Stat401B
Lab#9 Problem #1
Spring 2014
Analysis of Variance
Source
Model
Error
C. Total
DF
2
9
11
Sum of Squares
883.3333
220.3333
1103.6667
Mean Square
441.667
24.481
F Ratio
18.0408
Prob > F
0.0007*
Lack Of Fit
Source
Lack Of Fit
Pure Error
Total Error
DF
6
3
9
Sum of Squares
211.58333
8.75000
220.33333
Mean Square
35.2639
2.9167
F Ratio
12.0905
Prob > F
0.0330*
Max RSq
0.9921
Parameter Estimates
Term
Intercept
Temp
Conc
Estimate
-51.5
0.4533333
0.8666667
Std Error
19.94553
0.080798
0.403992
t Ratio
-2.58
5.61
2.15
Prob>|t|
0.0296*
0.0003*
0.0605
Residual by Temp
Residual Percent
Residual by Predicted Plot
Residual Percent
Residual by Conc
Normal Plot of Stud. Res.
Fit of the Full Model
Summary of Fit
RSquare
RSquare Adj
Root Mean Square Error
Mean of Response
Observations (or Sum Wgts)
0.971912
0.948505
2.27303
67.83333
12
Analysis of Variance
Source
Model
Error
C. Total
DF Sum of Squares
5
1072.6667
6
31.0000
11
1103.6667
Mean Square
214.533
5.167
F Ratio
41.5226
Prob > F
0.0001*
Lack Of Fit
Source
Lack Of Fit
Pure Error
Total Error
DF Sum of Squares
3
22.250000
3
8.750000
6
31.000000
Mean Square
7.41667
2.91667
F Ratio
2.5429
Prob > F
0.2318
Max RSq
Parameter Estimates
Term
Intercept
Temp
Conc
(Temp-225)*(Temp-225)
(Conc-20)*(Temp-225)
(Temp-225)*(Temp-225)*(Conc-20)
Residual by Predicted Plot
Estimate
-48.66667
0.4533333
0.9
-0.0112
-0.026
-0.00008
Std Error
10.58038
0.037118
0.321455
0.0021
0.009092
0.00063
t Ratio
-4.60
12.21
2.80
-5.33
-2.86
-0.13
Prob>|t|
0.0037*
<.0001*
0.0312*
0.0018*
0.0288*
0.9031
Fit of the Reduced Model
Summary of Fit
RSquare
RSquare Adj
Root Mean Square Error
Mean of Response
Observations (or Sum Wgts)
0.933555
0.908638
3.02765
67.83333
12
Analysis of Variance
Source
Model
Error
C. Total
DF Sum of Squares
3
1030.3333
8
73.3333
11
1103.6667
Mean Square
343.444
9.167
F Ratio
37.4667
Prob > F
<.0001*
Lack Of Fit
Source
Lack Of Fit
Pure Error
Total Error
DF Sum of Squares
5
64.583333
3
8.750000
8
73.333333
Mean Square
12.9167
2.9167
F Ratio
4.4286
Prob > F
0.1252
Max RSq
Parameter Estimates
Term
Intercept
Temp
Conc
(Temp-225)*(Temp-225)
Percent Residual
Residual by Predicted Plot
Estimate
-48
0.4533333
0.8666667
-0.0112
Std Error
12.2361
0.049441
0.247207
0.002797
t Ratio
-3.92
9.17
3.51
-4.00
Prob>|t|
0.0044*
<.0001*
0.0080*
0.0039*
Problem#2 (a)
80
Tensile Strength
75
70
65
60
2
1
4
3
Mix
Source
Mix
Error
C. Total
Test
DF Sum of Squares
3
448.15000
16
154.40000
19
602.55000
H 0 : µ=
µ=
µ=
µ4
1
2
3
Mean Square
149.383
9.650
H a : at least
at α = .05 .
vs.
reject the null hypothesis
F Ratio
15.4801
one inequality. Since p-value<.05,
(b)
The sample means are :
LSD=
y1 = 62.2 , y2 = 65.4 , y3 = 70.4 , y4 = 74.6
2.12 × 9.65 × 2 / 5 = 4.165
mix
1
2
62.2
65.4
------------
3
70.4
4
74.6
From JMP:
LSMeans Differences Student's t
α=0.050
t=2.11991
Level
4
A
3
B
2
C
1
C
Least Sq Mean
74.600000
70.400000
65.400000
62.200000
Levels not connected by same letter are significantly different.
(c)
W=
mix
4.05 × 9.65 × 1/ 5 = 5.626
1
2
3
4
62.2
65.4
70.4
74.6
----------------------------------
Prob > F
<.0001*
From JMP:
LSMeans Differences Tukey HSD
α= 0.050 Q= 2.86102 <<<<<<<Note :This Q value is different from the value from the table. i.e. 4.05/ 2 =2.864
Level
4
A
3
A B
2
B C
1
C
Least Sq Mean
74.600000
70.400000
65.400000
62.200000
Levels not connected by same letter are significantly different.
(d)
H 0 :1/ 2( µ1 + µ2 ) − 1/ 2( µ3 + µ4 ) =
0
Hypothesis
Contrast :
=
1 1 −1 −1
1
Estimate:
62.2 + 65.4 − 70.4 − 74.6 =
−17.4
ˆ 1 =
Standard Error=
=
V (ˆ 1 )
12 + 12 + 12 + 12
9.65
=
5
=
9.65 4 / 5 2.78
Thus the t-statistic is:
tc =
−17.4
= −6.26 and the percentile from the t-table is t.025,16 = 2.12
2.78
Thus R.R. is
| t |> 2.12 ; Thus the null hypothesis is rejected at α = .05 since
6.26 is in the rejection region.
From JMP output:
t-statistic = -6.26
p-value = <.0001
Reject null hypothesis at α = .05
Hand computations for the other three comparisons are not shown here but those
computations are similar to above and are summarized below.
Contrast
A
B
C
D
Est
s.e.
t
p-value
i)
1
1
-1
-1
-17.4
2.78
-6.26
<.0001
ii)
1
-1
1
-1
-7.4
2.78
-2.66
.017
iii)
1
-1
0
0
-3.2
1.96
-1.63
.1229
iv)
0
0
1
-1
-4.2
1.96
-2.14
.0483
The results from JMP are shown below:
Contrast
Test Detail
1
2
3
4
Estimate
Std Error
t Ratio
Prob>|t|
SS
0.5
0.5
-0.5
-0.5
-8.7
1.3892
-6.262
1.1e-5
378.45
0.5
-0.5
0.5
-0.5
-3.7
1.3892
-2.663
0.017
68.45
1
-1
0
0
-3.2
1.9647
-1.629
0.1229
25.6
0
0
1
-1
-4.2
1.9647
-2.138
0.0483
44.1
Summary Statement
The data show that there is a significant difference between the average tensile strength of asphalt for
the two aggregate types and two compaction methods, respectively. The average strength appear to be
higher for the kneading compaction method compared to the static compaction method . and higher
for the basalt aggregate type compared to the silicious aggregate type. However, there is no significant
difference in average strength between the two compaction methods when used with silicious
aggregate type, where as there is a significant difference between them when used with basalt
aggregate type.