DESIGN OF EXPERIMENTS

 

1.0 INTRODUCTION

A designed experiment is a statistical experiment and its subsequent analysis to determine if there are differences between variables acting on some outcome. It also determines which variables have the greatest impact on the outcome. An equation of the regression line may be developed so that the outcome or response can be predicted based on the variables that went into the process. The statistical technique called analysis of variance (ANOVA) is used in designed experiments to determine if there are significant differences between the variables. A good designed experiment will yield much information with relatively little data. Matrix algebra is used extensively in the analysis of experiments. There are many different types of designed experiments. Two types of experiments, the factorial design and Latin square design will be reviewed in this chapter.

 

2.0 FACTORIAL DESIGN

A factorial design is an arrangement in which all levels of each factor are combined with all levels of every other factor.

 

Example 1

An experiment is conducted to determine the effect of temperature and pressure on the yield of a product. The numbers in the cells represent the yield (response).

 

 

Temperature (x1)

Pressure (x2)

120o

160o

115 psi

6,8

6,10

125 psi

5,7

2,4

 

This is a 22 experiment. (Two factors at two levels each. The exponent indicates the number of factors). The temperature and pressure are the two factors with each having two settings or levels.

The objective of the factorial experiment is to determine:

 

2.1 The Linear Model (Regression Equation)

Y = B0 + B1x1 + B2x2 + B3x1x2 + e

Y

= yield or response

x1

= temperature

x2

= pressure

x1x2

= interaction of the two variables

e

= error or residual

 

B0, B1, B2, and B3 are coefficients of the variables in the equation. The equation is

Y = 6.0 - 1.5x1 - 0.5x2 - 1.0x1x2 + 1.25. The coefficients are found by matrix algebra and the various matrices that are used to compute the coefficients are shown for information on the next page. The explanation of matrix algebra techniques is beyond the scope of QReview.

 

2.2 Analysis of Variance (ANOVA) Table

 

Source SS d.f. MS F FCritical
Temperature (x1)

2

1

2

0.57

7.71

Pressure (x2)

18

1

18

5.15

7.71

Interaction (x1x2)

8

1

8

2.28

7.71

Residual

14

4

3.5

 

 

SS = Sum of Squares

d.f. = Degrees of Freedom

MS = Mean Square = SS/d.f. = Variance

F = F Ratio = (MSsource/MSresidual)

The critical value of F is found in the F table. Using a level of significance of .05, 1 d.f. for the numerator and 4 d.f. for the denominator, the critical value of F is 7.71. In this example the pressure has the greatest impact on the outcome, but since the F ratio for pressure does not exceed the critical F value, the impact is not statistically significant.

 

2.3 Coefficients for the Linear Model

 

Pressure (x1)

Coded

Temperature (x2)

Coded

Yield (y)

115

-1

120

-1

6

115

-1

160

1

6

125

1

120

-1

5

125

1

160

1

2

115

-1

120

-1

8

115

-1

160

1

10

125

1

120

-1

7

125

1

160

1

4

 

     

x0

x1

x2

x1x2

e

 

6

 

1

-1

-1

1

-1

 

6

 

1

-1

1

-1

-1

 

5

 

1

1

-1

-1

-1

Y =

2

X =

1

1

1

1

-1

 

8

 

1

-1

-1

1

1

 

10

 

1

-1

1

-1

1

 

7

 

1

1

-1

-1

1

 

4

 

1

1

1

1

1

 

 

1

1

1

1

1

1

1

1

 

-1

-1

1

1

-1

-1

1

1

Transpose = X' =

-1

1

-1

1

-1

1

-1

1

 

1

-1

-1

1

1

-1

-1

1

 

-1

-1

-1

-1

1

1

1

1

 

  8 0 0 0 0   48
  0 8 0 0 0   -12
(X'X) = 0 0 8 0 0 (X'Y) = -4
  0 0 0 8 0   -8
  0 0 0 0 8   10

 

 

.125

0

0

0

0

  6.0
 

0

.125

0

0

0

Coefficients -1.5
Inverse = (X'X)-1 =

0

0

.125

0

0

B=(X'X)-1(X'Y) = -0.5
 

0

0

0

.125

0

  -1.0
 

0

0

0

0

.125

  1.25

 

Therefore B0 = 6.0, B1 = -1.5, B2 = -.5, B3 = -1.0 and e = 1.25

 

2.4 Calculations for the ANOVA Table

 

Temperature x1

Pressure x2

X1,1

X1,2

S x2

X2,1

6,8

6,10

30

X2,2

5,7

2,4

18

S x1

26

22

48

Sx2 = 330

CM is called the correction for the mean.

Let CM = (Sx)2/nT,

CM = (48)2/8 = 288

 

SST = SS (Total) = Sx2 - CM = 330 - 288 = 42

 

SSt = SS (Temperature) = S(S columns)2/n(col) - CM = [(26)2 + (22)2]/4 - 288

= 290 - 288 = 2

 

SSp = SS (Pressure) = S(S rows)2/n(rows) - CM = [(30)2 + (18)2]/4 - 288

= 306 - 288 = 18

 

SSi = SS (Interaction) = S(S cells)2/n(cell) - SSt - SSp - CM

= [(14)2 + (12)2 +(16)2 + (6)2]/2 - 2 - 18 - 288 = 316 - 2 - 18 = 8

 

SSr = SS (Residual) = [SS (Total) - SS (Temperature) - SS (Pressure) - SS (Interaction)]

= 42 - 2 - 18 - 8 = 14

 

2.5 Degrees of Freedom

Total: d.f. = n - 1 = 8 - 1 = 7

Temperature: d.f. = No. levels - 1 = 2 - 1 = 1

Pressure: d.f. = No. levels - 1 = 2 - 1 = 1

Interaction: d.f. = No. levels - 1 = 2 - 1 = 1

Residual: Total d.f. - S d.f. of each factor = 7 - 3 = 4


3.0 LATIN SQUARE DESIGN

A Latin square is an experimental design in which each level of each factor is combined only once with each level of two other factors or variables. A Latin square design implies the presence of three qualitative variables: rows, columns and treatments. In a Latin square experiment, it is assumed that no interactions between the variables exist.

Example 2

The following is an experiment to study the effects of four different additives on the fuel consumption of automobiles when four different drivers drive four different cars. The letters A, B, C, and D represent the additives. The additives are the treatments. The numbers in parentheses are the yields in miles per gallon. This is a 4 X 4 Latin square experiment.

 

 

CAR

DRIVER

1 2 3 4 S x(rows)
I D (20) A (21) B (26) C (25) 92
II A (20) D (23) C (26) B (27) 96
III C (16) B (15) D (13) A (16) 60
IV B (20) C (17) A (15) D (20) 72
S x(columns) 76 76 80 88 320

Sx2 = 6696

Sx(TREATMENTS): A = 72, B = 88, C = 84, D = 76

 

3.1 The Linear Model (regression equation)

Y = Grand Mean + Driver effects + Car effects + additive effects + Residual (e)

Y = B0 + B1x1 + B2x2 + B3x3 + B4x4 + B5x5 + B6x6 + B7x7 + B8x8 + B9x9 + e

Y = B0 + driver differences + Car differences + Additive differences + e

The calculations for the coefficients of the linear model are left to the student. When the coefficients are computed and the linear model is set up, the coded values for any car, driver or additive are entered. The fuel consumption (Y) is then determined from the equation.

 

3.2 Analysis of Variance (ANOVA) Table

Source SS d.f. MS F FCritical
Cars 24

3

8 3.0 4.76
Drivers 216

3

72 27.0 4.76
Additives 40

3

13.33 5.0 4.76
Residual 16

6

2.67    
Total 296

15

     

For a = .05, the critical F value is 4.76. Therefore, there is a significant difference between drivers and also between additives. The drivers have the greatest impact on the outcome.

 

3.3 Calculations for the ANOVA Table

CM = (Sx)2/n = (320)2/16 = 6400

SST = SS (Total) = Sx2 - CM = 6696 - 6400 = 296

SSd = SS (Drivers) = S(S Rows)2/n(rows) - CM

= [(92)2 + (96)2 + (60)2 + (72)2]/4 - 6400 = 6616 - 6400 = 216

SSc = SS (Cars) = S(S Columns)2/n(columns) - CM

=[(76)2 + (76)2 + (80)2 + (88)2]/4 - 6400 = 6424 - 6400 = 24

SSa = SS (Additives) = S(S treatments)2/n(treatments) - CM

= [(72)2 + (88)2 + (84)2 + (76)2]/4 - 6400 = 6440 - 6400 = 40

SSr = SS (Residual or Error) = SST - SSc - SSd - SSa = 296 - 24 - 216 - 40 = 16

 

3.4 Degrees of Freedom

Total: d.f. = 16 - 1 = 15

Drivers: d.f. = 4 - 1 = 3

Cars: d.f. = 4 - 1 = 3

Additives: d.f. = 4 - 1 = 3

Residual: d.f = 15 - 3 - 3 - 3 = 6

 

4.0 GLOSSARY OF TERMS