IE341 Midterm

1. The effects of a 2 x 2 fixed effects factorial design are:

A effect = 20 B effect = 10 AB effect = 16 = 35 (a) Write the fitted regression model for this

design.

Y = 35 + 10(X1) + 5(X2) + 8(X1X2)

Plot the interaction effect. You need to get all 4 corners of the design

to do this. -- Y = 35 + 10(-1) + 5(-1) + 8(-1)(-1) = 28 +- Y = 35 + 10(+1) + 5(-1) + 8(+1)(-1) = 32 -+ Y = 35 + 10(-1) + 5(+1) + 8(-1)(+1) = 22 ++ Y = 35 + 10(+1) + 5(+1) + 8(+1)(+1) =

58

Y

1. (continued)

A and B interaction

B low

B low

B high

B high

20

30

40

50

60

- 1 0 1

level of A

response

2. The mean results for a completely balanced experiment are

With these cells means, compute orthogonal contrasts and their SS for:

(a) High Temperature vs the average of Low and Medium Temperature.

(b) Low Temperature vs Medium Temperature (c) High Temperature vs Medium temperature

Pressure

Temperature Low Medium High

Low 7 6 8

Medium 6 5 13

High 11 10 15

2. Before you can compute the contrasts, you need the temp means.

Low Medium High Mean 7 8 12 (a) contrast -1 -1 2 = 9n SS = n292 / n(6) = 81n/6 (b) contrast -1 1 0 = 1n SS = n212 / n(2) = n/2 (c) This contrast cannot be done because there are on

ly 2 df in the temperature factor and the first two contrasts have used these 2 df .

Note: Because I got the corrected version online so late, I didn’t deduct for errors in this problem having to do with multiplying by n.

3. The E(MS) for the effects in an experiment are:

(a) What type of design is this? This is a random effects 2-factor

factorial. (b) Set up all the F-tests.

A effect: F = MSA / MSAB B effect: F = MSB / MSAB AB effect: F = MSAB / MSE

22

222

222

2

)(

)(

)(

)(

nMSE

JnnMSE

KnnMSE

MSE

AB

B

A

E

4. 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? 24-1 fractional factorial

(b) What is its generator? D = ABC

(c) What is the design resolution? Because I = ABCD, design resolution

= IV (d) List all confounded effects.

A + BCD AB + CD B + ACD AC + BD C + ABD AD + BC D + ABC

(e) What is the defining relation of the complementary fraction?

I = -ABCD

5. Fabric strength produced by 4 types of loom is being studied. The data are:

Type 1 Type 2 Type 3 Type 4

30 30 35 25

30 35 35 25

25 25 40 20

35 30 30 30

5. (a) Do the ANOVA. Mean 1= 30 Mean 2= 30 Mean 3= 35 Mean 4= 25 Grand = 30 SS between = 4(0 + 0 + 25 + 25) = 200 MS between = 200 / 3 SS within = 50 + 50 + 50 + 50 = 200 MS within = 200 / 12 F = 200/3 * 12/200 = 4 p = 0.03459 Source SS df MS p Loom type 200 3 200/3 0.03459 Error 200 12 200/12 Total 400 15

5. (b) Plot the effect.

(c) For greatest fabric strength, which type of loom would you choose?

Loom Type 3

Plot of loom type

20

25

30

35

40

1 2 3 4

Loom type

response

6. 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 94 2 47 0.089 Operators 220 2 110 0.007 AB 80 4 20 0.354 Error 306 18 17 Total 700 26

6. (a) What type of ANOVA is this? A random effects two-way ANOVA (b) What proportion of variance is

due to looms, to operators, to interaction, and to error?

Looms:

Operators:

AB:

Error: σ2 = MSE = 17

3)3(3

20472

Kn

MSMS ABA

10)3(3

201102

Jn

MSMS ABB

13

17202

n

MSMS EAB

6. (continued) Total: A + B + AB + E 3 + 10 + 1 + 17 = 31 proportion due to Looms = 3/ 31 proportion due to Operators =

10/ 31 proportion due to interaction = 1

/ 31 proportion due to error = 17 / 31

7. 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? A randomized blocks design (b) Why is it designed this way? So that differences between the

three furnaces do not affect differences between stirring rates.

7 (c) Set up the experiment for the process engineer.

Furnace 1

Furnace 2

Furnace 3

Stir rate 1

Stir rate 2

Stir rate 3

8. 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.

8. Run Aperture

settingExposure

timeDevelop

timeMask

dimension

1 Small 20 30 Small

2 Large 20 30 Small

3 Small 30 30 Small

4 Large 30 30 Small

5 Small 20 45 Small

6 Large 20 45 Small

7 Small 30 45 Small

8 Large 30 45 Small

9 Small 20 30 Large

10 Large 20 30 Large

11 Small 30 30 Large

12 Large 30 30 Large

13 Small 20 45 Large

14 Large 20 45 Large

15 Small 30 45 Large

16 Large 30 45 Large

8. What kind of design did you create? A 24 factorial design with all fixed

effects

Note: Many of you did a 24-1 fractional

factorial. Since there is no restriction on the number of runs, it is not acceptable to design an experiment that is troublesome when you can design the full factorial and have no confounding at all.

However, if you did a fractional correctly and identified it as such, I took off only 5 instead of 10 points.

9. 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.

This is a Latin Square design.

9. (continued)

Day 1 Day 2 Day 3 Day 4 Day 5

Batch 1 A B C D E

Batch 2 B C D E A

Batch 3 C D E A B

Batch 4 D E A B C

Batch 5 E A B C D

A, B, C, D, E are the five ingredients under test.

10. What is the difference between a fixed effects model, a random effects model, and a mixed model?

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.

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.

11. What is the logic behind decomposition of the total SS in

ANOVA?

The logic is that the total SS includes variability due to one or more factors and variability due to error. We can separate these and test the effects of the factors.

12. What are two ways of dealing with nuisance variables in the designs we have studied?

We can deal with nuisance variables by blocking, by a Latin Square, or by a Graeco-Latin Square.

Why do they work? They all work because all levels of the

factors of interest are tested at each combination of levels of the nuisance factors.

13. Why is a single-replicate design undesirable?

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.

What is Cuthbert Daniel’s idea and how does it help in this situation?

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.

14. How do you check ANOVA model adequacy?

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.

15. What do you mean by a completely randomized design?

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.

Grade distribution95-100% 9 A+

90-94% 14 A0

85-89% 14 A-

80-84% 9 B+

75-79% 3 B0

70-74% 1 B-

65-69% 2 C+

60-64% 1 C0

55-59% 1 C-

<40% 1 D-

Question summaryQ Students who lost points Total points lost

1 4 15

2 6 31

3 19 75

4 6 6

5 18 30

6 47 (worst question) 329

7 16 56

8 34 (2nd worst) 193

9 7 30

10 16 50

11 12 53

12 14 60

13 21 81

14 22 87

15 14 63