Subject: BIA 654 Experiment Design II
Semester: Fall 2018

Date: 10/15/2018

This exercise was inspired by a real problem described in the article “Strategic Testing Stops Leaky
Litter Cartons in Their Tracks” (Packaging Digest, August 2001, Please find this paper in Canvas

to READ). The exercise resembles what the actual company did, but the data are not real.

The makers of “Cats Love It” cat litter are facing a serious problem. Retail customers are reporting

that cartons of the firm’s premium brand cat litter are leaking the product onto store shelves. The

company realizes that while cat lovers are used to cleaning stray sprays of litter tracked through the

house, they are not willing to put up with cartons that leak on the way home.

Management has determined that the problem is with the carton-sealing process. Cartons are filled

and sealed on a production line run by 20 workers. The company decides to perform 3-factor factorial

experiment. A run consists of filling and sealing 200 cartons. The factors to be tested and levels of

each are shown below.

Factor A is line speed with the minus level at 22 cartons per minute and the plus level at 30 cartons

per minute. Factor B is the pressure applied by the gluing machine, with the minus level being lower

pressure and the plus level being higher pressure. Factor C is the amount of glue used, with the plus

level being the current amount and the minus level being 40% less glue.

The response is the proportion(%) of cartons that leak, whose values are observed as 8, 45, 47, 10,

8, 40, 41, 8, listed in the standard order. For the questions below, you can do them either by hands

(recommended) or computer.

(a) Make the design matrix (with ‘+’ and ‘–’ symbols) that includes main effects and two- and threefactor

interactions.

(b) What are the estimated main effects of factor A and B? What is the estimated AC interaction?

(c) Display the effects graphically through main effects of C and interaction plots for BC and AC.

Answer:


A B C AB AC BC ABC Response
Standard Order Line
Speed

Pressure Amount of Glue
1 – – – + + + – 8
2 + – – – – + + 45
3 – + – – + – + 47
4 + + – + – – – 10
5 – – + + – – + 8
6 + – + – + – – 40
7 – + + – – + – 41
8 + + + + + + + 8
B)

Main effect Factor A=(45+10+40+8)/4-(8+47+41+8)/4=-0.25

B= (47+10+41+8)/4-(8+45+8+40)4=1.25

Interaction=(8+47+40+8)/4-(45+10+8+41)/4=-0.25

C) Sum(y+)/n+=(8+45+47+10)/4=27.5 and Sum(y-)/n-=(8+40+10+8)/4=24.5
10 8

8 8

27.5 24.5

41

47

45 40

+ +

–



+

– –

math Experiment案例math Experiment案例
(a) Suppose the three real estate appraisers independently examined each of five properties chosen at
random from a particular neighborhood and gave appraised values. The manager of the appraisal

service that employs them suspects that systematic differences among their methods of appraisal

are causing undesirable inconsistencies in appraised values provided by her service. The issue

is whether there are, in fact, systematic differences among the values reported by the three

appraisers. The data, APPRAISED VALUES (In thousands of dollars), are shown below.

Question: What kind of statistical analysis should you perform? Choose from: (Two sample ttest,

pairwise t-test, One-way ANOVA, One-way ANOVA with block, Two-way ANOVA, Factorial

design). If your answer is One-way ANOVA with block, which variable does play a role of block?

Appraiser

Property A B C

1 90 93 92

2 94 96 88

3 91 92 84

4 85 88 83

5 88 90 87

(b) Suppose each of 4 Chardonnay wines of the same vintage was judged by 5 judges. Each wine was

blinded and given to each judge in randomized order. The wines were scored on a 40-point scale,

with higher scores meaning better wine. Based on those scores, we want to test whether there is

a difference between the 4 Chardonnay wines in mean scores.

Question: What kind of statistical analysis should you perform? Choose: (Two sample t-test,

pairwise t-test, One-way ANOVA, One-way ANOVA with block, Two-way ANOVA, Factorial

design). If your answer is One-way ANOVA with block, which variable does play a role of block?

1

(c) A researcher was interested in whether an individual’s interest in politics was influenced by their

level of education and gender. Therefore, the dependent variable (response variable) was “interest

in politics” and the two independent variables were “gender” and “level of education.”

In particular, the researcher wanted to know whether there was an interaction between education

level and gender. Put another way, was the effect of level of education on interest in politics

different for males and females?

To answer this question, a random sample of 60 participants were recruited to take part in the

study–30 males and 30 females–equally split by level of education: school, college and university

(i.e., 10 participants in each group). Each participant in the study completed a questionnaire

that scored their interest in politics on a scale of 0 to 100, with higher scores indicating a greater

interest in politics.

Question: What kind of statistical analysis should you perform to answer for the original question?

(Two sample t-test, pairwise t-test, One-way ANOVA, One-way ANOVA with block, Two-way

ANOVA, Factorial design)

Answer:

a)one way anova c) two way anova

####building matrix for convenience

a <- matrix(

c(90,94,91,85,88,93,96,92,88,90,92,88,84,83,87),

nrow=5,

ncol=3)

A <- a[,1]

B <- a[,2]

C <- a[,3]

####combine data

combine <- data.frame(cbind(A,B,C))

combine

summary(combine)

stack.comb <- stack(combine)

stack.comb

propo <- c(1,2,3,4,5)

###anova

anva <- aov(values~ind,data = stack.comb)

summary(anva)

comb2 <- cbind(stack.comb,propo)

propo1 <- rep(propo,3)

comb2



anva2 <- aov(values~ind+propo1,data=stack.comb)

anva2



> anva

Call:

aov(formula = values ~ ind, data = stack.comb)



Terms:

ind Residuals

Sum of Squares   62.8     132.8

Deg. of Freedom     2        12



Residual standard error: 3.32666

Estimated effects may be unbalanced

> anva2

Call:

aov(formula = values ~ ind + propo1, data = stack.comb)



Terms:

ind propo1 Residuals

Sum of Squares  62.8   58.8      74.0

Deg. of Freedom    2      1        11



Residual standard error: 2.593699

Estimated effects may be unbalanced



B) two sample t test
> t.test1 <- t.test(a2,b2,alternative = “two.sided”, var.equal = FALSE)

> t.test2 <- t.test(a2,c2,alternative = “two.sided”, var.equal = FALSE)

> t.test3 <- t.test(b2,c2,alternative = “two.sided”, var.equal = FALSE)

> t.test1



Welch Two Sample t-test



data:  a2 and b2

t = -2.0327, df = 27.709, p-value = 0.05177

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-4.41809197  0.01809197

sample estimates:

mean of x mean of y

89.6      91.8



> t.test2



Welch Two Sample t-test



data:  a2 and c2

t = 2.391, df = 27.905, p-value = 0.02379

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

0.4007834 5.1992166

sample estimates:

mean of x mean of y

89.6      86.8



> t.test3



Welch Two Sample t-test



data:  b2 and c2

t = 4.4696, df = 27.303, p-value = 0.0001239

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

2.705863 7.294137

sample estimates:

mean of x mean of y

91.8      86.8





Recall the comparison between factorial design and one-factor-at-a-time (OFAT) approach considered in class: In the 22 factorial design (without replications), we conduct a total of 4 runs. The main effects of both factors (say A and B) are estimated by the difference between averages of two observations at the low and the high levels of each factor. Therefore, in order for the OFAT to achieve the same level of precision (that is, to obtain averages of two), one must have 2 runs at each experimental combination. Hence, for the 2 factor case, OFAT requires 2 × 3 = 6 total runs whereas 22 factorial design needs 2 2 = 4 runs. Now, suppose there are 5 factors (A, B, C, D, E), each having two-levels. We know 25 factorial design (without replications), we conduct a total of 32 runs. How many total runs are required in order for the OFAT approach to achieve the same precision? Explain why.
Answer:

Since this is 2^5 factorial OFAT approach needs 96 runs to achieve the similar result. In OFAT approach, we need to change one factor at a time. When change the a factor level, others remains the same. When it comes to 2 factorial design, it needs 4 runs while OFAT is 6. In Three factors it takes 2^3 runs, and 16 runs in OFAT. For 4 factorial design, it needs 2^4=16 runs and 40 runs for OFAT. So 2,4,6. 3,8,16. 4,16,40. We can conclude that it need 2^(n-1) *(n+1) times OFAT so 5 factors needs 96 runs.