Outline Solutions to Tutorial Sheet 1

Outline Solutions to Tutorial Sheet 1

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Outline Solutions to Tutorial Sheet 111. The corresponding Minitab plots and regression output are below. The relevant parts of theoutput are highlighted.400300200100100 150 200 250 300 350 400 450 500 550ReservRegression AnalysisThe regression equation isPasseng = 33.1 + 0.690 ReservPredictor Coef StDev T PConstant 33.07 19.93 1.66 0.128Reserv 0.69047 0.05553 12.43 0.000S = 26.26 R-Sq = 93.9% R-Sq(adj) = 93.3%Analysis of VarianceSource DF SS MS F PRegression 1 106666 106666 154.62 0.000Residual Error 10 6898 690Total 11 113564 Fitted Line PlotResiduals Versus the Fitted Values(response is Passeng) Regression PlotY = 33.0721 + 0.690468X40 R-Sq = 93.9 %3040020100300-10-20200-30-40100 200 300 400100Fitted Value100 150 200 250 300 350 400 450 500 550ReservResidualPassengPasseng2. (i) The scatter plot shows a strong positive linear relationship.85756555453525151 2 3 4 5Load (kN) (ii) Regression AnalysisThe regression equation isExtens (mm) = 2.07 + 15.9 Load (kN)Predictor Coef StDev T PConstant 2.070 2.590 0.80 0.455Load (kN 15.9363 0.7733 20.61 0.000S = 2.967 R-Sq = 98.6% R-Sq(adj) = 98.4%Analysis of VarianceSource DF SS MS F PRegression 1 3738.1 3738.1 424.66 0.000Residual Error 6 52.8 8.8Total 7 3790.92The regression is highly significant (P=0.000) and the R value tells us that 98.6% of ...

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Nombre de lectures 9
Langue English
Signaler un problème
Outline Solutions to Tutorial Sheet 11
1.
The corresponding Minitab plots and regression output are below. The relevant parts of the
output are highlighted.
Regression Analysis
The regression equation is
Passeng = 33.1 + 0.690 Reserv
Predictor
Coef
StDev
T
P
Constant
33.07
19.93
1.66
0.128
Reserv
0.69047
0.05553
12.43
0.000
S = 26.26
R-Sq = 93.9%
R-Sq(adj) = 93.3%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
1
106666
106666
154.62
0.000
Residual Error
10
6898
690
Total
11
113564
Fitted Line Plot
100
200
300
400
-40
-30
-20
-10
0
10
20
30
40
Fitted Value
Residual
Residuals Versus the Fitted Values
(response is Passeng)
550
500
450
400
350
300
250
200
150
100
400
300
200
100
Reserv
Passeng
100
150
200
250
300
350
400
450
500
550
100
200
300
400
Reserv
Passeng
Y = 33.0721 + 0.690468X
R-Sq = 93.9 %
Regression Plot
2. (i) The scatter plot shows a strong positive linear relationship.
(ii)
Regression Analysis
The regression equation is
Extens (mm) = 2.07 + 15.9 Load (kN)
Predictor
Coef
StDev
T
P
Constant
2.070
2.590
0.80
0.455
Load (kN
15.9363
0.7733
20.61
0.000
S = 2.967
R-Sq = 98.6%
R-Sq(adj) = 98.4%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
1
3738.1
3738.1
424.66
0.000
Residual Error
6
52.8
8.8
Total
7
3790.9
The regression is highly significant (
P=0.000
) and the
R
2
value tells us that 98.6% of the
observed variation in extension can be attributed to the different loads used. The slope (15.9)
tells us that if load is increased by 1kN, we would expect the extension to increase by 15.9mm.
The residual plot below looks random about zero so the model is reasonable.
(iii)
From the regression equation, predicted extensions are
41.8mm
,
97.5mm
and
161.1mm
respectively. The first is an interpolation (
X
= 2.5 is within the observed range of loads) so,
given the strength of the relationship, it is likely to be very accurate. The second is an
extrapolation (
X
= 6 is just outside the data range) and should be treated with caution. However
it may be fairly accurate since
R
2
is so large and it is only a
slight
extrapolation. The last is an
extrapolation well outside the range of data and could be wildly inaccurate – we have no idea
what the relationship is like for such values of load.
5
4
3
2
1
85
75
65
55
45
35
25
15
Load (kN)
Extens (mm)
15
25
35
45
55
65
75
85
-3
-2
-1
0
1
2
3
4
5
Fitted Value
Residual
Residuals Versus the Fitted Values
(response is Extens ()
3. (i)
The scatter plot confirms a linear relationship.
(ii)
The correlation coefficient is r =
0.976
and the 5% critical value for 12 pairs of values is
0.576
.
Thus since the calculated value is greater than the critical value, there really is a relationship
between nickel content and toughness.
(iii)
The regression equation is:
Toughness = 15.1 + 13.4 Nickel (%)
The required estimate is given by the slope (13.4). Thus we would expect toughness to increase
by
13.4 units
if nickel content is increased by 1 unit (1%).
When
Nickel(%)
= 2.2,
Toughness
= 15.1 + (13.4)(2.2) =
44.58
units.
4.
The various scatter diagrams are below. The relationship between
X
and
Y
is clearly non-linear.
The plot of
ln(Y)
against
X
is most linear.
3.5
3.0
2.5
65
55
45
Nickel (%)
Toughness
10
20
30
40
50
10
20
30
40
50
60
X
Y
10
20
30
40
50
2.2
3.2
4.2
X
ln(Y)
2.5
3.0
3.5
4.0
10
20
30
40
50
60
ln(X)
Y
2.5
3.0
3.5
4.0
2.2
3.2
4.2
ln(X)
ln(Y)
The correlation coefficients (which measure the extent of
linear
association) are
0.910
for
ln(Y)
against
X
,
0.774
for
Y
against
ln(X)
and
0.826
for
ln(Y)
against
ln(X)
. The fact that 0.910 is the
largest confirms that the best of the three transformations is to just log the Y values. Minitab oe
Excel then gives the regression line
ln(Y) = 1.84 + 0.0380 X
The slope of 0.0380 means that if
X
is increased by 1 unit, we would expect
ln(Y)
to increase by
0.0380 units.
When
X = 30,
estimated
ln(Y
) is 1.84 + (0.0380)(30) = 2.98
Thus, estimated
Y
is
=
98
.
2
e
19.7
units.
5.
R
2
values are 71.1% for diameter, 10.5% for penetration and 4.7% for temperature.
Diameter has easily the highest R
2
value and is therefore the best predictor. Over 70% of variations
in torque can be explained by the fact that the logs were of different diameters.
The multiple regression output is below with relevant parts highlighted.
Regression Analysis
The regression equation is
TORQUE = - 4.59 + 4.60 DIAMETER + 3.12 PENETRTN - 0.0474 TEMP
Predictor
Coef
StDev
T
P
Constant
-4.587
3.782
-1.21
0.239
DIAMETER
4.5958
0.4513
10.18
0.000
PENETRTN
3.1225
0.7986
3.91
0.001
TEMP
-0.04744
0.01809
-2.62
0.016
S = 3.317
R-Sq = 86.3%
R-Sq(adj) = 84.2%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
3
1384.37
461.46
41.95
0.000
Residual Error
20
220.01
11.00
Total
23
1604.38
Note that the overall significance (given as 0.000) means that the three variables
together
are
useful predictors of torque – variations in these three factors explain 86.3% of the variation in
torque. The
individual
significance values given at the top are all less than 0.05 which suggests that
all three variables are contributing something useful to the model.
Using the regression equation, the predicted torque is:
=
+
+
)
100
)(
0474
.
0
(
)
2
)(
12
.
3
(
)
6
)(
60
.
4
(
59
.
4
24.51
units.