STATISTICS 402 - Assignment 8
Solution
1. In order to increase the yield strength of steel, steel bars are heated to a critical
temperature in an oven for a specified amount of time, quenched in water and then
cooled in the air. The strength of the bar is then measured by subjecting it to testing
that destroys the bar. An experiment is performed to determine the effect of different
factors on the strength of steel. One factor is the temperature of the oven which has
two levels 1500 and 1600 degrees C. The second is the time the steel bar is heated at
that temperature, which has three levels 10 minutes, 20 minutes and 30 minutes.
a) What are the response, conditions and experimental material?
The response is the yield strength.
The conditions are 2 levels of oven temperature and 3 levels of time.
The experimental material consists of steel bars.
b) How many treatment combinations are there?
There are 6 treatment combinations.
c) If we have 4 replications of each treatment combination in a completely
randomized design, what is the size of the difference in temperature level means
that can be detected with Alpha = 0.05 and Beta = 0.05?
There will be 12 observations for each of the 2 temperatures. With Alpha =
0.05 and Beta = 0.05 the difference in temperature level means that can be
detected is 1.6 standard deviations.
d) If we have 4 replications of each treatment combination in a completely
randomized design, what is the size of the difference in heating time means that
can be detected with Alpha = 0.05 and Beta = 0.05?
There will be 8 observations for each of the 3 times. With Alpha = 0.05 and
Beta = 0.05 the difference in time level means that can be detected is between
2.0 and 2.5 standard deviations.
e) For a completely randomized design the treatments will be assigned at random to
the bars. How long will it take to run the experiment as a completely randomized
design?
There are 8 steel bars at heated for each of the times; 10 minutes, 20 minute
and 30 minutes so that is a total of 80 minutes plus 160 minutes plus 240
minutes, 480 minutes or 8 hours. This is the minimum amount of time
because you have to allow time for the oven to heat up and cool down when
the temperature is changed.
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f) If you are going to do a randomized complete block design why can’t you form
blocks by reusing?
Once a bar is heated the strength is measured and measuring the strength
destroys the bar so it cannot be used again.
g) You can use a randomized complete block design by sorting the steel bars
according to their initial strength (measured non-destructively). Give a partial
analysis of variance table for a randomized complete block design using four
blocks.
Source
Temperature
Time
Temperature*Time
Block
Error
C. Total
df
1
2
2
3
15
23
h) Using a randomized complete block design will not reduce the time it takes to run
the experiment because one steel bar at a time is heated at the assigned
temperature for the assigned time. To reduce the time it takes to run the
experiment three steel bars are heated in the oven at a randomly assigned
temperature, one bar, chosen at random, is removed after 10 minutes; a second
bar, chosen at random, is removed after 20 minutes and the third bar is removed
after 30 minutes. Explain why this is split plot (repeated measures) design. In
your explanation, you must answer the following questions. Assume you have 24
steel bars to use in the experiment.
This is a split plot/repeated measures design because there are two designs in
one experiment. There is a completely randomized design that is used to
assign the temperature level to groups of 3 steel bars (sorted on initial
strength to be more uniform). The times in the oven are randomly assigned
to steel bars within each group of three (the blocks). This is a randomized
complete block design.
i. What are the “whole plots”?
A whole plot is a group of 3 steel bars (sorted on initial strength to be
more uniform).
ii. What is the “whole plot” factor?
The whole plot factor is temperature.
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iii. How will random assignment be used for the whole plot factor? Be specific.
First sort the steel bars into groups of three based on initial strength
(measured non-destructively) so within each group the bars are more
uniform. There will be 8 groups. Assign 1500 degrees C at random to 4
of the groups and 1600 degrees C to the other 4 groups. This can be done
by having 8 cards, 4 with 1500 and 4 with 1600 on them. Shuffle the
cards and draw cards without replacement. The temperature on the card
will be assigned to the group of steel bars.
iv. What are the “sub plots”?
Subplots are individual bars within a group of three sorted to be more
uniform.
v. What is the “sub plot” factor?
The subplot factor is time in the oven.
vi. How will random assignment be used for the sub plot factor? Be specific.
For each group of three steel bars, number them with a unique integer 1,
2, or 3. Have three cards numbered 1, 2, and 3. Shuffle the cards. The
first card drawn (without replacement) is the number of the bar taken
out after 10 minutes. The second card drawn (without replacement) is
the number of the bar taken out after 20 minutes. The remaining bar is
taken out after 30 minutes. This procedure is repeated for each set of
three steel bars.
i) Construct a partial ANOVA table indicating sources of variation and degrees of
freedom. Also indicate how to construct the appropriate F tests for determining
the statistical significance of the model effects.
Source
Temperature
Groups[Temperature]
Time
Temperature*Time
Error
C. Total
df
1
6
2
2
12
23
F ratio
MSTemperature/MSGroups[Temperature]
MSTime/MSError
MSTemperature*Time/MSError
3
2. A psychology experiment is conducted on the effects of anxiety and muscular tension
on four different types of memory. There are 4 treatments (A, B, C, and D) with
different degrees of anxiety and muscular tension. Twelve students are chosen at
random from all students at a large university who have given their consent to
participate in psychology experiments. Treatments are assigned at random to the
students. Three students are assigned A, three students are assigned B, three students
are assigned C and three students are assigned D. Each student performs four types
of memory trials in random order. The random order is different for each student.
For each memory trial the number of memory errors is recorded for each student.
a) What is the response?
The response is the number of memory errors.
b) What are the experimental units?
The experimental units are students at a large university who have given
their consent to participate in psychology experiments.
c) What is the whole plot?
The whole plot is the student.
d) What is the whole plot, between subjects, factor?
The whole plot factor is the treatment (different degrees of anxiety and
muscular tension).
e) What is the sub plot?
The sub plot is also the student (reusing the student for each of the
memory trials).
f) What is the sub plot, within subjects, factor?
The sub plot factor is the different memory trials.
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g) A JMP data table is posted on the course web site. Use JMP to analyze these
data keeping in mind that this is a split plot (repeated measures) design. Be
sure to include plots of main effects and an interaction plot. Turn in the
computer output with your assignment.
Response: Errors
Summary of Fit
RSquare
RSquare Adj
Root Mean Square Error
Mean of Response
Observations (or Sum Wgts)
0.95981
0.921294
1.452966
10.33333
48
Analysis of Variance
Source
DF Sum of Squares
Model
23
1210.0000
Error
24
50.6667
C. Total
47
1260.6667
Mean Square
52.6087
2.1111
Test Denominator Synthesis
Source
Treatment
Memory Trial
Treatment*Memory Trial
Student[Treatment]&Random
MS Den
7.83333
2.11111
2.11111
2.11111
DF Den
8
24
24
24
Tests wrt Random Effects
Source
Treatment
Memory Trial
Treatment*Memory Trial
Student[Treatment]&Random
SS
164
950.167
33.1667
62.6667
MS Num
54.6667
316.722
3.68519
7.83333
F Ratio
24.9199
Prob > F
<.0001*
Denom MS Synthesis
Student[Treatment]&Random
Residual
Residual
Residual
DF Num
3
3
9
8
F Ratio
6.9787
150.0263
1.7456
3.7105
Prob > F
0.0127*
<.0001*
0.1328
0.0059*
Effect Details
Treatment
Effect Test
Sum of Squares
164.00000
F Ratio
6.9787
DF
3
Prob > F
0.0127*
Denominator MS Synthesis: Student[Treatment]&Random
5
Least Squares Means Table
Level
Least Sq Mean
Std Error
A
11.833333 0.80794664
B
8.500000 0.80794664
C
8.500000 0.80794664
D
12.500000 0.80794664
Mean
11.8333
8.5000
8.5000
12.5000
LSMeans Differences Tukey HSD
α=0.050 Q=3.20234
HSD = 3.20234(1.14261) = 3.659
LSMean[i] By LSMean[j]
Mean[i]-Mean[j]
A
B
C
D
Std Err Dif
Lower CL Dif
Upper CL Dif
A
0 3.33333 3.33333 -0.6667
0 1.14261 1.14261 1.14261
0 -0.3257 -0.3257 -4.3257
0 6.99236 6.99236 2.99236
B
-3.3333
0 1.8e-15
-4
1.14261
0 1.14261 1.14261
-6.9924
0 -3.659 -7.659
0.32569
0 3.65903 -0.341
C
-3.3333 -2e-15
0
-4
1.14261 1.14261
0 1.14261
-6.9924 -3.659
0 -7.659
0.32569 3.65903
0 -0.341
D
0.66667
4
4
0
1.14261 1.14261 1.14261
0
-2.9924 0.34097 0.34097
0
4.32569 7.65903 7.65903
0
Level
D
A
B
C
A
A B
B
B
Least Sq Mean
12.500000
11.833333
8.500000
8.500000
Levels not connected by same letter are significantly different.
Memory Trial
Effect Test
Sum of Squares
950.16667
F Ratio
150.0263
DF
3
Prob > F
<.0001*
Denominator MS Synthesis: Residual
6
Least Squares Means Table
Level
Least Sq Mean
Type 1
16.666667
Type 2
11.750000
Type 3
8.333333
Type 4
4.583333
Std Error
0.41943525
0.41943525
0.41943525
0.41943525
Mean
16.6667
11.7500
8.3333
4.5833
LSMeans Differences Tukey HSD
α=0.050 Q=2.75861
HSD = 2.75861(0.59317) = 1.636
LSMean[i] By LSMean[j]
Mean[i]-Mean[j]
Type 1 Type 2 Type 3 Type 4
Std Err Dif
Lower CL Dif
Upper CL Dif
Type 1
0 4.91667 8.33333 12.0833
0 0.59317 0.59317 0.59317
0 3.28034 6.69701 10.447
0 6.55299 9.96966 13.7197
Type 2
-4.9167
0 3.41667 7.16667
0.59317
0 0.59317 0.59317
-6.553
0 1.78034 5.53034
-3.2803
0 5.05299 8.80299
Type 3
-8.3333 -3.4167
0
3.75
0.59317 0.59317
0 0.59317
-9.9697 -5.053
0 2.11367
-6.697 -1.7803
0 5.38633
Type 4
-12.083 -7.1667
-3.75
0
0.59317 0.59317 0.59317
0
-13.72 -8.803 -5.3863
0
-10.447 -5.5303 -2.1137
0
Level
Type 1 A
Type 2
B
Type 3
C
Type 4
D
Least Sq Mean
16.666667
11.750000
8.333333
4.583333
Levels not connected by same letter are significantly different.
Treatment*Memory Trial
Effect Test
Sum of Squares
33.166667
F Ratio
1.7456
DF
9
Prob > F
0.1328
Denominator MS Synthesis: Residual
7
h) Are there some treatment effects that are different from zero? Report the
appropriate F- and P- values to support your answer.
Yes. F = 6.9787, P-value = 0.0127. The P-value is less than 0.05, so there
are some treatment effects different from zero.
i) Compute the HSD for comparing treatment means. Use the HSD to see which
treatments means are statistically different from other treatment means?
HSD = 3.20234(1.14261) = 3.659
Level
D
A
B
C
A
A B
B
B
Least Sq Mean
12.500000
11.833333
8.500000
8.500000
Levels not connected by same letter are significantly different.
Therefore, treatment D is statistically different from treatments B and C.
There are no statistically significant differences between treatments A
and D, or between treatments A, B or C.
j) Are there some memory trial effects that are different from zero? Report the
appropriate F- and P-values to support your answer.
Yes. F = 150.0263, P-value <0.0001. The P-value is less than 0.05, so there
are some treatment effects different from zero.
k) Compute the HSD for comparing memory trial means. Use the HSD to see
which memory trial means are statistically different from other memory trial
means?
HSD = 2.75861(0.59317) = 1.636
Level
Type 1 A
Type 2
B
Type 3
C
Type 4
D
Least Sq Mean
16.666667
11.750000
8.333333
4.583333
Levels not connected by same letter are significantly different.
Therefore all types of memory trials have mean number of memory
errors that are statistically different from those of all other types.
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l) Is there a statistically significant interaction between treatment and memory
trial? Report the appropriate F- and P-values to support your answer.
No. F = 1.7456, P-value = 0.1328. The P-value is greater than 0.05
therefore there is not a statistically significant interaction between the
treatments and the types of memory trials.
m) Comment on the residuals given on the next page. Remember that there are
two sets of residuals; one set for the whole plot and one set for the subplot.
Tell me what you see in the various plots (residuals vs. factor levels, Normal
quantile plot, box plot and histogram) and indicate what this tells you about
the Fisher conditions of equal standard deviations and normally distributed
errors.
Residuals for evaluating treatments:
The residuals for evaluating the treatments have variation that is quite
similar for each the treatments. The ratio of the largest standard
deviation to the smallest is less than 2. The equal standard deviation
condition is most likely met for evaluating differences among treatments.
The distribution of residuals is slightly skewed right. There are not many
residuals (only 8) so it is hard to tell, but the condition that errors be
normally distributed may be in doubt.
Residuals for evaluating memory trials:
The standard deviation for Type 1 is much larger than that for Type 2
(2.55 times as big). The equal standard deviation condition may not be
met. The distribution of residuals is skewed to the right indicating that
the condition that errors be normally distributed may not be met.
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Memory Errors for Different Treatments and Memory Trials
Level
A
B
C
D
Number
3
3
3
3
Mean
0
0
0
0
Whole Plot Residuals
Std Dev
1.44338
1.50000
0.90139
1.63936
Level
Type 1
Type 2
Type 3
Type 4
Number
12
12
12
12
Mean
0
0
0
0
Std Dev
1.54233
0.60302
0.99240
0.93744
Subplot Residuals
10