Homework #18
Score____________
/ 10
Name ______________
Text Exercise 12.2 Now, check some answers before submitting Homework #18:
(a) Construct the completed ANOVA table below.
SOURCE
df
SS
MS
f
4
24.7
6.175
4.914
Error
30
37.7
1.2567
Total
34
62.4
Treatments
(b)
5
(c) Answer this part by indicating what the f test statistic value is, what the
appropriate tabled f value is, what the P-value is, and whether or not multiple
comparison is necessary.
P < 0.01
f4,30;0.10 =
2.14
Since H0 is rejected, multiple comparison is necessary.
f4,30 = 4.914
Text Exercise 12.10
(a) Completely Randomized Design
(b)
The treatments are the different experiences with designing and evaluating
audit programs.
(c) Construct the completed ANOVA table below.
SOURCE
Groups
df
SS
MS
f
4.671
2
71.51
35.755
Error
42
321.47
7.654
Total
44
392.98
P-value
0.01 < P < 0.025
(d) Answer this part by summarizing the results (Step 4) of the f test to see if there
is sufficient evidence with  = 0.05 of at least one difference in mean hours
allocated to accounts receivable. Then, indicate whether or not multiple
comparison is necessary.
Since f2,42 = 4.671 and f2,42;0.05 = 3.23, we have sufficient evidence to reject H0 , We
conclude that there is at least one difference in mean hours allocated to accounts
receivable among the three groups (0.01 < P < 0.025). We need multiple comparison,
Additional HW Exercise 12.1
The mean breaking strength of rope is being studied for three types named Deluxe,
Econ, and Nogood. A 0.05 significance level is chosen for a hypothesis test to see if
there is any evidence that mean breaking strength is not the same for the rope types
Deluxe, Econ, and Nogood. Pieces of rope are randomly selected from each type,
and the breaking strengths in pounds of force are recorded as follows:
Deluxe
165 162 159 162
Econ
156 163 158
Nogood
151 154 160
(a) After defining the appropriate dummy variable(s), write a regression model for
the prediction of Y = “breaking strength” from “type of rope”.
1 for Deluxe
X1 =
1 for Econ
X2 =
0 otherwise
Y = 0 + 1X1 + 2X2 + 
or
0 otherwise
E(Y) = 0 + 1X1 + 2X2
(b) With the model defined in part (a), write the formula for the population mean
breaking strength for each type of rope.
mean for Deluxe = 0 + 1
mean for Econ = 0 + 2
mean for Nogood = 0
(c) Create an SPSS data file named ropedummy containing three variables named
X1, X2, and strength. Store the data in this SPSS file by entering each
breaking strength together with the corresponding values for each of the
dummy variables defined in part (a).
Use the Analyze > Regression > Linear options in SPSS to select the variable
strength for the Dependent slot, and select the variables X1 and X2 for the
Independent(s) section.
Title the output to identify the homework exercise (Additional HW Exercise
12.1 - part (c)), your name, today’s date, and the course number (Math 214).
Use the File > Print Preview options to see if any editing is needed before
printing the output. Attach the printed copy to this assignment before
submission.
(d) Write the least squares prediction equation for the model in part (a), and use
this equation to obtain the sample means for Deluxe, Econ, and Nogood.
The least squares prediction equation is strength = 155 + 7(X1) + 4(X2)
With X1 = 1 and X2 = 0, we find the mean for Deluxe = 155 + 7(1) + 4(0) = 162
With X1 = 0 and X2 = 1, we find the mean for Econ = 155 + 7(0) + 4(1) = 159
With X1 = 0 and X2 = 0, we find the mean for Nogood = 155 + 7(0) + 4(0) = 155
Title the output to identify the homework exercise (Additional HW Exercise
12.1 - part (e)), your name, today’s date, and the course number (Math 214).
Use the File > Print Preview options to see if any editing is needed before
printing the output. Attach the printed copy to this assignment before
submission.
(f) Summarize the results (Step 4) of the f test to see if there is sufficient evidence
that mean breaking strength is not the same for the rope types Deluxe, Econ,
and Nogood at the 0.05 level.
Since f2,7 = 3.419 and f2,7;0.05 = 4.74, we do not have sufficient evidence to reject H0 .
We conclude that mean breaking strength is the same for the rope types Deluxe,
Econ, and Nogood (0.05 < P < 0.10).
OR (P = 0.092)
Additional HW Exercise 12.1 - continued
(g) Based on the results in part (e), decide whether or not a multiple comparison
method is necessary. If no, then explain why not; if yes, then use Tukey’s
HSD multiple comparison method.
Since we have not rejected H0 , we do not need a multiple comparison method to
identify significant differences between means.
(h) Explain why three contiguous box plots are an appropriate graphical display for
the data, and use SPSS to create these. Title the output to identify the
homework exercise (Additional HW Exercise 12.1 - part (h)), your name,
today’s date, and the course number (Math 214). Attach a printed copy to this
assignment before submission.
Since the quantitative variable breaking strength is being compared for three rope
types, three contiguous box plots is an appropriate graphical display.
(i) Summarize the results (Step 4) of Levene’s f test to see if there is sufficient
evidence at the 0.05 level that the standard deviation of breaking strength is not
the same for the four types of rope. Then, decide whether or not we need to be
concerned about whether or not the assumption of equal standard deviations is
satisfied.
Since f2,7 = 0.951 and f2,7;0.05 = 4.74, we do not have sufficient evidence to reject H0 .
We conclude that the standard deviation of breaking strength is not significantly
different for the for the rope types Deluxe, Econ, and Nogood (0.10 < P).
OR (P = 0.431)
Since we have no evidence of a difference in standard deviation, we believe the
assumption of equal standard deviations is satisfied.
Additional HW Exercise 12.2
The mean penetration weight of water is being studied for four brands of paper
towels named Tuff, Econ, Cheep, and Super. A 0.05 significance level is chosen for
a hypothesis test to see if there is any evidence that mean penetration weight is not
the same for the brands Tuff, Econ, Cheep, and Super. Randomly selected
observations from each brand are observed, and the penetration weights in grams
are recorded as follows:
Tuff
112 104 110 108 112 114
Econ
120 116 122 114 112 118
Cheep
106 114 110 108 116 112
Super
120 110 112 112 114 116
(a) After defining the appropriate dummy variable(s), write a regression model for
the prediction of Y = “penetration weight” from “paper towel brand”.
1 for Tuff
X1 =
0 otherwise
1 for Econ
X2 =
Y = 0 + 1X1 + 2X2 + 3X3 + 
X
0 otherwise
or
1 for Cheep
X2 =
E(Y) = 0 + 1X1 + 2X2 +
0 otherwise
(b) With the model defined in part (a), write the formula for the population mean
penetration weight for each paper towel brand.
mean for Tuff = 0 + 1
mean for Econ = 0 + 2
mean for Cheep = 0 + 3
mean for Super =
0
Additional HW Exercise 12.3
Consider data for a one-way ANOVA where each of three treatments, labeled A, B,
and C, are applied to 4 subjects (that is, the data will consist of 3 random samples
each of size 4).
(a) Make up a data set for which the SSR will be 0 (zero), but the SSE will not be
0 (zero). (Hint: This means you need a data set for which there is no variation
between samples but there is some variation within samples.)
A
10 12 14 16
B
10 12 14 16
C
10 12 14 16
(Any data set where the numbers within
any sample are not all equal to each other
and the sample means are all equal to each
other is correct.)