Applied Statistics and Probability for Engineers, 5th edition
x t0.025,14 s 1 +
10 February 2010
1
1
xn +1 x + t0.025,14 s 1 +
n
n
1
1
xn +1 8.23 2.145(0.025) 1 +
15
15
8.17 xn +1 8.29
8.23 2.145(0.025) 1 +
95% two-sided confidence interval on mean rod
February 22, 2010
Applied Statistics and Probability for Engineers, 5th edition
d) Cooks distance values
1.36736
0.7536
0.7536
1.36736
0.0542
0.01917
0.03646
There are two influential points with Cooks distances greater than 1.
12-65. a) R 2 = 75.6%
b) As
February 22, 2010
Applied Statistics and Probability for Engineers, 5th edition
-R-SQ. (ADJ.) = 0.9893 SE = 0.039499
MAE = 0.028896 DurbWat = 1.869
Analysis of Variance for the Full Regression
Source
Sum of Squares
DF
Mean Square
F-Ratio
P-value
Model
2.7
Applied Statistics and Probability for Engineers, 5th edition
12-78.
February 22, 2010
y = 0 + 1 x1 + 2 x2 + 12 x12
y = 11.503 + 0.153x1 6.094 x2 0.031x12
0 for tool type 302
where x2 =
1 for tool type 416
Test of different slopes:
H 0 : 12 = 0
H1 : 12 0
February 22, 2010
Applied Statistics and Probability for Engineers, 5th edition
Analysis of Variance
Source
DF
SS
MS
F
P
2
11076473
5538237
99.67
0.000
Residual Error
37
2055837
55563
Total
39
13132310
Regression
Reject H0 and conclude regression model is
February 22, 2010
Applied Statistics and Probability for Engineers, 5th edition
e) y = 80.2 + 43x 5.92 x 2 + 0.425 x3
H 0 : 33 = 0
H1 : 33 0 = 0.05
t0 = 0.76
t.025,8 = 2.306
| t0 |> t0.025,8
/
Do not reject H0 and conclude that cubic term is not significa
Applied Statistics and Probability for Engineers, 5th edition
12-62. a) R 2 = 8.1%
b) Assumption of normality appears is not adequate.
c) The graphs indicate non-constant variance. Therefore, the model is not adequate.
12-58
February 22, 2010
Applied Stat
February 22, 2010
Applied Statistics and Probability for Engineers, 5th edition
P
P P
P K S S
A P C P B A S P P H H
Mallows
G G D G T E M V H G C G G F
Vars
R-Sq
R-Sq(adj)
C-p
S
F A V F G N I G T A T F A G
1
52.8
51.1
74.3
5.0233
1
49.3
47.5
81.6
5.2043
2
February 22, 2010
Applied Statistics and Probability for Engineers, 5th edition
Section 12-5
12-57. a)
The regression equation is
mpg = 49.9 - 0.0104 cid - 0.0012 rhp - 0.00324 etw + 0.29 cmp - 3.86
axle
+ 0.190 n/v
Predictor
Coef
SE Coef
T
P
Constant
49.
February 22, 2010
Applied Statistics and Probability for Engineers, 5th edition
12-48. The regression equation is
ARSNAILS = 0.001 + 0.00858 AGE - 0.021 DRINKUSE + 0.010 COOKUSE
Predictor
Coef
SE Coef
T
P
0.0011
0.9067
0.00
0.999
0.008581
0.007083
1.21
0.
February 22, 2010
Applied Statistics and Probability for Engineers, 5th edition
Anxiety
1
74.6
Unusual Observations
Obs
Age
Satisfaction
Fit
SE Fit
Residual
9
27.0
75.00
93.28
2.98
St Resid
-18.28
-2.87R
12-40. a) The Minitab result is shown below.
Regres
February 22, 2010
Applied Statistics and Probability for Engineers, 5th edition
H1 : 1 0
= 0.01
t0 =
=
1
se( 1 )
0.3339
= 0.49
0.6763
t0.005,9 3 = t0.005,6 = 3.707
| t0 |< t / 2,6 , Fail to reject H0, there is no enough evidence to conclude that 1 is sig
February 22, 2010
Applied Statistics and Probability for Engineers, 5th edition
Analysis of Variance
Source
DF
SS
MS
F
P
3
2373.59
791.20
191.09
0.000
Residual Error
28
115.93
4.14
Total
31
2489.52
Regression
Source
DF
Seq SS
Pct Comp
1
1614.43
Pct TD
1
5
Applied Statistics and Probability for Engineers, 5th edition
12-42
a) 0 t / 2, n p 2 c00
1.9122 t .025,7 se( 0 )
1.9122 (2.365)(10.055)
1.9122 23.78
25.6922 0 21.8678
1 t / 2 , n p 2 c
11
0.0931 t .025,7 se( 1 )
0.0931 (2.365)(0.0827)
0.0931 0.1956
Applied Statistics and Probability for Engineers, 5th edition
Do not reject H0
e) H 0 : 12 = 0
H1 : 12 0
= 0.01
SSR ( 12 | 1 , 2 ) = 29951.4 29787 = 163.9
f0 =
SSR 163.9
=
= 1.11
147
MS E
f 0.01,1,2 = 98.50
f 0 > f 0.01,1,2
/
Do not reject H0
f) 2 = 111.
February 22, 2010
Applied Statistics and Probability for Engineers, 5th edition
b)
Predicted Values for New Observations
New
Obs
Fit
SE Fit
1
44.6
21.9
99% CI
99% PI
(-36.7, 125.8)
(-112.8, 202.0)
Values of Predictors for New Observations
New
Contrast
Obs
Applied Statistics and Probability for Engineers, 5th edition
February 22, 2010
12-59. a) The Mintab output follows. The proportion of total variability explained by this model is:
R2 =
SS R 2373.59
=
= 0.95
SST 2489.52
b) Normal Probability Plot: Some mo
February 22, 2010
Applied Statistics and Probability for Engineers, 5th edition
Regression Analysis: Rating Pts versus Pct Comp, Pct TD, Pct Int
The regression equation is
Rating Pts = 2.99 + 1.20 Pct Comp + 4.60 Pct TD - 3.81 Pct Int
Predictor
Coef
SE Co
February 22, 2010
Applied Statistics and Probability for Engineers, 5th edition
Residuals Versus the Fitted Values
(response is y)
0.3
0.2
Residual
0.1
0.0
-0.1
-0.2
-0.3
11
12
13
14
15
16
Fitted Value
e) Normality assumption is reasonable.
Normal Probabi
Applied Statistics and Probability for Engineers, 5th edition
February 22, 2010
b) Minimum Cp (1.1) model:
W = 96.5 + 0.527 SV - 0.125 HR_P - 0.0847 BB_P + 0.0257 SO_P
Based on the graphs below, the model assumptions are not violated
12-109
Applied Statis
March 23, 2010
Applied Statistics and Probability for Engineers, 5th edition
b) P-value = 0.489. The variability due to random error is SSE = 0.146.
c) The Barley diet has the highest average protein content and lupins the lowest.
Scatter plot of mean pr
March 23, 2010
Applied Statistics and Probability for Engineers, 5th edition
Residuals Versus the Fitted Values
(response is STRENGTH)
Residual
10
0
-10
75
85
95
Fitted Value
Residuals Versus AIRVOIDS
Normal Probability Plot of the Residuals
(response is
Applied Statistics and Probability for Engineers, 5th edition
3
1.5550
3.4450
3.4450
0.4750
0.4750
-2.0250
2.3650
4
1.5550
2.3650
March 23, 2010
-0.1350
There are significant differences between levels 1 and 3, 4; 2 and 3, 4; and
3 and 4.
13-23. Fisher's
March 23, 2010
Applied Statistics and Probability for Engineers, 5th edition
Scatterplot of Mean Strength vs Cotton Percentage
22
Mean Strength
20
18
16
14
12
10
15
20
25
Cotton Percentage
30
35
Individual 95% CIs For Mean
Based on Pooled StDev
Level
N
Me