Download 1..1) A computer ANOVA output is shown below. Fill in the blanks

1..1) A computer ANOVA output is shown below.
Fill in the blanks.
You may give bounds on the P-value.
One-way ANOVA
2)
Source
DF
SS
MS
F
P
Factor
4
987.71
246.93
33.09
< 0.0001
Error
25
186.53
7.46
Total
29
1174.24
A computer ANOVA output is shown below.
Fill in the blanks.
You may give bounds on the P-value.
One-way ANOVA
2.
(a) Fill in the blanks.
Source
DF
SS
MS
F
P
Factor
3
36.15
12.05
1.21
0.3395
Error
16
159.89
9.99
Total
19
196.04
You may give bounds on the P-value.
Completed table is:
Source
DF
SS
MS
F
P
Treatment
4
1020.56
255.14
30.14
< 0.00001
Block
5
323.82
64.765
7.65
<0.001
Error
20
169.33
8.466
Total
29
1513.71
(b) How many blocks were used in this experiment?
Six blocks were used.
(c) What conclusions can you draw?
The treatment effect is significant; the means of the five treatments are not all equal.
3.
a) The assumption of normality is necessary to test the claim. According to the normal probability
plots, the
assumption of normality does not appear to be violated. This is evident from the fact that the
data appear
to fall along a straight line.
Probability Plot of EX5-81V1
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
99.58
1.529
25
0.315
0.522
Percent
80
70
60
50
40
30
20
10
5
1
95
96
97
98
99
100 101
EX5-81V1
102
103
104
Probability Plot of EX5-81V2
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
Percent
80
70
60
50
40
30
20
10
5
1
100
102
104
106
EX5-81V2
108
110
112
b) 1) the parameters of interest are the variances of resistance of products, 12 , 22
2) H0: 12  22
3) H1: 12  22
4)  = 0.05
5) The test statistic is
f0 
s12
s22
6) Reject H0 if f0 < f0.975,24,34 where f0.975,24,34 =
1
1

 0.459
f0.025,34 ,24 2.18
or f0 > f0.025,24,34 where f0.025,24,34 =2.07
7) s1 = 1.53
n1 = 25
s2 =1.96
n2 = 35
104.9
1.930
33
0.328
0.506
f0 
(153
. )2
(196
. )2
 0.609
8) Since 0.601 > 0.459, do not reject H0 and conclude the variances are not significantly different
at
 = 0.05.
c)
Two-Sample T-Test and CI: vendor1, vendor2
Two-sample T for vendor1 vs vendor2
N
Mean StDev
vendor1 25 99.58 1.53
vendor2 33 104.93 1.93
SE Mean
0.31
0.34
Difference = mu (vendor1) - mu (vendor2)
Estimate for difference: -5.351
95% upper bound for difference: -4.567
T-Test of difference = -5 (vs <): T-Value = -0.75
Both use Pooled StDev = 1.7695
P-Value = 0.229
DF = 56
Since p-value > 0.05, we cannot reject the null hypothesis . That is we cannot prove that the mean
resistance of vendor 2 is at least 5 higher than that of vendor 1.
4.
a),
d = 0.667
b)
sd = 2.964, n = 12
95% confidence interval:
 s 
 s 
d  t  / 2,n 1 d    d  d  t  / 2,n 1 d 
 n
 n
 2.964 
 2.964 
0.667  2.201
   d  0.667  2.201

 12 
 12 
1.216  d  2.55
indication that
c) According to the normal probability plots, the assumption of normality does not appear to be
since the data fall approximately along a straight line.
Normal Probability Plot
.999
.99
.95
Probability
violated
Since zero is contained within this interval, we are 95% confident there is no significant
one design language is preferable.
.80
.50
.20
.05
.01
.001
-5
0
5
diff
A vera ge : 0 .66 66 67
S tDe v: 2 .9 644 4
N : 12
A nde rso n-Darlin g N orm al ity Te st
A -Sq ua red : 0. 31 5
P -Va lue : 0.5 02
5.
The effective life of insulating fluids at an accelerated load of 35 kV is being studied. Test data have
been obtained for four types of fluids. The results from a completely randomized experiment were as follows:
Fluid Type
1
2
3
4
17.6
16.9
21.4
19.3
18.9
15.3
23.6
21.1
(a) Is there any indication that the fluids differ?
Life (in h) at 35 kV Load
16.3
17.4
18.6
17.1
19.4
18.5
16.9
17.5
20.1
19.5
20.5
18.3
21.6
20.3
22.3
19.8
Use  = 0.05.
Since the P-value = 0.05, there is a difference in means at  = 0.05.
Minitab Output
One-way ANOVA: Life(in h) versus Fluid Type
Source
DF
SS
MS
F
P
Fluid Type 3 29.92 9.97 3.10 0.050
Error
20 64.29 3.21
Total
23 94.21
(b) Which fluid would you select, given that the objective is long life?
Descriptive Statistics: Life
Variable
Life
Fluid
Type
1
2
3
4
Mean
18.650
17.950
20.950
18.900
StDev
1.952
1.854
1.879
1.441
Minitab Output
Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons among Levels of Fluid Type
Individual confidence level = 98.89%
Fluid Type = 1 subtracted from:
Fluid
Type
Lower Center Upper
2
-3.598 -0.700 2.198
3
-0.598
2.300 5.198
4
-2.648
0.250 3.148
+---------+---------+---------+--------(---------*--------)
(---------*--------)
(---------*--------)
+---------+---------+---------+---------6.0
-3.0
0.0
3.0
Fluid Type = 2 subtracted from:
Fluid
Type
Lower Center Upper
3
0.102 3.000 5.898
4
-1.948
0.950 3.848
+---------+---------+---------+--------(---------*---------)
(--------*---------)
+---------+---------+---------+---------6.0
-3.0
0.0
3.0
Fluid Type = 3 subtracted from:
Fluid
Type
Lower Center Upper
4
-4.948 -2.050 0.848
+---------+---------+---------+--------(--------*---------)
+---------+---------+---------+---------6.0
-3.0
0.0
3.0
Fluid Type 3.
Descriptive Statistics shows that Fluid Type 3 has the longest life. From the result of Tukey’s method, we
found that, although there is no significant difference between Fluid Types 1, 3 and 4, there is a significant
difference between Fluid Types 2 and 3, with Fluid Type 3 having a longer lifetime.
(c) Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied?
Residual Plots for Life
N orm a l P roba bil ity P l ot
V ers us Fits
99
3.0
R e sidu a l
Pe rc e n t
90
50
10
1
-5.0
1.5
0.0
-1.5
-3.0
-2.5
0.0
R e sidu a l
2.5
5.0
18
19
20
F itte d Va lu e
His togra m
V ers us Order
3.0
3.6
R e sidu a l
F re qu e n c y
4.8
2.4
1.5
0.0
-1.5
1.2
0.0
21
-3.0
-2.4
-1.2
0.0
1.2
R e sidu a l
2.4
2
4
6
8 10 12 14 16 18 20 22 24
O bse rva tion O rde r
There is nothing unusual in the residual plots.
6..
An article in the Fire Safety Journal (“The Effect of Nozzle Design on the Stability and Performance of
Turbulent Water Jets,” Vol. 4, August 1981) describes an experiment in which a shape factor was determined
for several different nozzle designs at six levels of jet efflux velocity. Interest focused on potential differences
between nozzle designs, with velocity considered as a nuisance variable. The data are shown below:
Jet Efflux Velocity (m/s)
Nozzle
Design
1
2
3
4
5
11.73
0.78
0.85
0.93
1.14
0.97
14.37
0.80
0.85
0.92
0.97
0.86
16.59
0.81
0.92
0.95
0.98
0.78
20.43
0.75
0.86
0.89
0.88
0.76
23.46
0.77
0.81
0.89
0.86
0.76
28.74
0.78
0.83
0.83
0.83
0.75
(a) Does nozzle design affect the shape factor?
variance, using  = 0.05.
Compare nozzles with a scatter plot and with an analysis of
Two-way ANOVA: Shape versus Nozzle Design, block
Source
DF
SS
Nozzle Design 4 0.102180
block
5 0.062867
Error
20 0.057300
Total
29 0.222347
S = 0.05353
R-Sq = 74.23%
Since p-value< 0.05,
MS
F
P
0.0255450 8.92 0.000
0.0125733 4.39 0.007
0.0028650
R-Sq(adj) = 62.63%
nozzle design has a significant effect on shape factor.
(b) Analyze the residual from this experiment.
The plots shown below do not give any indication of serious problems. Three is some indication of a mild
outlier on the normal probability plot and on the plot of residuals versus the predicted velocity.
Residual Plots for Shape
N orm a l P roba bility P l ot
V ers us Fits
99
0.10
R e sidu a l
Pe rc e n t
90
50
0.00
-0.05
10
1
0.05
-0.10
-0.10
-0.05
0.00
0.05
R e sidu a l
0.10
0.7
0.8
His togra m
0.9
F itte d Va lu e
1.0
V ers us Order
8
R e sidu a l
F re qu e n c y
0.10
6
4
2
0
0.05
0.00
-0.05
-0.10
-0.05
0.00
0.05
R e sidu a l
0.10
2
4
6
8 10 12 14 16 18 20 22 24 26 28 30
O bse rva tion O rde r
Individual Value Plot of RESI1
0.15
0.10
R ES I1
0.05
0.00
-0.05
-0.10
1
2
3
Noz z le De s ign
4
(c) Which nozzle designs are different with respect to shape factor
Descriptive Statistics: Shape
Nozzle
Design
Mean
StDev
1
0.78167 0.02137
2
0.8533 0.0372
3
0.9017 0.0422
4
0.9433 0.1136
5
0.8133 0.0866
Tukey 95.0% Simultaneous Confidence Intervals
Response Variable Shape
All Pairwise Comparisons among Levels of Nozzle Design
Nozzle Design = 1 subtracted from:
Variable
Shape
Nozzle
Design
2
3
4
5
Lower
-0.02077
0.02757
0.06923
-0.06077
Center Upper
0.07167 0.1641
0.12000 0.2124
0.16167 0.2541
0.03167 0.1241
-----+---------+---------+---------+(-----*-----)
(-----*-----)
(-----*-----)
(-----*-----)
-----+---------+---------+---------+-0.15
0.00
0.15
0.30
Nozzle Design = 2 subtracted from:
Nozzle
Design
3
4
5
Lower
Center
Upper -----+---------+---------+---------+-0.0441 0.04833 0.14077
(-----*-----)
-0.0024 0.09000 0.18243
(-----*-----)
-0.1324 -0.04000 0.05243
(-----*-----)
5
-----+---------+---------+---------+-0.15
0.00
0.15
0.30
Nozzle Design = 3 subtracted from:
Nozzle
Design
4
5
Lower
Center
Upper
-0.0508 0.04167 0.134100
-0.1808 -0.08833 0.004100
-----+---------+---------+---------+(-----*-----)
(-----*-----)
-----+---------+---------+---------+-0.15
0.00
0.15
0.30
Nozzle Design = 4 subtracted from:
Nozzle
Design
5
Lower Center
-0.2224 -0.1300
Upper -----+---------+---------+---------+-0.03757 (-----*-----)
-----+---------+---------+---------+-0.15
0.00
0.15
0.30
위의 결과로부터, Nozzle 1과 3, 1과4,
4와 5 가 서로 다르다.