STATISTICS 402 - Assignment 7
The Federal Aviation Administration does an experiment on the ability to detect targets on a radar screen. Two factors are manipulated, the amount of background noise or “ground clutter” (levels
Low, Medium and High) and the type of filter placed over the radar screen (Type 1 and Type 2).
Treatments (A, B, C, D, E and F) are made by factorial crossing of the levels of ground clutter with the type of filter. For each of the treatment combinations a signal representing an aircraft is introduced and the intensity increased until the radar operator is able to detect the signal. A lower intensity is desirable because it means the radar operator can detect a weaker signal. A total of 36 observations are collected, six for each treatment combination.
1) Suppose the experiment is run as completely randomized design. a) Explain how to assign treatments to experimental units for the experiment if it is run as a completely randomized design?
Treatment combinations are assigned completely at random. To do this put 36 chips in a bag; 6 chips with each of A, B, C, D, E, and F printed on them. Mix thoroughly. Draw from the bag, without replacement. The letter on the chip indicates the treatment combination which will be assigned to that run of the experiment.
The data for a completely randomized experiment are given below.
A B C D E F
Type 1, L Type1, M Type 1, H Type 2, L Type 2, M Type 2, H
94
102
106
98
96
98
101
106
105
96
100
98
113
112
108
98
110
104
79
78
86
72
78
75
87
90
97
80
86
88
93
91
95
83
90
91 b) Analyze the data and give a complete analysis of variance table including all sources of variation, degrees of freedom, sums of squares, mean squares, F statistics and P-values.
Response: Intensity
Analysis of Variance
Source
Filter
Clutter
Filter*Clutter
Treatment
Error
C. Total
DF Sum of Squares Mean Square
1 2601.0 2601.0
2
2
5
30
35
666.0
96.0
3363.0
677.0
4040.0
333.0
48.0
672.6
22.567
F Ratio
115.2585
14.7563
2.1270
29.8050
Prob > F
<.0001*
<.0001*
0.1368
<.0001*
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c) Are there statistically significant treatment effects? Support your answer statistically.
Yes. F = 29.8050, P-value < 0.0001. The small P-value indicates that there are some statistically significant treatment effects. d) Are there statistically significant factor effects? Support your answer statistically.
Filter yes. F = 115.2585, P-value < 0.0001. The small P-value indicates that there are some statistically significant effects due to type of filter.
Clutter yes. F = 14.7563, P-value < 0.0001. The small P-value indicates that there are some statistically significant effects due to the amount of ground clutter. e) Is there statistically significant interaction between the factors? Support your answer statistically.
No. F = 2.127, P-value = 0.1368. The P-value is larger than 0.05 which indicates that the interaction between type of filter and amount of ground clutter is not statistically significant. f) Is there a “best” combination of type of filter and amount of background noise or “ground clutter”? Support your answer statistically.
Yes. The HSD for treatment means is 3.0416(2.74267) = 8.34. An absolute difference between treatment sample means greater than 8.34 is statistically significant. Treatment D: Type 2 filter and a Low amount of ground clutter gives the smallest sample mean intensity (78.0) and this sample mean intensity is significantly different from all other treatment sample means (the next lowest is
Treatment E: Type 2 filter with Medium amount of ground clutter, 88.0).
2) Different operators have different abilities to detect signals so operators contribute to variation in the intensity values. Suppose the experiment is run as a randomized complete block design with six operators. a) Explain how to assign treatments to experimental units for the experiment if it is run as a randomized complete block design? Be sure to answer this question within the context given above.
Reuse operators so that each operator experiences all six treatment combinations.
The order each operator experiences the six treatment combinations is randomized.
To do this put 6 chips in a bag. Each chip has a letter; A, B, C, D, E or F. Have each operator choose chips without replacement and have the operator perform the runs in that order. The randomization is done separately for each operator.
The data for a randomized complete block experiment are given on the next page. Note: this is the same data as given in problem 1 except for the inclusion of the Operator variable.
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A B C D E F
Operator Type 1, L Type1, M Type 1, H Type 2, L Type 2, M Type 2, H
1
2
94
102
101
106
113
112
79
78
87
90
93
91
3
4
5
6
106
98
96
98
105
96
100
98
108
98
110
104
86
72
78
75
97
80
86
88
95
83
90
91 b) Analyze the data and give a complete analysis of variance table including all sources of variation, degrees of freedom, sums of squares, mean squares, F statistics and P-values.
Response: Intensity
Analysis of Variance
Source
Filter
Clutter
Filter*Clutter
Treatment
Operator
Error
C. Total
DF Sum of Squares Mean Square
1 2601.0 2601.0
2
2
5
5
25
35
666.0
96.0
3363.0
468.0
209.0
4040.0
333.0
48.0
672.6
93.6
8.36
F Ratio
311.1244
39.8325
5.7416
80.4545
11.1962
Prob > F
<.0001*
<.0001*
0.0089*
<.0001*
<.0001* c) Are there statistically significant treatment effects? Support your answer statistically.
SSTreatment = 2601.0 + 666.0 + 96.0 = 3363.0 or 3831.0 – 468.0 = 3363.0
MSTreatment = 3363.0/5 = 672.6
F = 672.6/8.36 = 80.45
P-value < 0.0001
The P-value is so small (<0.05) there are some statistically significant treatment effects. d) Are there statistically significant factor effects? Support your answer statistically.
Filter yes. F = 311.1244, P-value < 0.0001. The small P-value indicates that there are some statistically significant effects due to type of filter.
Clutter yes. F = 39.8325, P-value < 0.0001. The small P-value indicates that there are some statistically significant effects due to the amount of ground clutter.
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e) Is there statistically significant interaction between the factors? Support your answer statistically.
Yes. F = 5.7416, P-value = 0.0089. The P-value is smaller than 0.05 which indicates that the interaction between type of filter and amount of ground clutter is statistically significant. f) Is there a “best” combination of type of filter and amount of background noise or “ground clutter”? Support your answer statistically.
Yes. The HSD for treatment means is 3.08179(1.66933) = 5.14. An absolute difference between treatment sample means greater than 5.14 is statistically significant. Treatment D: Type 2 filter and a Low amount of ground clutter give the smallest sample mean intensity (78.0) and this sample mean intensity is significantly different from all other treatment sample means (the next lowest is Treatment E:
Type 2 filter with Medium amount of ground clutter, 88.0).. g) Has blocking on Operator been effective? Support your answer statistically.
Yes. Blocking on Operator is effective if it reduces the MSError. The MSError was
22.567 for the completely randomized experiment and 8.36 for the randomized complete block experiment. Blocking on Operator reduced the MSError. This reduction in the MSError made the interaction between type of filter and amount of ground clutter statistically significant (when it wasn’t before).
3) The setup for each experimental treatment takes time so that only six treatments can be done in one day. There are differences from day to day, in addition to the differences from operator to operator, that may affect the intensity values. a) Given this information, why would a Latin Square design be appropriate?
A Latin Square design would be appropriate because there are two blocking variables: Operator and Day. By using the Latin Square design the variation from operator to operator and day to day can be accounted for separately from the error term in the analysis of variance.
The data for a Latin square experiment are given below. Note: this is the same data as given in problem 2 except for the inclusion of the Day variable.
Day 1 Day 2 Day 3 Day 4 Day 5 Day 6
Operator 1 A
94
C
113
B
101
E
87
F
93
D
79
Operator 2 B
106
A
102
E
90
D
78
C
112
F
91
Operator 3 C
108
B
105
F
95
A
106
D
86
E
97
Operator 4 D
72
F
83
A
98
B
96
E
80
C
98
Operator 5
Operator 6
F
90
E
88
E
86
D
75
D
78
C
104
C
110
F
91
A
96
B
98
B
100
A
98
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b) Analyze the data and give a complete analysis of variance table including all sources of variation, degrees of freedom, sums of squares, mean squares, F statistics and P-values.
Response: Intensity
Analysis of Variance
Source
Filter
Clutter
Filter*Clutter
Treatment
Operator
Day
Error
C. Total
DF Sum of Squares Mean Square
1 2601.0 2601.0
2
2
5
5
5
20
35
666.0
96.0
3363.0
468.0
9.667
199.333
4040.0
333.0
48.0
672.6
93.6
1.933
9.967
F Ratio
260.9699
33.4114
4.8161
67.4849
9.3913
0.1940
Prob > F
<0.0001*
<0.0001*
0.0196*
<.0001*
<0.0001*
0.9613 c) Are there statistically significant treatment effects? Support your answer statistically.
SSTreatment = 2601.0 + 666.0 + 96.0 = 3363
or 3840.667 – 468.0 – 9.667 = 3363
MSTreatment = 3363/5 = 672.5
F = 672.6/9.967 = 67.48
P-value < 0.0001
The P-value is so small (<0.05) there are some statistically significant treatment effects. d) Has including the second blocking factor, Day, been effective? Support your answer statistically.
No. Including the second blocking factor has not been effective because the
MSError was 8.36 for the randomized complete block experiment and 9.967 for the
Latin Square experiment. Including the blocking on Day actually increased the
MSError. This increase in the MSError did not affect the final analysis as both factors are still statistically significant as is the interaction between type of filter and amount of ground clutter. The Type 2 filter and a Low amount of ground clutter is still the best treatment combination giving the lowest mean intensity (78.0) which is significantly different from all other treatment combinations.
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