14 14 11 11 11 12 11 13 12.44 of 3 12 13 13 10 8 11 12 8 10 10.78 Stool 4 10 12 9 9 7 10 11 7 8 9.22 10. The strength of concrete used in commercial construction tends to vary from one batch to another. Consequently, small test cylinders of concrete sampled from a batch are “cured” for periods up to about 28 days in temperature- and moisture-controlled environments before strength measure- ments are made. Concrete is then “bought and sold on the basis of strength test cylinders” ASTM C 31 Standard Test x i a 5 .05 2 .13 2 .33 2 .53 2 .33 2 .10 2 .04 2 .27 2 .04 2 .27 2 .14 2 .20 2 .07 2 .40 Method for Making and Curing Concrete Test Specimens in the Field. The accompanying data resulted from an experi- ment carried out to compare three different curing methods with respect to compressive strength MPa. Analyze this data. Batch Method A Method B Method C 1 30.7 33.7 30.5 2 29.1 30.6 32.6 3 30.0 32.2 30.5 4 31.9 34.6 33.5 5 30.5 33.0 32.4 6 26.9 29.3 27.8 7 28.2 28.4 30.7 8 32.4 32.4 33.6 9 26.6 29.5 29.2 10 28.6 29.4 33.2 11. For the data of Example 11.5, check the plausibility of assumptions by constructing a normal probability plot of the residuals and a plot of the residuals versus the predicted val- ues, and comment on what you learn. 12. Suppose that in the experiment described in Exercise 6 the five houses had actually been selected at random from among those of a certain age and size, so that factor B is ran- dom rather than fixed. Test versus using a level .01 test.

13. a.

Show that a constant d can be added to or subtracted from each without affecting any of the ANOVA sums of squares. b. Suppose that each is multiplied by a nonzero constant c. How does this affect the ANOVA sums of squares? How does this affect the values of the F statistics F A and F B ? What effect does “coding” the data by have on the conclusions resulting from the ANOVA pro- cedures? 14. Use the fact that with to show that , so that is an unbiased estimator for 15. The power curves of Figures 10.5 and 10.6 can be used to obtain type II error for the F test in two-factor ANOVA. For fixed values of , the quantity is computed. Then the figure corre- sponding to is entered on the horizontal axis at the value the power is read on the vertical axis from the curve labeled , and . a. For the corrosion experiment described in Exercise 2, find b when , and . Repeat for , and . b. By symmetry, what is b for the test of H 0B versus H aB in Example 11.1 when , and ? s 5 .3 b 1 5 .3, b 2 5 b 3 5 b 4 5 2 .1 s 5 4 a 1 5 6, a 2 5 0, a 3 5 a 4 5 2 3 s 5 4 a 1 5 4, a 2 5 0, a 3 5 a 4 5 2 2 b 5 1 2 power n 2 5 I 2 1J 2 1 f , n 1 5 I 2 1 f 2 5 JI ga i 2 s 2 a 1 , a 2 , c , a I b 5 P a i . a ˆ i 5 X i 2 X E X i 2 X 5 a i ga i 5 gb j 5 E X ij 5 m 1 a i 1 b j y ij 5 cx ij 1 d x ij x ij H a : s B 2 . H : s B 2 5 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook andor eChapters. Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 11.2 Two-Factor ANOVA with K ij . 1 In Section 11.1, we analyzed data from a two-factor experiment in which there was one observation for each of the IJ combinations of factor levels. The were assumed to have an additive structure with . Additivity means that the difference in true average responses for any two levels of the factors is the same for each level of the other factor. For example, , independent of the level j of the second factor. This is shown in Figure 11.1a, in which the lines connecting true average responses are parallel. Figure 11.1b depicts a set of true average responses that does not have addi- tive structure. The lines connecting these are not parallel, which means that the difference in true average responses for different levels of one factor does depend on the level of the other factor. When additivity does not hold, we say that there is inter- action between the different levels of the factors. The assumption of additivity in Section 11.1 allowed us to obtain an estimator of the random error variance s 2 MSE that was unbiased whether or not either null hypothesis of interest was true. When for at least one pair, a valid estimator of s 2 can be obtained with- out assuming additivity. Our focus here will be on the case , so the num- ber of observations per “cell” for each combination of levels is constant. Fixed Effects Parameters and Hypotheses Rather than use the themselves as model parameters, it is customary to use an equivalent set that reveals more clearly the role of interaction. m ij ’ s K ij 5 K . 1 i, j K ij . 1 m ij ’ s m ij 2 m i rj 5 m 1 a i 1 b j 2 m 1 a i r 1 b j 5 a i 2 a i r m ij 5 m 1 a i 1 b j , ga i 5 gb j 5 m ij ’ s NOTATION 11.7 m 5 1 IJ g i g j m ij m i 5 1 J g j m ij m j 5 1 I g i m ij Thus m is the expected response averaged over all levels of both factors the true grand mean, is the expected response averaged over levels of the second factor when the first factor A is held at level i, and similarly for . m j m i DEFINITION 11.8 from which 11.9 m ij 5 m 1 a i 1 b j 1 g ij g ij 5 m ij 2 m 1 a i 1 b j 5 b j 5 m j 2 m 5 the effect of factor B at level j a i 5 m i 2 m 5 the effect of factor A at level i interaction between factor A at level i and factor B at level j The model is additive if and only if all . The are referred to as the inter- action parameters. The a i s are called the main effects for factor A, and the b j ’s are the main effects for factor B. Although there are , and in addition to m, the conditions for any i, and for any j [all by virtue of 11.7 and 11.8] imply that only IJ of these new parameters are independ- ently determined: m, of the of the b j ’s, and of the g ij ’s. I 2 1J 2 1 a i ’s J 2 1 I 2 1 ⌺ i g ij 5 ga i 5 0, gb j 5 0, ⌺ j g ij 5 IJ g ij ’s I a i ’s, J b j ’s g ij ’s g ij ’s 5 0 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook andor eChapters. Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.