Two-Factor Analysis of Variance

14.3 Two-Factor Analysis of Variance

To present general formulas for the analysis of variance of a two-factor experiment using repeated observations in a completely randomized design, we shall consider the case of n replications of the treatment combinations determined by a levels of factor A and b levels of factor B. The observations may be classified by means of a rectangular array where the rows represent the levels of factor A and the columns represent the levels of factor B. Each treatment combination defines a cell in our array. Thus, we have ab cells, each cell containing n observations. Denoting the kth observation taken at the ith level of factor A and the jth level of factor B by y ijk , Table 14.1 shows the abn observations.

Table 14.1: Two-Factor Experiment with n Replications

A 1 2 ···

b Total Mean

The observations in the (ij)th cell constitute a random sample of size n from a population that is assumed to be normally distributed with mean μ ij and variance σ 2 . All ab populations are assumed to have the same variance σ 2 . Let us define

566 Chapter 14 Factorial Experiments (Two or More Factors) the following useful symbols, some of which are used in Table 14.1:

Y ij. = sum of the observations in the (ij)th cell, Y i.. = sum of the observations for the ith level of factor A, Y .j. = sum of the observations for the jth level of factor B, Y ... = sum of all abn observations, y ¯ ij. = mean of the observations in the (ij)th cell, y ¯ i.. = mean of the observations for the ith level of factor A, y ¯ .j. = mean of the observations for the jth level of factor B, y ¯ ... = mean of all abn observations.

Unlike in the one-factor situation covered at length in Chapter 13, here we are assuming that the populations, where n independent identically distributed ob- servations are taken, are combinations of factors. Also we will assume throughout that an equal number (n) of observations are taken at each factor combination. In cases in which the sample sizes per combination are unequal, the computations are more complicated but the concepts are transferable.

Model and Hypotheses for the Two-Factor Problem

Each observation in Table 14.1 may be written in the form y ijk =μ ij +ǫ ijk ,

where ǫ ijk measures the deviations of the observed y ijk values in the (ij)th cell from the population mean μ ij . If we let (αβ) ij denote the interaction effect of the ith level of factor A and the jth level of factor B, α i the effect of the ith level of factor A, β j the effect of the jth level of factor B, and μ the overall mean, we can write

μ ij =μ+α i +β j + (αβ) ij ,

and then

y ijk =μ+α i +β j + (αβ) ij +ǫ ijk ,

on which we impose the restrictions

The three hypotheses to be tested are as follows:

1. H ′ 0 :α 1 =α 2 =···=α a = 0,

H ′ 1 : At least one of the α i is not equal to zero.

1 : At least one of the β j is not equal to zero.

14.3 Two-Factor Analysis of Variance 567

3. H ′′′

0 : (αβ) 11 = (αβ) 12 = · · · = (αβ) ab = 0,

1 : At least one of the (αβ) ij is not equal to zero. We warned the reader about the problem of masking of main effects when inter-

H ′′′

action is a heavy contributor in the model. It is recommended that the interaction test result be considered first. The interpretation of the main effect test follows, and the nature of the scientific conclusion depends on whether interaction is found. If interaction is ruled out, then hypotheses 1 and 2 above can be tested and the interpretation is quite simple. However, if interaction is found to be present the interpretation can be more complicated, as we have seen from the discussion of the drying time and temperature in the previous section. In what follows, the structure of the tests of hypotheses 1, 2, and 3 will be discussed. Interpretation of results will be incorporated in the discussion of the analysis in Example 14.1.

The tests of the hypotheses above will be based on a comparison of independent estimates of σ 2 provided by splitting the total sum of squares of our data into four components by means of the following identity.

Partitioning of Variability in the Two-Factor Case

Theorem 14.1: Sum-of-Squares Identity

Symbolically, we write the sum-of-squares identity as SST = SSA + SSB + SS(AB) + SSE,

where SSA and SSB are called the sums of squares for the main effects A and

B, respectively, SS(AB) is called the interaction sum of squares for A and B, and SSE is the error sum of squares. The degrees of freedom are partitioned according to the identity

abn − 1 = (a − 1) + (b − 1) + (a − 1)(b − 1) + ab(n − 1).

Formation of Mean Squares

If we divide each of the sums of squares on the right side of the sum-of-squares identity by its corresponding number of degrees of freedom, we obtain the four statistics

S 2 SSA

2 SSB

2 SS(AB)

(a − 1)(b − 1)

ab(n − 1) All of these variance estimates are independent estimates of σ 2 under the condition

that there are no effects α i ,β j , and, of course, (αβ) ij . If we interpret the sums of

568 Chapter 14 Factorial Experiments (Two or More Factors) squares as functions of the independent random variables y 111 ,y 112 ,...,y abn , it is

not difficult to verify that

E(S 2 3 SS(AB) )=E

=σ 2 + n

(αβ) 2 ij ,

(a − 1)(b − 1)

(a − 1)(b − 1) i=1 j=1

ab(n − 1)

from which we immediately observe that all four estimates of σ 2 are unbiased when

0 ,H 0 , and H 0 are true.

To test the hypothesis H 0 ′ , that the effects of factors A are all equal to zero, we compute the following ratio:

F-Test for

f = Factor A 1 1

which is a value of the random variable F ′ 1 having the F-distribution with a − 1 and ab(n−1) degrees of freedom when H 0 is true. The null hypothesis is rejected at the α-level of significance when f 1 >f α [a − 1, ab(n − 1)].

Similarly, to test the hypothesis H ′′

0 that the effects of factor B are all equal to zero, we compute the following ratio:

F-Test for

Factor B

which is a value of the random variable F 2 ′′ having the F-distribution with b − 1 and ab(n − 1) degrees of freedom when H 0 is true. This hypothesis is rejected at the α-level of significance when f 2 >f α [b − 1, ab(n − 1)].

Finally, to test the hypothesis H ′′′

0 , that the interaction effects are all equal to zero, we compute the following ratio:

F-Test for

Interaction

which is a value of the random variable F 3 having the F-distribution with ′′′ (a − 1)(b − 1) and ab(n − 1) degrees of freedom when H 0 is true. We con- clude that, at the α-level of significance, interaction is present when f 3 >

f α [(a − 1)(b − 1), ab(n − 1)].

As indicated in Section 14.2, it is advisable to interpret the test for interaction before attempting to draw inferences on the main effects. If interaction is not sig- nificant, there is certainly evidence that the tests on main effects are interpretable. Rejection of hypothesis 1 on page 566 implies that the response means at the levels

14.3 Two-Factor Analysis of Variance 569 of factor A are significantly different, while rejection of hypothesis 2 implies a simi-

lar condition for the means at levels of factor B. However, a significant interaction could very well imply that the data should be analyzed in a somewhat different manner—perhaps observing the effect of factor A at fixed levels of factor

B, and so forth. The computations in an analysis-of-variance problem, for a two-factor experi- ment with n replications, are usually summarized as in Table 14.2.

Table 14.2: Analysis of Variance for the Two-Factor Experiment with n Replications Source of

Computed Variation

Sum of

Degrees of

Mean

f Main effect:

Squares Freedom

f 2 = s 2 2 Two-factor

interactions: s AB 2 SS(AB) (a − 1)(b − 1) s 2

f 3 = 3 s 2 Error

3 = SS(AB) (a−1)(b−1)

SSE

ab(n − 1)

s 2 = SSE ab(n−1)

Example 14.1: In an experiment conducted to determine which of 3 different missile systems is preferable, the propellant burning rate for 24 static firings was measured. Four dif- ferent propellant types were used. The experiment yielded duplicate observations of burning rates at each combination of the treatments.

The data, after coding, are given in Table 14.3. Test the following hypotheses: ′ (a) H 0 : there is no difference in the mean propellant burning rates when different ′′ missile systems are used, (b) H 0 : there is no difference in the mean propellant ′′′ burning rates of the 4 propellant types, (c) H 0 : there is no interaction between

the different missile systems and the different propellant types.

Table 14.3: Propellant Burning Rates Missile

Propellant Type

Solution : ′ 1. (a) H :α 1 =α 2 =α 3 = 0.

(b) H 0 :β 1 =β 2 =β 3 =β 4 = 0.

570 Chapter 14 Factorial Experiments (Two or More Factors)

(c) H ′′′

1 : At least one of the α i is not equal to zero. (b) H ′′

2. (a) H ′

1 : At least one of the β j is not equal to zero. (c) H ′′′

1 : At least one of the (αβ) ij is not equal to zero. The sum-of-squares formula is used as described in Theorem 14.1. The analysis

of variance is shown in Table 14.4. Table 14.4: Analysis of Variance for the Data of Table 14.3

Source of

Sum of

Degrees of

Mean Computed

Missile system

Propellant type

The reader is directed to a SAS GLM Procedure (General Linear Models) for analysis of the burning rate data in Figure 14.2. Note how the “model” (11 degrees of freedom) is initially tested and the system, type, and system by type interac- tion are tested separately. The F-test on the model (P = 0.0030) is testing the accumulation of the two main effects and the interaction.

0 and conclude that different missile systems result in different mean propellant burning rates. The P -value is approximately 0.0169. (b) Reject H ′′

(a) Reject H ′

0 and conclude that the mean propellant burning rates are not the same for the four propellant types. The P -value is approximately 0.0010.

(c) Interaction is barely insignificant at the 0.05 level, but the P -value of approx- imately 0.0513 would indicate that interaction must be taken seriously.

At this point we should draw some type of interpretation of the interaction. It should be emphasized that statistical significance of a main effect merely implies that marginal means are significantly different. However, consider the two-way table of averages in Table 14.5.

Table 14.5: Interpretation of Interaction

b 1 b 2 b 3 b 4 Average

It is apparent that more important information exists in the body of the table— trends that are inconsistent with the trend depicted by marginal averages. Table

14.5 certainly suggests that the effect of propellant type depends on the system

14.3 Two-Factor Analysis of Variance 571

The GLM Procedure

Dependent Variable: rate

Sum of

Source

Pr > F Model

DF Squares

Mean Square

F Value

Corrected Total

R-Square Coeff Var

Root MSE

rate Mean

Pr > F system

DF Type III SS

Mean Square

F Value

10.75 0.0010 system*type

2.97 0.0512 Figure 14.2: SAS printout of the analysis of the propellant rate data of Table 14.3.

being used. For example, for system 3 the propellant-type effect does not appear to be important, although it does have a large effect if either system 1 or system

2 is used. This explains the “significant” interaction between these two factors. More will be revealed subsequently concerning this interaction.

Example 14.2: Referring to Example 14.1, choose two orthogonal contrasts to partition the sum of squares for the missile systems into single-degree-of-freedom components to be used in comparing systems 1 and 2 versus 3, and system 1 versus system 2.

Solution : The contrast for comparing systems 1 and 2 with 3 is

w 1 =μ 1. +μ 2. − 2μ 3. .

A second contrast, orthogonal to w 1 , for comparing system 1 with system 2, is given by w 2 =μ 1. −μ 2. . The single-degree-of-freedom sums of squares are

Notice that SSw 1 + SSw 2 = SSA, as expected. The computed f-values corre- sponding to w 1 and w 2 are, respectively,

Compared to the critical value f 0.05 (1, 12) = 4.75, we find f 1 to be significant. In fact, the P-value is less than 0.01. Thus, the first contrast indicates that the

572 Chapter 14 Factorial Experiments (Two or More Factors) hypothesis

H 0 : (μ 1. +μ 2. )=μ 3.

is rejected. Since f 2 < 4.75, the mean burning rates of the first and second systems are not significantly different.

Impact of Significant Interaction in Example 14.1

If the hypothesis of no interaction in Example 14.1 is true, we could make the general comparisons of Example 14.2 regarding our missile systems rather than separate comparisons for each propellant. Similarly, we might make general com- parisons among the propellants rather than separate comparisons for each missile system. For example, we could compare propellants 1 and 2 with 3 and 4 and also propellant 1 versus propellant 2. The resulting f-ratios, each with 1 and 12 degrees of freedom, turn out to be 24.81 and 7.39, respectively, and both are quite significant at the 0.05 level.

From propellant averages there appears to be evidence that propellant 1 gives the highest mean burning rate. A prudent experimenter might be somewhat cau- tious in drawing overall conclusions in a problem such as this one, where the f-ratio for interaction is barely below the 0.05 critical value. For example, the overall evi- dence, 31.60 versus 29.85 on the average for the two propellants, certainly indicates that propellant 1 is superior, in terms of a higher burning rate, to propellant 2. However, if we restrict ourselves to system 3, where we have an average of 28.85 for propellant 1 as opposed to 28.10 for propellant 2, there appears to be little or no difference between these two propellants. In fact, there appears to be a stabilization of burning rates for the different propellants if we operate with sys- tem 3. There is certainly overall evidence which indicates that system 1 gives a higher burning rate than system 3, but if we restrict ourselves to propellant 4, this conclusion does not appear to hold.

The analyst can conduct a simple t-test using average burning rates for system

3 in order to display conclusive evidence that interaction is producing considerable difficulty in allowing broad conclusions on main effects. Consider a comparison of propellant 1 against propellant 2 only using system 3. Borrowing an estimate of

σ 2 from the overall analysis, that is, using s 2 = 1.24 with 12 degrees of freedom, we have

which is not even close to being significant. This illustration suggests that one must

be cautious about strict interpretation of main effects in the presence of interaction.

Graphical Analysis for the Two-Factor Problem of Example 14.1

Many of the same types of graphical displays that were suggested in the one-factor problems certainly apply in the two-factor case. Two-dimensional plots of cell means or treatment combination means can provide insight into the presence of

14.3 Two-Factor Analysis of Variance 573 interactions between the two factors. In addition, a plot of residuals against fitted

values may well provide an indication of whether or not the homogeneous variance assumption holds. Often, of course, a violation of the homogeneous variance as- sumption involves an increase in the error variance as the response numbers get larger. As a result, this plot may point out the violation.

Figure 14.3 shows the plot of cell means in the case of the missile system propellant illustration in Example 14.1. Notice how graphically (in this case) the lack of parallelism shows through. Note the flatness of the part of the figure showing the propellant effect for system 3. This illustrates interaction among the factors. Figure 14.4 shows the plot of residuals against fitted values for the same data. There is no apparent sign of difficulty with the homogeneous variance assumption.

Figure 14.3: Plot of cell means for data of Example 14.1. Numbers represent missile systems.

Figure 14.4: Residual plot of data of Example 14.1.

574 Chapter 14 Factorial Experiments (Two or More Factors)

Example 14.3: An electrical engineer is investigating a plasma etching process used in semicon- ductor manufacturing. It is of interest to study the effects of two factors, the C 2 F 6 gas flow rate (A) and the power applied to the cathode (B). The response is the etch rate. Each factor is run at 3 levels, and 2 experimental runs on etch rate are made for each of the 9 combinations. The setup is that of a completely randomized design. The data are given in Table 14.6. The etch rate is in A ◦ /min.

Table 14.6: Data for Example 14.3

Power Supplied

C 2 F 6 Flow Rate

801 The levels of the factors are in ascending order, with level 1 being low level and

level 3 being the highest. (a) Show an analysis of variance table and draw conclusions, beginning with the

test on interaction. (b) Do tests on main effects and draw conclusions. Solution : A SAS output is given in Figure 14.5. From the output we learn the following.

The GLM Procedure

Dependent Variable: etchrate

Sum of

Pr > F Model

Source

DF Squares

Mean Square

F Value

Corrected Total

R-Square Coeff Var

Root MSE

etchrate Mean

Pr > F c2f6

Source

DF Type III SS

Mean Square

F Value

<.0001 c2f6*power

Figure 14.5: SAS printout for Example 14.3. (a) The P-value for the test of interaction is 0.4485. We can conclude that there

is no significant interaction.

(b) There is a significant difference in mean etch rate for the 3 levels of C 2 F 6 flow rate. Duncan’s test shows that the mean etch rate for level 3 is significantly

Exercises 575 higher than that for level 2 and the rate for level 2 is significantly higher than

that for level 1. See Figure 14.6(a). There is a significant difference in mean etch rate based on the level of power to the cathode. Duncan’s test revealed that the etch rate for level 3 is sig- nificantly higher than that for level 2 and the rate for level 2 is significantly higher than that for level 1. See Figure 14.6(b).

Duncan Grouping

Mean

N c2f6

Duncan Grouping

Mean N power

Figure 14.6: SAS output, for Example 14.3. (a) Duncan’s test on gas flow rate; (b) Duncan’s test on power.

Exercises

14.1 An experiment was conducted to study the ef- until it was tested. The results, in milligrams of ascor- fects of temperature and type of oven on the life of bic acid per liter, were recorded. Use a 0.05 level of a particular component. Four types of ovens and significance to test the hypothesis that 3 temperature levels were used in the experiment. (a) there is no difference in ascorbic acid contents

Twenty-four pieces were assigned randomly, two to among the different brands of orange juice concen- each combination of treatments, and the following re-

trate;

sults recorded. (b) there is no difference in ascorbic acid contents for

the different time periods; Temperature ( F)

Oven

O 1 O 2 O 3 O 4 (c) the brands of orange juice concentrate and the 500

number of days from the time the juice was blended

until it was tested do not interact. 550

Time (days)

42.8 53.2 40.4 47.6 Using a 0.05 level of significance, test the hypothesis Sealed-Sweet

48.0 47.0 48.5 43.3 (a) different temperatures have no effect on the life of

Minute Maid

48.2 49.6 45.2 47.6 the component;

(b) different ovens have no effect on the life of the com- 14.3 Three strains of rats were studied under 2 envi- ponent;

ronmental conditions for their performance in a maze (c) the type of oven and temperature do not interact.

test. The error scores for the 48 rats were recorded. Strain

Mixed Dull constituted frozen orange juice concentrate stored in a

14.2 To ascertain the stability of vitamin C in re-

Environment

Bright

33 83 101 94 refrigerator for a period of up to one week, the study

Free

36 14 33 56 Vitamin C Retention in Reconstituted Frozen Orange

41 76 122 83 Juice was conducted by the Department of Human Nu-

22 58 35 23 trition and Foods at Virginia Tech. Three types of

60 89 136 120 frozen orange juice concentrate were tested using 3 dif-

Restricted

35 126 38 153 ferent time periods. The time periods refer to the num-

83 110 64 128 ber of days from when the orange juice was blended

576 Chapter 14 Factorial Experiments (Two or More Factors)

Five different muscles

Use a 0.01 level of significance to test the hypothesis that

4: middle deltoid (a) there is no difference in error scores for different

1: anterior deltoid

5: triceps environments;

2: pectorial major

3: posterior deltoid

(b) there is no difference in error scores for different were tested on each of 3 subjects, and the experiment strains;

was carried out 3 times for each treatment combina- (c) the environments and strains of rats do not inter- tion. The electromyographic data, recorded during the act.

serve, are presented here.

Muscle 14.4 Corrosion fatigue in metals has been defined as

1 2 3 4 5 the simultaneous action of cyclic stress and chemical

Subject

1 32 5 58 10 19 attack on a metal structure. A widely used technique

59 1.5 61 10 20 for minimizing corrosion fatigue damage in aluminum

38 2 66 14 23 involves the application of a protective coating. A

study conducted by the Department of Mechanical En- 2 63 10 64 45 43 gineering at Virginia Tech used 3 different levels of hu-

60 9 78 61 61 midity

Low: 20–25% relative humidity 54 43 29 46 85 Medium: 55–60% relative humidity

47 42 23 55 95 High: 86–91% relative humidity

Use a 0.01 level of significance to test the hypothesis that

and 3 types of surface coatings (a) different subjects have equal electromyographic

measurements;

Uncoated: no coating (b) different muscles have no effect on electromyo- Anodized: sulfuric acid anodic oxide coating

graphic measurements; (c) subjects and types of muscle do not interact.

Conversion: chromate chemical conversion coating The corrosion fatigue data, expressed in thousands of

14.6 An experiment was conducted to determine cycles to failure, were recorded as follows:

whether additives increase the adhesiveness of rubber products. Sixteen products were made with the new

Relative Humidity additive and another 16 without the new additive. The Coating

observed adhesiveness was as recorded below. 361 469

Temperature ( ◦ C) Uncoated

Without Additive

With Additive

(a) Perform an analysis of variance with α = 0.05 to 4.2 3.5 3.6 3.9 test for significant main and interaction effects.

(b) Use Duncan’s multiple-range test at the 0.05 level Perform an analysis of variance to test for significant main and interaction effects. of significance to determine which humidity levels result in different corrosion fatigue damage.

14.7 The extraction rate of a certain polymer is 14.5 To determine which muscles need to be sub- known to depend on the reaction temperature and jected to a conditioning program in order to improve the amount of catalyst used. An experiment was con- one’s performance on the flat serve used in tennis, a ducted at four levels of temperature and five levels of study was conducted by the Department of Health, the catalyst, and the extraction rate was recorded in Physical Education and Recreation at Virginia Tech. the following table.

Exercises 577 Amount of Catalyst

(c) Do secondary tests that will allow the engineer to 0.5%

learn the true impact of cutting speed. 50 C 38 45 57 59 57 (d) Show a plot that graphically displays the interac-

tion effect.

60 C 44 56 70 73 61

43 57 69 72 58 14.10 Two factors in a manufacturing process for an 70 ◦ C 44 56 70 73 61 integrated circuit are studied in a two-factor experi-

47 60 67 61 59 ment. The purpose of the experiment is to learn their 80 ◦ C 49 62 70 62 53 effect on the resistivity of the wafer. The factors are

47 65 55 69 58 implant dose (2 levels) and furnace position (3 levels). Experimentation is costly so only one experimental run Perform an analysis of variance. Test for significant is made at each combination. The data are as follows.

main and interaction effects.

Position 14.8 In Myers, Montgomery, and Anderson-Cook

Dose

1 15.5 14.8 21.3 (2009), a scenario is discussed involving an auto

2 27.2 24.9 26.1 bumper plating process. The response is the thickness It is to be assumed that no interaction exists between

of the material. Factors that may impact the thickness these two factors. include amount of nickel (A) and pH (B). A two-factor (a) Write the model and explain terms. experiment is designed. The plan is a completely ran- (b) Show the analysis-of-variance table. domized design in which the individual bumpers are assigned randomly to the factor combinations. Three (c) Explain the 2 “error” degrees of freedom. levels of pH and two levels of nickel content are involved (d) Use Tukey’s test to do multiple-comparison tests on in the experiment. The thickness data, in cm × 10 −3 ,

furnace position. Explain what the results show. are as follows: Nickel Content

14.11 A study was done to determine the impact of (grams)

pH

5 5.5 6 two factors, method of analysis and the laboratory do- ing the analysis, on the level of sulfur content in coal.

Twenty-eight coal specimens were randomly assigned

to 14 factor combinations, the structure of the experi-

mental units represented by combinations of seven lab-

88 69 oratories and two methods of analysis with two speci-

mens per factor combination. The data, expressed in

72 percent of sulfur, are as follows: (a) Display the analysis-of-variance table with tests for

Method both main effects and interaction. Show P-values.

1 2 (b) Give engineering conclusions.

Laboratory

0.105 0.105 0.108 learned from the analysis of the data?

What have you

0.122 0.127 0.124 (c) Show a plot that depicts either a presence or an

0.112 0.109 0.111 absence of interaction.

0.096 0.110 0.097 14.9 An engineer is interested in the effects of cut-

0.119 0.116 0.122 ting speed and tool geometry on the life in hours of

a machine tool. Two cutting speeds and two different (The data are taken from G. Taguchi, “Signal to geometries are used. Three experimental tests are ac- Noise Ratio and Its Applications to Testing Material,” complished at each of the four combinations. The data Reports of Statistical Application Research, Union of are as follows.

Japanese Scientists and Engineers, Vol. 18, No. 4, Tool

Cutting Speed

(a) Do an analysis of variance and show results in an

analysis-of-variance table.

(b) Is interaction significant? If so, discuss what it (a) Show an analysis-of-variance table with tests on in-

means to the scientist. Use a P-value in your con- teraction and main effects.

clusion.

(b) Comment on the effect that interaction has on the (c) Are the individual main effects, laboratory, and test on cutting speed.

method of analysis statistically significant? Discuss

578 Chapter 14 Factorial Experiments (Two or More Factors) what is learned and let your answer be couched in

treatment influence magnesium uptake. the context of any significant interaction.

(d) Fit the appropriate regression model with treat- (d) Do an interaction plot that illustrates the effect of

ment as a categorical variable. Include interaction interaction.

in the model.

(e) Do a test comparing methods 1 and 2 at laboratory (e) Is interaction significant in the regression model? 1 and do the same test at laboratory 7. Comment on what these results illustrate.

14.14 Consider the data set in Exercise 14.12 and an- swer the following questions.

14.12 In an experiment conducted in the Civil Engi- neering Department at Virginia Tech, growth of a cer- (a) Both factors, copper and time, are quantitative in tain type of algae in water was observed as a function

nature. As a result, a regression model may be of of time and the dosage of copper added to the water.

interest. Describe what might be an appropriate model using x = copper content and x

The data are as follows. Response is in units of algae. 1 2 = time.

Fit the model to the data, showing regression coef-

ficients and a t-test on each. Copper

Time in Days

5 12 18 (b) Fit the model

0.34 0.36 0.23 Y=β 0 +β 1 x 1 +β 2 x 2 +β 12 x 1 x 2

0.22 0.31 0.25 and compare it to the one you chose in (a). Which is more appropriate? Use R 2 as a criterion. 3 0.20 0.30 0.27 adj

0.24 0.30 0.25 14.15 The purpose of the study The Incorporation of (a) Do an analysis of variance and show the analysis-

a Chelating Agent into a Flame Retardant Finish of a of-variance table.

Cotton Flannelette and the Evaluation of Selected Fab- ric Properties, conducted at Virginia Tech, was to eval-

(b) Comment concerning whether the data are suffi- uate the use of a chelating agent as part of the flame cient to show a time effect on algae concentration. retardant finish of cotton flannelette by determining its

(c) Do the same for copper content. Does the level of effect upon flammability after the fabric is laundered copper impact algae concentration?

under specific conditions. There were two treatments (d) Comment on the results of the test for interaction. at two levels. Two baths were prepared, one with car-

How is the effect of copper content influenced by boxymethyl cellulose (bath I) and one without (bath time?

II). Half of the fabric was laundered 5 times and half was laundered 10 times. There were 12 pieces of fab-

14.13 In Myers, Classical and Modern Regression ric in each bath/number of launderings combination. with Applications (Duxbury Classic Series, 2nd edition, After the washings, the lengths of fabric that burned 1990), an experiment is described in which the Envi- and the burn times were measured. Burn times (in ronmental Protection Agency seeks to determine the seconds) were recorded as follows: effect of two water treatment methods on magnesium

Bath II uptake. Magnesium levels in grams per cubic centime-

Launderings

Bath I

ter (cc) are measured, and two different time levels are 5 13.7 23.0 15.7 6.2 5.4 25.5 5.0 15.8 14.8 4.4 5.0 3.3

incorporated into the experiment. The data are as fol- 14.0 29.4 9.7 16.0 2.5 1.6 lows:

10 27.2 16.8 12.9 18.2 8.8 14.5 Time (hr)

14.2 27.4 11.5 17.7 18.3 9.9 (a) Do an interaction plot. What is your impression?

(b) Do an analysis of variance and show tests for the (a) Perform an analysis of variance. Is there a signifi- main effects and interaction.

cant interaction term?

(c) Give scientific findings regarding how time and (b) Are there main effect differences? Discuss.

14.4 Three-Factor Experiments 579