IE341 Problems

1.Nuisance effects can be known or unknown. (a) If they are known, what are the ways you can deal with them? (b) What happens if they are unknown? Solution (a) If the known nuisance effects can be controlled, we can use blocking, Latin squares, or Graeco-Latin squares. If the known nuisance effects are uncontrollable, we use ANCOVA. (b) If the nuisance effects are unknown, they are absorbed into error. That is why randomization is so important. It randomizes out the effects of unknown nuisance sources and does not allow them to concentrate their effect in just one level of a factor.

2. If you have an uncontrollable nuisance variable that you suspect affects the response variable, (a) what kind of analysis should you do? (b) What happens if you don’t do this type of analysis? Solution (a) You should do ANCOVA. (b) If you ignore the nuisance variable, its effects are included in error, thus making a less powerful test for the factor(s) of interest.

3.We are testing 3 different types of glue for tensile strength when joining parts together. But the tensile strength is also affected by the thickness of the glue applied. Our data are: (a) What kind of analysis should you do? (b) Explain what role each of the following plays in the design: - glue type - tensile strength - glue thickness Glue Type 1Glue Type 2Glue Type 3 StrengthThicknessStrengthThicknessStrengthThickness

3.Solution (a) You should do ANCOVA to adjust the strength measure for the effect of thickness. (b) glue type is the factor under test tensile strength is the response variable glue thickness is the covariate

4.The surface finish on parts made by 3 machines is being studied. Each machine has three operators, and each operator tests two specimens. Because the machines are in different locations, different operators are used for each machine. The data are: (a) What kind of design is this? (b) What makes it this kind of design? Operator Machine 1Machine 2Machine Rep Rep

4.Solution (a) This is a nested design. (b) It is nested because the three operators under machine 1 are not the same as those under machine 2 or those under machine 3. Each set of operators belongs to its own machine.

5.An experiment is performed to test the effect of temperature (2 levels) and heating time (3 levels) on the strength of steel. During the experiment, the oven is heated to one of the two temperatures and 3 specimens are placed in the oven. After 10 minutes, one specimen is removed. After 20 minutes, a second specimen is removed. After 30 minutes, the third specimen is removed. Then the temperature is changed to the other level and the process is repeated. Three shifts were used to collect the data, shown in the next slide. (a) What kind of experiment is this? Why? (b) What role does each of the following variables play in this design: shift temperature time

5 (continued) (c) Set up the ANOVA table for this kind of design showing the variables that can be tested for significance and the error term used to test them. (no computations) 1500˚F1600˚F Shift 110 min min min6162 Shift 210 min min min5969 Shift 310 min min min7169

5.Solution (a) this is a split-plot experiment because there is a restriction on randomization of the time variable. (b) shift is a blocking factor temperature is the whole-plot factor time is the split-plot or sub-plot factor

5.Solution (continued) (c) Source SS df MS p Whole plot Shift (A) Temperature (B) x AB (whole-plot error) Split-Plot Time (C) x AC (error for test of C) BC x ABC (error for test of BC)

6.Four different feed rates were studied in a test of a machine to produce parts for aircraft. From prior experience, the engineer has learned that changing the feed rate will not change the average dimension of the parts, but it might change the variability. He makes 4 production runs at each of 4 feed rates, measures the standard deviation, and these are his data. What must he do to analyze these data? Why? Production Run Feed rateRun 1Run 2Run 3Run

6.Solution In order to test the effect of the feed rate on the standard deviation of the part measurement, the experimenter must first transform the data to -ln(sd) because the raw standard deviations are not normally distributed, but their logs are. Now he can run the ANOVA.

7. Four different designs are being studied for a computer circuit to see which has the least noise. There are 4 replicates. The data are (a) What type of analysis would you suggest? (b) Why? (c) Do the first step of the analysis. DesignNoise observed

7. Solution (a)The best type of analysis is a Kruskal- Wallis rank analysis because the data are clearly not normal. (b) The first step is to get the ranks: levelvalueorderrank

8.An engineer trying to optimize his process has run a first-order model in two variables with the following results. Source SS df MS p Regression Residual Interaction Pure Quad Error Total (a) What would you recommend that he do next? Why? (b) What kind of design would you suggest?

8.Solution (a) He should use a second-order model because the pure quadratic term is significant in his first-order model. (b) He should probably use a central composite design (CCD).

9.(a) Provide a central composite design (not a sketch) for two variables. (b) Explain under what circumstances it is useful.

9.Solution (a) (b) The CCD is useful for second-order RSM designs. X1X

10.The canonical form of the fitted RSM model is (a) What do you know about the stationary point? Why? (b) What do you know about the sensitivity of the response to the two variables? Solution (a) The stationary point is a maximum because both eigenvalues are negative. (b) The response is more sensitive to X 2 because its |eigenvalue| is greater than that for X 1.

11.What makes mixture experiments different from other RSM experiments? Solution Mixture experiments are different because their levels are proportions and thus must sum to 1.0. This makes their levels dependent on one another. For other kinds of experiments, the levels of the factors are independent of one another.

12. (a) Describe the kind of design that is ordinarily used for mixture experiments. (b) What are its disadvantages and how do you overcome them? Solution (a) Mixture experiments use simplex designs. (b) The lattice simplex has either pure mixtures or combinations of all but one of the ingredients. To allow for all ingredients, the simplex centroid design adds center points.

13. What are two ways of determining whether the response at the stationary point is a maximum, a minimum, or a saddle point? Solution 1. Examine the contour plots. 2. Examine the eigenvalues in the canonical equation. If all are negative, the stationary point is a maximum. If all are positive, it is a minimum. If they are of mixed sign, it is a saddle point.

14. A mixture experiment has resulted in the following polynomial that is a good fit to the data. (a) If you are looking for a maximum response, what would you choose? Why? (b) What would you avoid? Why? Solution (a) Choose the X 1 X 2 mixture because it has the largest coefficient. If you want a pure blend, you would select X 3 because it has the largest coefficient of the pure blends.

14. Solution (continued) (b) The combination to avoid is X 2 X 3 because its coefficient is negative and thus this combination reduces the response rather than increasing it.

15. A textile mill has a problem. The strength of the cloth produced has too much variability. They wonder if it’s due mostly to the looms or to the operators, so they decide to study it. From the large number of looms they have, 3 looms are chosen randomly. Also 3 operators are chosen at random from all the operators in the plant. 3 replicates are run for each combination of loom and operator. The ANOVA table is Source SS df MS p Looms Operators AB Error Total

15. (continued) (a)What type of ANOVA is this? (b)Given the E(MS) table below, what proportion of variance is due to looms, to operators, to interaction, and to error? Solution (a) It is a random effects ANOVA because the levels of both factors were chosen at random.

15. Solution (continued) Proportion (b) σ 2 = MS(E) = 17 17/31 σ τ 2 = MS(A) – MS(AB) = = 3 3/31 Kn 3*3 σ β 2 = MS(B) – MS(AB) = = 10 10/31 Jn 3*3 σ τβ 2 = MS(AB) – MS(E) = = 1 1/31 n 3 Total σ 2 = = 31

16. Four factors are to be used in a manufacturing process for integrated circuits to improve yield. A is aperture setting (small, large), B is exposure time (20 sec, 30 sec), C is development time (30 sec, 45 sec), D is mask dimension (small, large). You are the statistical consultant for the firm and you are asked to design the experiment. You’d better do it or the boss will be angry and you know what that means. What kind of design did you create?

16.Solution Design is a 2 4 factorial design RunAperture setting Exposure time Develop time Mask dimension 1Small2030Small 2Large2030Small 3 30 Small 4Large30 Small Small 6Large2045Small Small 8Large3045Small Large 10Large2030Large 11Small30 Large 12Large30 Large 13Small2045Large 14Large2045Large 15Small3045Large 16Large3045Large