4.3.1 Fitted Effects for 2-Factor Studies

Factor A → I levels

Factor B → J levels

Equal replication, n replicates, in each treatment group.

→ Balanced design.

Problem 8: A = charge size with I = 2 levels of 2.5 ml and 5.0 ml

B = propellant with J = 3 levels of lighter fluid, gasoline and

carburetor fluid

Propellant

Lighter Carburetor

Charge Fluid Gasoline Fluid

58 50 76 79 90 86

2.5 43 49 84 73 79 82

59 71 86

5.0 65 59 96 101 107 102

61 68 94 91 91 95

67 87 97

First, plot the data!

As always the first step is to plot the data.

Checking for

· Effects of factors

o Main effects

o Interactions

· Outliers

· Changes in variances

If we have only 2 factors this is relatively easy.

It helps to offset values from the same treatment so they do not overlap.

One way to accomplish this would be to let the horizontal axis for A be

A = A + 0.1*(B-1)

Notation: = sample mean with A at level i and B at level j

Propellant

j = 1 j = 2 j = 3

Charge Lighter Gasoline Carburetor

i = 1 2.5 ml

i = 2 5.0 ml

= sample mean for A at level i = sample mean for B at level j

overall mean of all values

Propellant

Charge Lighter Gasoline Carburetor

2.5 ml =53.8 =76.6 =84.6 =71.67

5.0 ml =64.0 =93.8 =98.4 =85.4

=58.9 =85.2 =91.5 =78.53

Since these data are balanced we can find either by

· averaging all 15 values for the 2.5 ml charge

· averaging the averages

If the number of values in each mean had not been equal, we would need to take a weighted average of these averages to find the overall average of all points.

Since these data are balanced we can find either by

· averaging all 30 values

· averaging the averages

For factorial data, effects of factors are described as

· main effects

· interactions

The main effect of B = lighter fluid describes on average how much distance is gained or lost using lighter fluid compared to the overall average.

Lighter fluid mean = = 58.9 Overall mean = = 78.53

Effect of lighter fluid = 58.9 – 78.53 = -19.63 = b1

On average (over all charges) lighter fluid propelled balls 19.63 feet than the overall average.

Effect of gasoline = 85.2 – 78.53 = 6.67 = b2

Effect of carb fluid = 91.5 – 78.53 = 12.97 = b1

Check: -19.63 + 6.67 + 12.97 = 0.01 = 0 except for round-off.

Deviations from mean sum to zero.

The fitted main effect for factor B at level j is

bj =

The fitted main effect for factor A at level i is

ai =

Effect of charge 2.5 = 71.67 – 78.53 = -6.87 = a1

Effect of charge 5.0 = 85.4 – 78.53 = 6.87 = a2

Check: -6.87 + 6.87 = 0.

A = Propellant

Charge LF Gasoline Carb F Mean

2.5 ml 71.67 a1 = 71.67 – 78.53

5.0 ml 85.40 a2 = 81.40 – 78.53

Mean 58.9 85.2 91.5 =78.53

b1 = 58.9 – 78.53 b2 = 85.2 – 78.53 b3 = 91.5 – 78.53

Interactions check the extent to which main effects are consistent at different levels of the other factor.

Are the propellant effects the same for each charge?

Are the charge effects the same at each propellant?

In this case the effects of the propellants are similar for both charges.

The increase in distance going from lighter to gas or gas to carburetor fluid is pretty similar for both charges.

The lines are fairly parallel, particularly in view of variability in plot above.

A situation with absolutely no interaction would have exactly parallel lines.

The interaction between factor A and B is denoted AB or A*B.

The corresponding effect sizes for interactions are abij, analogous to ai and bj.

abij measures the extent to which deviates from a fit with parallel lines.

To fit parallel lines, the fitted values depend only on main effects, no interaction terms.

For parallel profiles

Fitting parallel lines:

Fitted value for a=1 and b=1, charge 2.5 and propellant lighter fluid

78.53 – 6.87 – 19.63 = 52.03

Compared to the overall average, we lose

· 6.87 feet using charge 2.5

· 19.63 feet using lighter fluid

The deviation of =53.8 from this parallel lines, no interaction fit is

ab11 = 53.8 – 52.03 = 1.77

Using charge 1 and propellant 1 went 1.77 feet farther than expected than predicted with a no interaction model.

ab11 = added boost for that particular combination of charge and propellant.

Charge Lighter Gasoline Carburetor

2.5 ml =53.8 =76.6 =84.6

5.0 ml =64.0 =93.8 =98.4

=78.53

Computing the abij effects:

Propellant

Charge LF Gas C F

2.5 53.8 – (78.53 – 6.87 – 19.63) 76.6 – (78.53 – 6.87 + 6.67) etc a1 = – 6.87

5.0 64.0 – (78.53 + 6.87 – 19.63) 93.8 – (78.53 + 6.87 + 6.67 etc a2 = + 6.87

b1 = – 19.63 b2 = + 6.67 etc

For the propellant example

a1 = -6.87

a2 = 6.87

b1 = -19.63

b2 = 6.67

b3 = 12.97

Charge / Propellant / PARALLEL
i / j / / ai / bj / FIT / MEAN / abij
1 / 1 / 78.53 / -6.87 / -19.63 / 52.03 / 53.8 / 1.77
1 / 2 / 78.53 / -6.87 / 6.67 / 78.33 / 76.6 / -1.73
1 / 3 / 78.53 / -6.87 / 12.97 / 84.63 / 84.6 / -0.03
2 / 1 / 78.53 / 6.87 / -19.63 / 65.77 / 64 / -1.77
2 / 2 / 78.53 / 6.87 / 6.67 / 92.07 / 93.8 / 1.73
2 / 3 / 78.53 / 6.87 / 12.97 / 98.37 / 98.4 / 0.03

= 1.77

= -1.73

= -0.03

= -1.77

= 1.73

= 0.03

The fitted lines from a no interaction model are parallel.

The ab interaction effects measure how far each mean is from this parallel fit with no interaction.

Residuals

Section 4.3 Residuals and ANOVA Tables

The fitted value for each group is the mean for that treatment group

The model for the measured y value involves main effects and interactions

true population effect of ith level of A

true population effect of jth level of B

true population interaction effect

The residuals are

Plotting residuals

In addition to the initial plot, plot residuals to check for

· Equal variances

· Drift over time

· Normality

Plot residuals

· Versus predicted values (treatment group means)

· Versus run order from 1 to a∙b∙n

Also plot residuals versus

· A levels

· B levels

For only two factors, the plot of all original data points shows pretty well how for example variances change with the factors. For factorial studies with more than 2 factors, we can’t plot all original data points one plot, so it becomes more essential to plot residuals versus the factors.

Normal plot or plots

· Residuals.

o For all residuals together if variances appear similar throughout.

o For subsets of data with similar variances. For example if we have smaller variances for breaking strengths of one brand of bungee cord, we would draw separate normal plots on the same graph for each brand.

Possibly increasing variance with larger values.

Possibly increasing variance