STATISTICS 402 - Assignment 5
Solution
1. An experiment will be performed on individuals with elevated serum cholesterol. We
want to see if the type of drug (Lipitor® or Zocor®) has an effect on the change in
serum cholesterol (serum cholesterol level before treatment – serum cholesterol level
after three months of treatment). A second factor, the amount of drug (0 mg/day, 10
mg/day and 20 mg/day), will also be investigated in a two factor completely
randomized experiment.
a) What is the response?
Change in serum cholesterol (serum cholesterol level before treatment –
serum cholesterol level after three months of treatment).
b) What are the experimental units?
The experimental units are individuals with elevated serum cholesterol.
c) The experimenter will use factorial crossing to create the treatment combinations.
How many treatment combinations will there be in the experiment? List all the
treatment combinations.
There will be 6 treatments.
0 mg Lipitor®, 0 mg Zocor®,
5 mg Lipitor®, 5 mg Zocor®,
10 mg Lipitor®, 10 mg Zocor®.
d) The experimenter would like to be able to detect a difference in treatment
population means of 1.4 standard deviations with Alpha = 0.05 and Beta = 0.10.
How many experimental units will the experimenter need?
The experimenter will need 18 experimental units for each of the 6
treatments for a total of 108 individuals with elevated serum cholesterol.
e) With this number of units, what size difference in factor level population means
can be detected when Alpha=0.05 and Beta=0.10?
A difference of between 0.6 and 0.7 standard deviations for drug means.
A difference of between 0.8 and 0.9 standard deviations for amount means.
f) Because of budget constraints, only 6 units are available for each treatment
combination. How does this choice affect the size of the detectable difference in
treatment population means? in factor level population means? Use Alpha=0.10
and Beta=0.10.
With 6 units for each treatment, one can detect a 2.5 standard deviation
difference in treatment populations means, a 1 standard deviation difference
in drug population means and a 1.4 standard deviation difference in amount
population means.
1
g) Explain how you would randomly assign treatments to the experimental units.
Include a table that indicates the random assignment of treatments for this
experiment to experimental units. Remember the budget constraints in f).
Number the 36 individuals with elevated serum cholesterol with a unique
number between 0 and 36. Use JMP to do the random assignment. Enter
Lipitor® in 18 rows and Zocor® in 18 rows of a column named Drug. In a
second column named Amount have six rows with 0 mg, six rows with 10 mg
and six rows with 20 mg for each drug. In a third column labeled Individual
use Column – Formula – Random – Col Shuffle. For each row, the Drug and
Amount are assigned to the individual whose number is in the Individual
column.
Drug
Lipitor®
Lipitor®
Lipitor®
Lipitor®
Lipitor®
Lipitor®
Lipitor®
Lipitor®
Lipitor®
Lipitor®
Lipitor®
Lipitor®
Lipitor®
Lipitor®
Lipitor®
Lipitor®
Lipitor®
Lipitor®
Amount
0 mg
0 mg
0 mg
0 mg
0 mg
0 mg
10 mg
10 mg
10 mg
10 mg
10 mg
10 mg
20 mg
20 mg
20 mg
20 mg
20 mg
20 mg
Individual
10
2
15
31
33
8
6
23
13
26
24
18
32
17
14
19
36
5
Drug
Zocor®
Zocor®
Zocor®
Zocor®
Zocor®
Zocor®
Zocor®
Zocor®
Zocor®
Zocor®
Zocor®
Zocor®
Zocor®
Zocor®
Zocor®
Zocor®
Zocor®
Zocor®
Amount
0 mg
0 mg
0 mg
0 mg
0 mg
0 mg
10 mg
10 mg
10 mg
10 mg
10 mg
10 mg
20 mg
20 mg
20 mg
20 mg
20 mg
20 mg
Individual
22
27
30
34
21
4
9
1
25
3
7
11
12
29
28
35
16
20
2. An experiment is performed by researchers in kinesiology on the effects of step
height and stepping frequency on the change in heart rate. Forty eight individuals are
recruited from students majoring in kinesiology. They all give their consent to
participate in the experiment. Each individual is asked to step up onto a platform and
step down repeatedly. The height of the platform will be either 15 cm or 30 cm. The
rate at which a participant steps will be 14, 21, 28 or 35 steps per minute. A
combination of platform height and stepping frequency will be assigned to each
participant completely at random so that 6 participants are assigned each
combination. Participants step up and down off the platform for 5 minutes. The
change in heart rate (heart rate after stepping – heart rate before stepping) is
calculated for each participant.
2
a) What is the response?
The response is the change in heart rate (heart rate after stepping – heart
rate before stepping).
b) What are the conditions?
The conditions are step height (15 cm and 30 cm) and stepping frequency
(14, 21, 28 and 35 steps per minute).
c) How many treatments are there?
There are 2x4 = 8 treatments.
d) What are the experimental units?
The experimental units are students majoring in kinesiology.
e) What is an outside variable that is controlled in this experiment?
The amount of time spent stepping is the same. All students step for 5
minutes.
f) What is an outside variable that is not controlled in this experiment?
The fitness level of each of the students is not controlled. Some may be very
fit and some may not be as fit.
g) Describe, in some detail, what contributes to chance, random, error variation in
this experiment.
In general, differences among experimental units that are treated the same
contribute to chance, random error variation. For this experiment,
differences in the fitness level of students assigned a particular combination
of step height and stepping frequency will contribute to chance, random,
error variation.
h) Give the full model describing the relationship between the response, conditions
and random error. Be sure to define all the terms in the model within the context
of the problem.
𝒀 = 𝝁 + 𝝉𝒊𝒋 + 𝜺
𝒀 = 𝝁 + 𝜶𝒊 + 𝜷𝒋 + 𝜶𝜷𝒊𝒋 + 𝜺
Y is the change in heart rate.
𝝁 is the overall population mean.
𝝉𝒊𝒋 is the treatment effect for step height, i, and stepping frequency, j.
𝜶𝒊 is the step height effect of step height, i.
𝜷𝒋 is the stepping frequency effect of stepping frequency, j.
𝜶𝜷𝒊𝒋 is the interaction effect for step height, i, and stepping frequency, j.
𝜺 is the random error.
3
A JMP data table is posted on the course website. Use JMP to analyze the data.
i) Estimate the treatment effects.
Mean change in heart rate is 34.6875.
The means for the 6 treatments are given in the following table.
Height=15 cm
Height=30 cm
Frequency=14 Frequency=21 Frequency=28 Frequency=35
12.00
19.50
28.83
39.50
19.50
37.00
49.33
71.83
The estimated treatment effects are given in the following table.
Height=15 cm
Height=30 cm
Frequency=14 Frequency=21 Frequency=28 Frequency=35
–22.6875
–15.1875
–5.8542
4.8125
–15.1875
2.3125
14.6458
37.1458
j) Test the hypothesis that all the treatment effects are zero against the alternative
that at least one is not zero. Be sure to give an appropriate null and alternative
hypothesis. Report the value of the appropriate test statistic, P-value, decision,
reason for the decision and a conclusion within the context of the problem.
𝑯𝟎 : 𝒂𝒍𝒍 𝒕𝒉𝒆 𝝉𝒊𝒋 = 𝟎
𝑯𝑨 : 𝒔𝒐𝒎𝒆 𝝉𝒊𝒋 ≠ 𝟎
F = 30.6498, P-value < 0.0001
Because the P-value is so small (< 0.05) we reject the null hypothesis and
conclude that some of the combinations of step height and stepping
frequency affect the change in heart rate.
k) Estimate the step height effects.
Height=15 cm
Height=30 cm
Estimated Effect
24.9583 – 34.6875 = –9.7292
44.4167 – 34.6875 = +9.7292
l) Test the hypothesis that the effects of step height are zero against the alternative
that at least one is not zero. Be sure to give an appropriate null and alternative
hypothesis using the notation from your answer to h). Report the value of the
appropriate test statistic, P-value, decision, reason for the decision and a
conclusion within the context of the problem.
𝑯𝟎 : 𝒂𝒍𝒍 𝒕𝒉𝒆 𝜶𝒊 = 𝟎
𝑯𝑨 : 𝒔𝒐𝒎𝒆 𝜶𝒊 ≠ 𝟎
F = 61.7012, P-value < 0.0001
Because the P-value is so small (< 0.05) we reject the null hypothesis and
conclude that the two step heights affect the change in heart rate differently.
4
m) Where are the statistically significant differences in step height sample means?
Support you answer statistically.
Because there are only two step heights, 15 cm has a statistically different
mean change in heart rate than 30 cm. The F test in l) supports this
conclusion.
n) Estimate the stepping frequency effects.
Estimated Effect
15.75 – 34.6875 = –18.9375
28.25 – 34.6875 = –6.4375
39.0833 – 34.6875 = +4.3958
55.6667 – 34.6875 = +20.9792
Frequency=14
Frequency=21
Frequency=28
Frequency=35
o) Test the hypothesis that all the stepping frequency effects are zero against the
alternative that at least one is not zero. Be sure to give an appropriate null and
alternative hypothesis using the notation from your answer to h). Report the value
of the appropriate test statistic, P-value, decision, reason for the decision and a
conclusion within the context of the problem.
𝑯𝟎 : 𝒂𝒍𝒍 𝒕𝒉𝒆 𝜷𝒋 = 𝟎
𝑯𝑨 : 𝒔𝒐𝒎𝒆 𝜷𝒋 ≠ 𝟎
F = 46.6892, P-value < 0.0001
Because the P-value is so small (< 0.05) we reject the null hypothesis and
conclude that some of the stepping frequencies affect the change in heart
rate.
p) Where are the statistically significant differences in stepping frequency sample
means? Support you answer statistically.
𝑯𝑺𝑫 = 𝟐. 𝟔𝟖𝟎𝟒𝟐(𝟑. 𝟓𝟎𝟑𝟐𝟕) = 𝟗. 𝟑𝟗, so any differences in stepping
frequency sample means greater than 9.39 are deemed statistically
significant.
Level
35
28
21
14
A
B
C
D
Mean
55.6667
39.0833
28.2500
15.7500
Levels not connected with the same letter are significantly different.
Therefore each stepping frequency has a sample mean change in heart rate
that is different from all of the other stepping frequencies.
5
q) Are there any interaction effects that are different from zero? Be sure to give an
appropriate null and alternative hypothesis using the notation from your answer to
h). Report the value of the appropriate test statistic, P-value, decision, reason for
the decision and a conclusion within the context of the problem.
𝑯𝟎 : 𝒂𝒍𝒍 𝒕𝒉𝒆 𝜶𝜷𝒊𝒋 = 𝟎
𝑯𝑨 : 𝒔𝒐𝒎𝒆 𝜶𝜷𝒊𝒋 ≠ 𝟎
F = 4.2599, P-value < 0.0106
Because the P-value is small (< 0.05) we reject the null hypothesis and
conclude that there is statistically significant interaction between step height
and stepping frequency.
r) Construct an interaction plot. Comment on the plot and what it tells you about the
interaction between the two factors. Be specific and be sure your answer deals
with the context of this experiment.
As stepping frequency increases, the average change in heart rate tends to
increase. As height increases, the average change in heart rate tends to
increase. However, the effect of height is greater when the stepping
frequency is 35 steps per minute compared to 14 steps per minute. This is an
indication of interaction. For 21 steps per minute and 28 steps per minute
the effect of height is about the same.
s) Based on the analysis of these data, what combination of step height and stepping
frequency would you recommend if someone wanted to increase their heart rate
the most?
Using the 35 steps per minute frequency and 30 cm step height produced a
mean increase in heart rate of 71.83 beats per minute. The HSD for
comparing treatment combination means is 3.19651(4.95438) = 15.84.
Because the mean for 35 steps per minute frequency and 30 cm step height is
more than 15.84 larger than the any of the other treatment combination
means the differences between this treatment combination and all others are
statistically significant.
6
Therefore I would recommend 35 steps per minute frequency and a 30 cm
height in order for kinesiology majors similar to those who participated in
the experiment to increase their heart rate the most after 5 minutes of
stepping.
t) Construct plots of residuals versus the levels of the two factors. Describe the plots
and indicate what this tells you about the Fisher conditions necessary for the
analysis of variance.
Level
15
30
Level
14
21
28
35
n
24
24
Mean Std Dev
0 5.94845
0 9.62711
n
12
12
12
12
Mean Std Dev
0 4.8006
0 9.7211
0 10.0280
0 7.0475
The amount of variation within each step height is not too different. The
ratio of the largest standard deviation to the smallest is less than 2. The
amount of variation for each stepping frequency differs. The ratio of the
largest standard deviation to the smallest is greater than 2. The equal
standard deviation condition may not be met.
7
u) Construct plots of the distribution of residuals. Describe each of the plots in the
distribution of residuals. Indicate what this tells you about the Fisher conditions
necessary for the analysis of variance.
Residual
The histogram looks fairly symmetric (there may be a slight skew to the
right). The box plot looks fairly symmetric (there may be a slight skew to the
right, with the mean slightly bigger than the median). The points on the
normal quantile plot follow the diagonal (normal model) line fairly well. It
appears that the errors could be normally distributed.
8
Turn in the JMP output that you have used to answer the questions.
Response: Change
Summary of Fit
RSquare
RSquare Adj
Root Mean Square Error
Mean of Response
Observations (or Sum Wgts)
0.842859
0.815359
8.58123
34.6875
48
Analysis of Variance
Source
DF Sum of Squares
Model
7
15798.813
Error
40
2945.500
C. Total
47
18744.313
Effect Tests
Source
Frequency
Height
Frequency*Height
DF
3
1
3
Mean Square
2256.97
73.64
Sum of Squares
10314.229
4543.521
941.063
F Ratio
30.6498
Prob > F
<.0001*
F Ratio
46.6892
61.7012
4.2599
Prob > F
<.0001*
<.0001*
0.0106*
Effect Details
Frequency
Least Squares Means Table
Level
Least Sq Mean
Std Error
14
15.750000
2.4771876
21
28.250000
2.4771876
28
39.083333
2.4771876
35
55.666667
2.4771876
Mean
15.7500
28.2500
39.0833
55.6667
9
LSMeans Differences Tukey HSD
α=0.050 Q=2.68042 HSD = 2.68042(3.50327) = 9.39
LSMean[i] By LSMean[j]
Mean[i]-Mean[j]
14
Std Err Dif
Lower CL Dif
Upper CL Dif
14
0
0
0
0
21
12.5
3.50327
3.10976
21.8902
28
23.3333
3.50327
13.9431
32.7236
35
39.9167
3.50327
30.5264
49.3069
21
28
35
-12.5
3.50327
-21.89
-3.1098
0
0
0
0
10.8333
3.50327
1.44309
20.2236
27.4167
3.50327
18.0264
36.8069
-23.333
3.50327
-32.724
-13.943
-10.833
3.50327
-20.224
-1.4431
0
0
0
0
16.5833
3.50327
7.19309
25.9736
-39.917
3.50327
-49.307
-30.526
-27.417
3.50327
-36.807
-18.026
-16.583
3.50327
-25.974
-7.1931
0
0
0
0
Level
35
28
21
14
Least Sq
Mean
55.666667
39.083333
28.250000
15.750000
A
B
C
D
Levels not connected by same letter are significantly different.
Height
Least Squares Means Table
Level
Least Sq Mean
Std Error
15
24.958333
1.7516361
30
44.416667
1.7516361
Mean
24.9583
44.4167
10
Frequency*Height
Least Squares Means Table
Level
Least Sq Mean
Std Error
14,15
12.000000
3.5032723
14,30
19.500000
3.5032723
21,15
19.500000
3.5032723
21,30
37.000000
3.5032723
28,15
28.833333
3.5032723
28,30
49.333333
3.5032723
35,15
39.500000
3.5032723
35,30
71.833333
3.5032723
LS Means Plot
Level
35,30
28,30
35,15
21,30
28,15
21,15
14,30
14,15
A
B
B C
B C
C D
D E
D E
E
Least Sq
Mean
71.833333
49.333333
39.500000
37.000000
28.833333
19.500000
19.500000
12.000000
Levels not connected by same letter are significantly different.
11
LSMeans Differences Tukey HSD
α=0.050 Q=3.19651 HSD = 3.19651(4.95438) = 15.84
LSMean[i] By LSMean[j]
Mean[i]-Mean[j]
14,15
Std Err Dif
Lower CL Dif
Upper CL Dif
14,15
0
0
0
0
14,30
7.5
4.95438
-8.3367
23.3367
21,15
7.5
4.95438
-8.3367
23.3367
21,30
25
4.95438
9.1633
40.8367
28,15
16.8333
4.95438
0.99663
32.67
28,30
37.3333
4.95438
21.4966
53.17
35,15
27.5
4.95438
11.6633
43.3367
35,30
59.8333
4.95438
43.9966
75.67
14,30
21,15
21,30
28,15
28,30
35,15
35,30
-7.5
4.95438
-23.337
8.3367
0
0
0
0
3.6e-15
4.95438
-15.837
15.8367
17.5
4.95438
1.6633
33.3367
9.33333
4.95438
-6.5034
25.17
29.8333
4.95438
13.9966
45.67
20
4.95438
4.1633
35.8367
52.3333
4.95438
36.4966
68.17
-7.5
4.95438
-23.337
8.3367
-4e-15
4.95438
-15.837
15.8367
0
0
0
0
17.5
4.95438
1.6633
33.3367
9.33333
4.95438
-6.5034
25.17
29.8333
4.95438
13.9966
45.67
20
4.95438
4.1633
35.8367
52.3333
4.95438
36.4966
68.17
-25
4.95438
-40.837
-9.1633
-17.5
4.95438
-33.337
-1.6633
-17.5
4.95438
-33.337
-1.6633
0
0
0
0
-8.1667
4.95438
-24.003
7.67004
12.3333
4.95438
-3.5034
28.17
2.5
4.95438
-13.337
18.3367
34.8333
4.95438
18.9966
50.67
-16.833
4.95438
-32.67
-0.9966
-9.3333
4.95438
-25.17
6.50337
-9.3333
4.95438
-25.17
6.50337
8.16667
4.95438
-7.67
24.0034
0
0
0
0
20.5
4.95438
4.6633
36.3367
10.6667
4.95438
-5.17
26.5034
43
4.95438
27.1633
58.8367
-37.333
4.95438
-53.17
-21.497
-29.833
4.95438
-45.67
-13.997
-29.833
4.95438
-45.67
-13.997
-12.333
4.95438
-28.17
3.50337
-20.5
4.95438
-36.337
-4.6633
0
0
0
0
-9.8333
4.95438
-25.67
6.00337
22.5
4.95438
6.6633
38.3367
-27.5
4.95438
-43.337
-11.663
-20
4.95438
-35.837
-4.1633
-20
4.95438
-35.837
-4.1633
-2.5
4.95438
-18.337
13.3367
-10.667
4.95438
-26.503
5.17004
9.83333
4.95438
-6.0034
25.67
0
0
0
0
32.3333
4.95438
16.4966
48.17
-59.833
4.95438
-75.67
-43.997
-52.333
4.95438
-68.17
-36.497
-52.333
4.95438
-68.17
-36.497
-34.833
4.95438
-50.67
-18.997
-43
4.95438
-58.837
-27.163
-22.5
4.95438
-38.337
-6.6633
-32.333
4.95438
-48.17
-16.497
0
0
0
0
12