1

Re-expressing Data Chapter 6 – Normal Model

–What if data do not follow a Normal model?

Chapters 8 & 9 – Linear Model–What if a relationship between

two variables is not linear?

2

Re-expressing Data Re-expression is another

name for changing the scale of (transforming) the data.

Usually we re-express the response variable, Y.

3

Goals of Re-expression Goal 1 – Make the distribution

of the re-expressed data more symmetric.

Goal 2 – Make the spread of the re-expressed data more similar across groups.

4

Goals of Re-expression Goal 3 – Make the form of a

scatter plot more linear. Goal 4 – Make the scatter in

the scatter plot more even across all values of the explanatory variable.

5

Ladder of PowersPower: 2Re-expression:Comment: Use on left skewed

data.

2y

6

Ladder of PowersPower: 1Re-expression:Comment: No re-expression.

Do not re-express the data if they are already well behaved.

y

7

Ladder of PowersPower: ½ Re-expression:Comment: Use on count data

or when scatter in a scatter plot tends to increase as the explanatory variable increases.

y

8

Ladder of PowersPower: “0” Re-expression: Comments: Not really the “0”

power. Use on right skewed data. Measurements cannot be negative or zero.

ylog

9

Ladder of PowersPower: –½, –1 Re-expression: Comments: Use on right

skewed data. Measurements cannot be negative or zero. Use on ratios.

yy

1,

1

10

Goal 1 - Symmetry Data are obtained on the time

between nerve pulses along a nerve fiber.

Time is rounded to the nearest half unit where a unit is of a second.

– 30.5 represents

th

501

sec 61050530 ..

11

.01

.05

.10

.25

.50

.75

.90

.95

.99

-3

-2

-1

0

1

2

3

Nor

mal

Qua

ntile

Plo

t

20

40

60

Cou

nt0 10 20 30 40 50 60 70

Time ( sec)th

501

12

Time – Nerve Pulses Distribution is skewed right. Sample mean (12.305) is much

larger than the sample median (7.5).

Many potential outliers. Data not from a Normal model.

13

.01

.05

.10

.25

.50

.75

.90

.95

.99

-3

-2

-1

0

1

2

3

Nor

mal

Qua

ntile

Plo

t

10

20

30

40

Cou

nt0 1 2 3 4 5 6 7 8 9

Sqrt(Time)

14

.01

.05

.10

.25

.50

.75

.90

.95

.99

-3

-2

-1

0

1

2

3

Nor

mal

Qua

ntile

Plo

t

10

20

30

Cou

nt-1 0 1 2 3 4 5

Log(Time)

15

Summary Time – Highly skewed to the

right. Sqrt(Time) – Still skewed right. Log(Time) –Fairly symmetric

and mounded in the middle.– Could have come from a Normal

model.

16

Goal 3 – Straighten Up What is the relationship

between the temperature of coffee and the time since it was poured?–Y, temperature ( oF)–X, time (minutes)

17

80

90

100

110

120

130

140

150

160

170

180

190

200Te

mp

0 10 20 30 40 50 60

Time (min)

Bivariate Fit of Temp By Time (min)

18

Cooling Coffee There is a general negative

association – as time since the coffee was poured increases the temperature of the coffee decreases.

19

Linear Model

100

110

120

130

140

150

160

170

180

190

Tem

p (F

)

-10 0 10 20 30 40 50 60

Time (min)

20

Linear Model Fit Summary

– Predicted Temp = 176.7 – 1.56*Time

– On average, temperature decreases 1.56 oF per minute.

– R2 = 0.99, 99% of the variation in temperature is explained by the linear relationship with time.

21

Plot of Residuals

-5

-4

-3

-2

-1

0

1

2

3

4

5R

esid

ual

-10 0 10 20 30 40 50 60

Time (min)

22

Curved Pattern There is a clear pattern in the

plot of residuals versus time.–Under predict, over predict,

under predict. The linear fit is very good,

but we can do better.

23

4.5

4.6

4.7

4.8

4.9

5

5.1

5.2

5.3

5.4

5.5

Log(

Tem

p)

-10 0 10 20 30 40 50 60

Time (min)

Linear Fit

Bivariate Fit of Log(Temp) By Time (min)

24

Log(Temp) by Time Summary

– Predicted Log(Temp) = 5.1946 –0.0114*Time

–On average, log temperature decreases 0.0114 log(oF) per minute.

25

Plot of Residuals

-0.010

-0.005

0.000

0.005

0.010

Res

idua

l

-10 0 10 20 30 40 50 60

Time (min)

26

Interpretation There is a random scatter of

points around the zero line. The linear model relating

Log(Temp) to Time is the best we can do.

27

Original Scale? Predicted Log(Temp) = 5.1946 –

0.0114*Time Predicted Temp =

180.3*e–0.0114*Time

– Predicted temp at time=0, 180.3 oF– The predicted temp in one more minute

is the predicted temp now multiplied by e–0.0114 = 0.98866

28

JMP Method 1

–Create a new column in JMP, Log(Temp): Cols – Formula –Transcendental – Log.

29

JMP Method 1 (continued)

–Fit Y by XY – Log(Temp)X – Time

–Fit Linear

30

JMP Method 2

–Fit Y by XY – TempX – Time

–Fit SpecialTransform Y – Log