Outline Solutions to Tutorial Sheet 1

Graham Crocker

Description

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 ...

Informations

• Publié par: Idwa
• Nombre de lectures: 9
• Langue: English

Extrait

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.