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.
- Factor -
A variable or attribute that influences or is suspected of influencing the characteristic being investigated.
- Level -
The values of a factor being examined in an experiment.
- Treatment -
One set of levels for all factors employed in a given experiment.
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:
- The factors, if any, that have a significant impact on the yield
- The variable or factor, if any, that has the greatest impact on the yield
- The linear model (regression line) for the yield or response (Y)
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 |
S
x2 = 330CM is called the correction for the mean.
Let CM = (
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 |
S
x2 = 6696S
x(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 = 6400SST = SS (Total) =
Sx2 - CM = 6696 - 6400 = 296SSd = 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
This is a technique for determining whether
there are significant differences between two or more sample averages. Usually the t test
is used to test for differences between two sample averages and analysis of variance is
used to test differences between three or more.
The Latin square is used when one primary factor is under
investigation and the results may be affected by two other experimental variables. It is
assumed that no interactions exist. Each treatment occurs once in every row and once in
every column. The number of treatments must equal the number of rows and number of
columns.
Noise reducing and volume shifters.