M25- Growth & Transformations 1 Department of ISM, University of Alabama, 1992-2003 Lesson Objectives: Recognize exponential growth or decay. Use log(Y.

M25- Growth & Transformations 1 Department of ISM, University of Alabama, 1992-2003

Lesson Objectives:

• Recognize exponential growth or decay.

• Use log(Y ) to construct the prediction equation.

• Reverse the process to get predicted values from log(Y ) models back in terms of Y.


L

wt

h

n e ar

g r o

i

E x p o n en

ti

a

l


Stuff $100 in a mattress each month,then after X months you will haveY = 0 + 100 X dollars.

This is linear growth; ZERO interest.

X

Y

Example 1


Stuff $1000 in a savings acct.that pays 10% interest each year,

then after X years you will haveY = 1000 ( 1.10 ) dollars.

X

This is exponential growth.

X

Y

Example 2


Linear growth increases by a fixed amount in each time period; Exponential growth increases by afixed percentage of the previoustotal.


If YY grows exponentially grows exponentially as X increases,

X

Y

then log log YY grows linearly grows linearly as X increases.

X

log Y


logb X = YY bYY = X

Properties of logarithms:

1. logbase 1 = 0

2. logb XY = logb X + logb Y

3. logb Xp = p logb X

A logarithm is an exponent.A logarithm is an exponent.


logb X = Y

Review of logarithms:

bY = X

log5 125 = 3 53 = 125

log101000 = 3 103 = 1000

ln X = natural log, or log base “e”e = 2.7182818

ln 1000 = 6.907 e6.907 = 1000


Why do we care about logarithms?

Back to the matress.Back to the matress.$1000. at 10% per year:$1000. at 10% per year:

ln Y = ln [1000 ( 1.10 )X]

= ln [1000] + ln( 1.10 )X

= ln [1000] + X ln( 1.10 )

= a + b X i.e., a straight linestraight line.

Not linear

equation!Not linear

equation!

Y = 1000 ( 1.10 )X

This IS a linearequation!

This IS a linearequation!

X

048

12

Y

110

1001000

500

1000

4 8 12 X-axis

Y

log10 Y

4 8 12 X

3

2

1

log Y

Example 3


If X = 6, log10 Y = 0 + .25 6= 1.51.5

If log10 Y = 1.5,

Y =

log10 Y = 0 + .25 X

4 8 12 X

3

2

1

log Y

Example 3


If X = 10, log10 Y = 0 + .25 10=

Y =4 8 12 X

3

2

1

log Y

Example 3

M25 Expon growth & Transforms 14 Department of ISM, University of Alabama, 1992-2003

Data

TransformationsData

Transformations


Ex: Z-scores, inches to cm, oC to oF temperature

The basic shape of the data distribution does not change.

Linear transformations of Y and/or X

do not affect r.

do not change the pattern of the relationship.


transform a skewed distributioninto a symmetric distribution,

straighten a nonlinear relationshipbetween two variables,

remove non-constant variance,

Nonlinear transformations can be used to:


Lesson Objectives: Learn how to recognize whena straight line is NOT the best fitthe pattern of the data.

Learn how to transform one or both of the variables to create a linear pattern.

Learn to use the transformed model to get estimates back in terms of the original Y variable.


What do we do if the relationship between Y

and X is not linear?

Always scatterplot the data first!

If the relationship is linear, then the model may produce reasonable estimates.


“Curved lines” can be straightened out by changing the form of a variable:

1.1. Replace “Replace “XX” with “” with “square root of square root of XX””

2.2. Replace “Replace “XX” with “” with “log log XX””

3.3. Replace “Replace “XX” with “” with “1/1/XX”, its inverse.”, its inverse.

Each step Each step downdown this list this list increasesincreasesthe “change in the bend of the line.”the “change in the bend of the line.”


““New New XX ” = ” = XX

ppp = 1p = 1

p = .5p = .5

p = -1p = -1

p = #p = #

p = 2p = 2

Square root

Inverse or reciprocal

logarithm

Changing the power, changes the bend:Changing the power, changes the bend:

Each step Each step downdown this list this list increasesincreasesthe “change in the bend of the line.”the “change in the bend of the line.”


X

X

ln X

1/X

X

X

ln X

1/X

YY

YY

ln Yln Y

1/Y1/Y

Y

Y

ln Y

1/Y

X

Y

Original patternOriginal pattern

Original patternOriginal pattern Original patternOriginal pattern

YY = a + b = a + b11X + bX + b22XX22

Rules of Engagement


b2 > 0

b2 < 0

or


Y = Federal expenditures on social insurance, in millions. X = Year

X 1960 1965 1970 1975 1980Y 14,307 21,807 45,246 99,715 191,162

a. plotb. fix data if necessaryc. get prediction equationd. predict for 2000.

Example 4


40

80

120

160

200

‘60 ‘65 ‘70 ‘75 ‘80

Example 4, continued


X

19601965197019751980

Y

14,30721,80745,24699,715

191,162

log10Y

4.1564.3394.6564.9995.281

log10 Y =



Y

40

80

120

160

200

‘60 ‘65 ‘70 ‘75 ‘80 ‘60 ‘65 ‘70 ‘75 ‘80

4.0

4.4

4.8

5.2

log Y



For 2000,

log10 Y = -110.04 + .05824 (2000)= ____________

log10Y = -110.04 + .05824 X

Y = 106.44006.4400

=

This is an exponent.


Example 4, in MinitabExample 4, in MinitabGraph

Plot …

Title

ScatterplotScatterplot


198019701960

200000

100000

0

Year

Exp

end

Social Ins. Fed Expenditures (millions $)

Plot shows severe curve.

Example 4, in MinitabExample 4, in Minitab ScatterplotScatterplot


Stat

Regression

Fitted Line Plot …

YY = a + b = a + bXX

Example 4, in MinitabExample 4, in Minitab RegressionRegression


198019701960

200000

100000

0

Year

Exp

end

S = 30940.5 R-Sq = 86.6 % R-Sq(adj) = 82.2 %

Expend = -16931302 + 8632.36 Year

Regression Plot

Straight linedoes not fitthe data verywell.Future yearsFuture yearswill be severelywill be severelyunderestimated!underestimated!

Same plot as before,with regression line overlayed.

Example 4, in MinitabExample 4, in Minitab RegressionRegression


Stat

Regression


loglog1010YY = a + b = a + bXX

Example 4, in MinitabExample 4, in Minitab

This boxcontrolsthe “scale”of the plot.


198019701960

200000

100000

0

Year

Exp

end

S = 0.0497219 R-Sq = 99.1 % R-Sq(adj) = 98.8 %

log(Expend) = -110.042 + 0.0582374 Year

Regression Plot

Result from equation must be “un-logged”: y = 10log(Expend)

Advantage: Can see the “new curved line” drawn through the original data.Disadvantage: Hard to tell if the fit is “good enough”.

Advantage: Can see the “new curved line” drawn through the original data.Disadvantage: Hard to tell if the fit is “good enough”.

Example 4, in MinitabExample 4, in Minitab “Logscale” boxNOT checked:Axes are stillY and X, butcurve is basedon the “log Y”.


Stat

Regression



Example 4, in MinitabExample 4, in Minitab

The boxIS checked.


198019701960

200000

150000

100000

80000

60000

40000

30000

20000

15000

Year

Exp

end

S = 0.0497219 R-Sq = 99.1 % R-Sq(adj) = 98.8 %

log(Expend) = -110.042 + 0.0582374 Year

Regression Plot

10,000

50,000

Advantage: Easier to see that the curve has been straightened.Disadvantage: Harder to read the scale.

Advantage: Easier to see that the curve has been straightened.Disadvantage: Harder to read the scale.

Results must be “un-logged”: y = 10log(Expend)

Example 4, in MinitabExample 4, in Minitab “Logscale” boxIS checked:Axes are “log Y”and X, but values onthe “log Y” scaleare “un-logged.”

5.2

4.8

4.6

4.4

4.2

4.0

Log scale

Un-Logged Y scale

M24 Std Error & r-square 35 Department of ISM, University of Alabama, 1992-2003

How helpful is “engine size” for estimating “mpg”?

Example 5

Continued . .

. .

Continued . .

. .


Analysis DiaryStep Y X s r-sqr Comments

1 mpg displace 2.880 54.6%

Slope of “displacement” in not zero; but plot indicates a curvedpattern.Transform a variable and re-run.

Example 5 “mpg_city” versus engine “displacement”

2 to be done in next section. RecallRecall


How helpful is engine size for estimating mpg?

Regression Analysis

The regression equation ismpg_city = 29.3 - 0.0480 displace

113 cases used 4 cases contain missing valuesPredictor Coef StDev T P

Constant 29.2651 0.7076 41.36 0.000displace 0.047967 0.004154 -11.55 0.000

S = 2.880 R-Sq = 54.6% R-Sq(adj) = 54.2%

Analysis of VarianceSource DF SS MS F PRegression 1 1106.1 1106.1 133.33 0.000Error 111 920.8 8.3Total 112 2026.9

displacement in cubic in.mpg_city in ??? Data in Car89 Data

P-value: a measure of the likelihoodthat the true coefficient is “zero.”

Example 5


mpg_city vs. displacementmpg_city vs. displacement

35025015050

35

30

25

20

15

displace

mpg

_city

S = 2.88 Is this a good fit? The data pattern appears curved; we can do better!

Example 5

Step 1YY = a + b = a + bXX

X

X

ln X

1/X

X

X

ln X

1/X

YY

YY

ln Yln Y

1/Y1/Y

Y

Y

ln Y

1/Y

X

Y



YY = a + b = a + b11X + bX + b22XX22

Rules of Engagement


b2 > 0

b2 < 0

or


350250150 50

40

30

20

displace

mp

g_

city

S = 0.0527122 R-Sq = 58.9 % R-Sq(adj) = 58.5 %

log(mpg_city) = 1.47973 - 0.0009579 displace

Regression Plot

mpg_city vs. displacementmpg_city vs. displacementExample 5

Step 2

log Y

X


“Logscale” boxIS checked:






2 log Y X 58.9%

Still curved, in same direction.


300200150100 90 80 70 60

40

30

20

displace

mp

g_

city

S = 0.0462544 R-Sq = 68.3 % R-Sq(adj) = 68.0 %

log(mpg_city) = 2.20505 - 0.404870 log(displace)

Regression Plot


Step 3

log Y

log X

loglog1010YY = a + b log = a + b log1010XX

Both “Logscale”boxes checked:






2 log Y X 58.9%

Still curved, same direction.

3 log Y log X 68.3%

Better fit possible on left end?


350250150 50

35

25

15

displace

mp

g_

city

S = 2.34002 R-Sq = 70.3 % R-Sq(adj) = 69.7 %

+ 0.0003536 displace**2

mpg_city = 40.6380 - 0.185449 displace

Regression Plot


Step 4

Y

X

Try:Y = a + b1X+ b2X2






2 log Y X 58.9%


3 log Y log X 68.3%


4 Y = a +b1X+b2X2 70.3%

Better fit; BUT illogical!Try inverse of Y.


Calc

Calculator …

Name of “New Y variable.”Name of “New Y variable.”


“right sideof the equation”

1/’mpg_city’

To change the functional form of a variablein Minitab:

List of namesof functions:


350250150 50

0.07

0.06

0.05

0.04

0.03

displace

1/m

pg

_c

S = 0.0054520 R-Sq = 61.3 % R-Sq(adj) = 60.9 %

1/mpg_c = 0.0312912 + 0.0001042 displace

Regression Plotmpg_city vs. displacementmpg_city vs. displacementExample 5

Step 5Try:1/Y = a + b1X

1/ Y

X

Went too far.






2 log Y X 58.9%


3 log Y log X 68.3%


4 Y = a +b1X+b2X2 70.3%

Better fit; BUT illogical!Try inverse of Y.

5 1/ Y X 61.3%

Too far; bent in otherdirection; NOT a good fit.etc.



Final model:Final model:


s = 0.04625 R-Sq = 68.3%Estimate “mean mpg_city” for displacement = 150.

Log10 150.0 =

log(mpg_city) = 2.2051 - 0.4049 ( _______ ) = _______

mpg_city = = 21.086 mpg21.086 mpg.________


300200150100 90 80 70 60

40

30

20

displace

mp

g_

city

S = 0.0462544 R-Sq = 68.3 % R-Sq(adj) = 68.0 %


Regression Plot


Back to Step 3

log Y

log X

loglog1010YY = a + b log = a + b log1010XX

Both “Logscale”boxes checked:

Recall

150

21.09


Example:

MPG vs HP for32 Car Models

Example 6


230210190170150130110907050

30

25

20

15

HP

GPM

Scatterplot of MPG vs Horsepower

for 32 Car Models

Non-Linear Relationship

Example 6

Step 1

Y

X

X

X

ln X

1/X

X

X

ln X

1/X

YY

YY

ln Yln Y

1/Y1/Y

Y

Y

ln Y

1/Y

X

Y



YY = a + b = a + b11X + bX + b22XX22

Rules of Engagement


b2 > 0

b2 < 0

or


0.0150.0100.005

30

25

20

15

1/HP

GPM

Plot of MPG vs 1/HP for 32 Car Models

Example 6

Step 4

Y

1/X


MPG = a + b 1HP

Suggests a model of the form:

Example 6


Example:

Price vs Weight for109 Car Models

Example 7


400030002000

60000

50000

40000

30000

20000

10000

0

WEIGHT

ECI

RP

Plot of Price vs Weight for 109 Car Models

Example 7

Step 1

Y

X


400030002000

60000

50000

40000

30000

20000

10000

0

WEIGHT

ECI

RP

Plot of Price vs Weight for 109 Car Models

Nonlinear with Nonconstant Variance

Example 7

Step 1

Y

X

X

X

ln X

1/X

X

X

ln X

1/X

YY

YY

ln Yln Y

1/Y1/Y

Y

Y

ln Y

1/Y

X

Y



YY = a + b = a + b11X + bX + b22XX22

Rules of Engagement


b2 > 0

b2 < 0

or


Non-Constant Variance

The variation of the Y values increases as X changes.

Generally,transform the Y variable first.

“Log Y” is a reasonable start.


400030002000

0.0002

0.0001

0.0000

WEIGHT

ECI

RP/1

Plot of 1/Price vs Weight for 109 Car Models

Constant Variance, but still nonlinear Transform WEIGHT

Example 7

Step 3

1/ Y

X

X

X

ln X

1/X

X

X

ln X

1/X

YY

YY

ln Yln Y

1/Y1/Y

Y

Y

ln Y

1/Y

X

Y



YY = a + b = a + b11X + bX + b22XX22

Rules of Engagement


b2 > 0

b2 < 0

or


0.00060.00050.00040.00030.0002

0.0002

0.0001

0.0000

1/WEIGHT

ECI

RP/1

Plot of 1/Price vs 1/Weight for 109 Car Models

Linear with constant variance! (outliers)

Example 7

Step 5

1/ Y

1/X



1Weight

= a + b 1

Priceor

Price = 1

a + b 1

Weight

Example 7




Example:

Sales vs Assets for80 Fortune 500 Companies

in 1986


50000400003000020000100000

50000

40000

30000

20000

10000

0

ASSETS

SELAS

Plot of Sales vs Assets for 80 Fortune 500 Companies

Example 8

Many small values,few large values;

compress both scales.

Step 1

Y

X


4.53.52.5

5

4

3

2

LogAssts

selaSgoL

Plot of Log(Sales) vs Log(Assets) for 80 Fortune 500 Companies

Example 8 Step 2

Use brushingto identifythese points.

Treat thetwo groupsseparately?

log Y

log X



log(Sales) = a + b log(Assets)

or

Sales = 10a

Assetsb

Example 8


WarningsWarnings

Data transformations do NOT work if there is no relationship in the original plot.

The transformations discussed (square root, log, reciprocal, etc.) are one-bend transformations.

Pattern having more than one bend need a different fix.


The Endof regression analysis,

for now . . . .

M25- Growth & Transformations 1 Department of ISM, University of Alabama, 1992-2003 Lesson Objectives: Recognize exponential growth or decay. Use log(Y.

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