1. Solutions

2. 1.The tensile strength of concrete produced by 4 mixer levels is being studied with 4 replications. The data are: Compute the MS due to mixers and plot the means. For the greatest tensile strength, which mixer would you choose? Mixer 1 Mixer 2Mixer 3Mixer 4 Rep 120304525 Rep 220354525 Rep 325 5020 Rep 415304030

3. Solution 1. To get the MS due to mixers, we need both the grand mean and the mixer means. The means are: Mixer 1 mean = 20 Mixer 3 mean = 30 Mixer 3 mean = 45 Mixer 4 mean = 25 Grand mean = 30 Mixer 3 provides the greatest tensile strength

4. 2.For the data in problem 1, compute MS(error) and do the ANOVA table. Solution

5. Solution (continued) 2.The F ratio = 466.67/16.67 = 28 or = 4*350 * 12 = 28 3 200 The ANOVA table is Source SS df MS p Mixers 1400 3 466.67 < 0.00001 Error 200 12 16.67 Total 1600 15

6. 3. Here are the cell and marginal means from an experiment. According to the ANOVA table, the A and B main effects and the AB interaction are all significant. If your objective is to minimize the response, which values of A and B would you select. Explain. B low B highA marg A low551535 A high253530 B marg402532.5

7. Solution 3. Since the AB interaction is significant, only the combination means shown in the interaction plot are interpretable. The minimum response is for combination A low, B high.

8. 4. A textile mill has a large number of looms to weave its cloth. At random, 3 looms are chosen to see if the looms are meeting the standard cloth output. The data are What type of ANOVA is this? Do the ANOVA to see if there is any real variability among the looms. Loom 1Loom 2Loom 3 Rep 112 11 Rep 2141312 Rep 3161413

9. Solution 4. This is a random effects 1-way ANOVA. To get the between-loom SS, we need the mean for each loom and the grand mean. The means are: Loom 1 = 14, Loom 2 = 13, Loom 3 = 12 Grand mean = 13 SS(between)=3[(14-13) 2 +(13-13) 2 +(12-13) 2 ]=6 MS(between)= SS(between) / df(between) = 6/(3-1) = 3 SS(within) = (12-14) 2 +(14-14) 2 +(16-14) 2 + (12-13) 2 +(13-13) 2 +(14-13) 2 + (11-12) 2 +(12-12) 2 +(13-12) 2 = 12 MS(within) = SS(within) / df(within) = 12/6 = 2 F = 3 / 2, which is not significant. The loom-to-loom variability has not been shown to be > 0.

10. 5. The E(MS) for the A effect is (a) What type of factor is this? (b) How many factors are in the model? (c) How would you set up the F-test?

11. Solution 5. (a) this is a random effects factor because all of its components are variances. (b) There are 2 factors in the model because there is only one interaction term in the E(MS). (c) The F-test is MS A / MS AB because to test for the A effect, you must get rid of the error and AB terms.

12. 6. Because of limited resources, an experiment with 4 factors at 2 levels each is placed in an 8-run design. (a) What kind of design is it? (b) What is its generator? (c) What is the design resolution? (d) List all confounded effects. (e) What is the defining relation of the complementary fraction?

13. Solution 6. (a) This is a 2 4-1 fractional factorial (b) The generator is D=ABC (c) The design is resolution I V (d) The confounded effects are A + BCD AB + CD B + ACD AC + BD C + ABD AD + BC D + ABC (e) The defining relation for the complemenary fraction is I = -ABCD.

14. 7. Three different circuit designs are being studied to find the least amount of noise present. The data are (a) How many replications does this experiment have? (b) Do the ANOVA. (c) Find the best circuit design. Circuit Design Noise present A231 B567 C435

15. Solution 7. (a) The design has 3 replications. (b) The means are Circuit A: 2 Circuit B: 6 Circuit C: 4 Grand Mean: 4 SS(error)= (2-2) 2 + (3-2) 2 + (1-2) 2 +(5-6) 2 + (6-6) 2 + (7-6) 2 +(4-4) 2 + (3-4) 2 + (5-4) 2 = 6 MS(error) = SS(error)/df(error) = 6/6 = 1

16. Solution (continued) 7. SS(circuits) = 3[(2-4) 2 +(6-4) 2 +(4-4) 2 ] = 24 MS(circuits)= SS(circuits)/df(circuits) = 24/2 = 12 Source SS df MS p Circuits 24 2 12 <0.008 Error 6 6 1 Total 30 8 (c) There is evidence of a difference among circuits. The best circuit design is A.

17. 8. The effects of a 2 x 2 fixed effects factorial design are: A effect = 1 B effect = -9 AB effect = -29 = 30.5 (a) Write the fitted regression model for this design. (b) Plot the interaction effect.

18. Solution 8.The fitted regression model is To plot the interaction effect, we need the means at all 4 corners of the design, that is (- 1,-1), (+1,-1), (-1,+1), and (+1,+1). Substituting into the regression equation we get

19. Solution (continued) 8. The interaction plot is

20. 9.A factory produces grain refiners in 3 different furnaces, each of which has its own unique operating characteristics. Each furnace can be run at 3 different stirring rates. The process engineer knows that stirring rate affects the grain size of the product, so he decides to run an experiment testing the three stirring rates on his 3 furnaces. (a) What type of design is this? Why? (b) Set up the experiment.

21. Solution 9. (a) This is a randomized blocks design. Each stirring rate is run on all three furnaces so that furnace effects will not contaminate the stirring rate effect. (b) The experiment is Furnace Stirring rateABC 5 rpmobs 10 rpmobs 15 rpmobs

22. 10. The effect of 5 different ingredients on reaction time is being studied. Each batch of material is large enough for only 5 runs. Moreover, only 5 runs can be made in a day. Design the experiment.

23. Solution 10.This is a Latin Square design, where A, B, C, D, E are the five ingredients under study.

24. 11. The yield of a chemical process is being studied using 5 batches of raw material and 5 acid concentrations, which are nuisance factors. Two factors are suspected to be important: standing time (5 levels) and type of catalyst (5 levels). (a) What kind of experiment is this? (b) Design the experiment.

25. Solution 11.This is a Graeco-Latin Square design. In this design, Roman letters are the five standing times and Greek letters are the five types of catalyst. Batches of raw material Acid concentration 12345 1AαBγCεDβEδ 2BβCδDαEγAε 3CγDεEβAδBα 4DδEαAγBεCβ 5EεAβBδCαDγ

26. 12. The mean results for a completely balanced experiment are With these cells means, compute orthogonal contrasts and their SS for: (a) High Pressure vs the average of Low and Medium Pressure. (b) Low Pressure vs Medium Pressure (c) Medium Pressure vs High Pressure Pressure TemperatureLowMediumHigh Low 768 Medium 6513 High 111015

27. Solution 12.To use orthogonal contrasts, we need Main effect means. These means for pressure are Low Level: (7+6+11)/3 = 8 Medium Level: (6+5+10)/3 = 7 High Level: (8+13+15)/3 =12 (a)The appropriate orthogonal contrast is n[(-1)(Low) +(-1)(Med) +(2)High]= n[(-1)(8) +(-1)(7) +(2)(12)] = 9n Its SS =

28. Solution (continued) 12.(b) The contrast is n[(1)(Low) +(-1)(Med) +(0)High]= n[(1)( 8 ) + (-1)(7) +(0)(12)]= 1n Its SS = ( c) This contrast is unavailable, given the other two, because Pressure has 3 levels and thus only 2 df. Each contrast uses up 1 df so after two contrasts, no more are available using the orthogonal contrast method.

29. 13. A semiconductor engineer is studying the effect of lamination temperature (55˚C and 75˚C), lamination time (10 seconds and 25 seconds), lamination pressure (5 tn and 10 tn), and firing temperature (1580˚C and 1620˚C) on the curvature of the substrates produced. He must finish his study in one day so he can do only 8 runs. He is not very good at experimental design and doesn’t know how to do it. You are his statistical consultant and he asks you to design an experiment for him. Design the experiment and explain what problems he will have after he gets his results.

30. Solution 13. The statistical consultant suggests a 2 4-1 fractional factorial design.

31. Solution (continued) 13. The statistician explains that when the engineer gets his results, he won’t be sure of exactly what they really are because the experiment is confounded by the following aliases: A+BCD C+ABD AB+CD AD+BC B+ACD D+ABC AC+BD But the statistical consultant says that some of the interactions might make no sense in terms of the process and the engineer might be able to rule them out, so he recommends that he do the design.

32. 14.What do you mean by a completely randomized design? Solution 14. A completely randomized design is one in which all sources of bias are removed by (1) randomly assigning the experimental units to the various treatment levels and (2) doing the experimental runs in random order.

33. 15. What is the difference between a fixed effects model, a random effects model, and a mixed model? Solution 15. A fixed effects model is one in which we are testing the effects of treatments in terms of their means, and the conclusions apply only to those treatments under test.

34. Solution (continued) 15. A random effects model is one in which all factors are random factors, that is, there are many levels of each factor, but only J of them are randomly selected for use in the experiment. In a random effects model, we are interested in estimating the variance components, that is the portion of the total variance due to each of the factors. A mixed model is one is which at least one of the factors is random.

35. 16. What is the purpose of blocking in a design? Solution 16. Blocking is used to remove the effects of a nuisance factor, which is known and controllable, by having all levels of all other factors tested in each block. Removing the effect of a nuisance factor reduces the residual error and makes the design more powerful.

36. 17. What are two ways of dealing with nuisance variables in the design we have studied? Why do they work? Solution 17. We can deal with nuisance variables by blocking, by a Latin Square, or by a Graeco-Latin Square. They all work because all levels of the factors of interest are tested at each combination of levels of the nuisance factors.

37. 18. What is the logic behind decomposition of the total SS in ANOVA? Solution 18. The logic is that the total SS includes variability due to the factor(s) and variability due to error. We can separate these and look at the effects of the factors under test.

38. 19. Why is a single-replicate design undesirable? How does Cuthbert Daniel’s idea help in this situation? Solution 19. A single replicate design does not allow for a pure error term to test effects of factors. Instead we must use higher- order interactions as error under the assumption that they are not significant.

39. Solution (continued) 19. Daniel’s approach allows us to look at all effects by plotting them on normal probability paper. If the effects fall on a line, they behave like error and are not significant. This gives us a way of looking at all effects and choosing those that fall off the line to include in the model. All effects along the line are used as residual.

40. 20. How do you check model adequacy? Solution 20. Checking model adequacy involves residual plots. Normality is checked by plotting residuals on normal probability paper and making sure they all fall along a line. Constant error variance is checked by plotting residuals against fitted values and making sure they all have about the same spread with no outliers.