Regression Correlation Background Defines relationship between two variables X and Y R ranges from...

Post on 03-Jan-2016

220 views 3 download

Transcript of Regression Correlation Background Defines relationship between two variables X and Y R ranges from...

Regression

Correlation BackgroundDefines relationship between two

variables X and YR ranges from

-1 (perfect negative correlation)0 (No correlation)

+1 (perfect positive correlation)

R=.689

Regression

Correlation BackgroundR2 Indicates reduction in error knowing X and Predicting Y R2 ranges from 0 (No reduction in error)1 (complete reduction in error)

R2=.474

Regression

ExamplesPredicting height from G.P.A.

R2 = 0 (Knowing height does not help predict G.P.A – best guess is always mean G.P.A.)

R2 = 1 (Knowing height in CM completely predicts height in Inches)

Regression

Real world examples are somewhere in between

Predicting height from weightR2 = .36 (Knowing height

somewhat helps predict weight)

Regression

But how do we figure out HOW to make that prediction given one of the variables?

Regression

Need background concept of slope

How much does Y change for a given change in X?

All lines have R=1

0

2

4

6

8

10

12

14

16

18

20

0 1 2 3 4 5 6 7 8 9

Y=X

Y=2X

Y=X/2

Regression

-20

-15

-10

-5

0

5

10

15

20

0 1 2 3 4 5 6 7 8 9

Y=-X

Y=-2X

Y=-X/2

All lines have R=-1

Regression

Need background concept of INTERCEPT

What is Y when X=0?

All lines have Same Slope but different intercept

-5

0

5

10

15

20

25

0 1 2 3 4 5 6 7 8 9

Y=2XY=2X+5Y=2X-3

Regression

Unique line is defined by Slope and Y-Intercept

Y=bX+a

b=slopea=Y-Interecpt

-7

-4

-1

2

5

8

11

14

17

20

0 1 2 3 4 5 6 7 8 9

Y=?x+?Y=?x+?Y=?x+?

Regression

Predicting depression from loneliness

Y= BDI Depression X= Loneliness

Y=2X+2

-7

-4

-1

2

5

8

11

14

17

20

0 1 2 3 4 5 6 7 8 9

Y=?x+?Y=?x+?Y=?x+?

Regression

Predicted vs. Actual R=1, R2=1No Error

Never happens like this in real world

-1

2

5

8

11

14

17

20

0 1 2 3 4 5 6 7 8 9

ActualDepressionscore

PredictedDepressionscore

Actual scores don’t fit on a line perfectly

Actual scores

0

3

6

9

12

15

18

21

24

27

1 2 3 4 5 6 7 8 9 10

Actual scores

Some possible solutions?Error is

Sum of (Predicted Y-Actual Y)2

0

3

6

9

12

15

18

21

24

27

0 1 2 3 4 5 6 7 8 9

Y=2x+4 (Error=50)

Y=1.5X+6(Error=85.25)

Actual scores

Where is the line with smallest error?

Least Squares Regression Line

Actual scores

0

3

6

9

12

15

18

21

24

27

1 2 3 4 5 6 7 8 9 10

Actual scores

Where is the line with smallest error?

Least Squares Regression Line

Calc slope=b=

Σ (X-X)(Y-Y)

----------------------------------------------------------

Σ (X-X)(X-X)

=2.13 with this data

Where is the line with smallest error?

Least Squares Regression Line

Calc y intercept = a Y- (b)(X)

=4 with this data

So Least squares regression line isY=2.13X+4

Where is the line with smallest error?

Least Squares Regression Line

0

3

6

9

12

15

18

21

24

27

1 2 3 4 5 6 7 8 9 10

Actual scores

Y=2.133X+4

How good is our prediction?Sum of (Predicted Y-Actual Y)2

X Score Actual Y score Predicted Y score Squared Error

0 5 4.00 1.00

1 7 6.13 0.75

2 8 8.27 0.07

3 11 10.40 0.36

4 8 12.53 20.55

5 15 14.67 0.11

6 17 16.80 0.04

7 22 18.93 9.40

8 18 21.07 9.40

9 25 23.20 3.24

4.5 13.6 44.93

Can we standardize this for an average Error?

Yes: Standard error of the estimate

Like a standard deviation

Gives average precition error per score

Standard error of the estimate = SQRT(SSresidual/Npairs-2)

In this example = SQRT(44.9/10-2)=SQRT(44.9/8)=2.36

Chi-square (χ2)

Non Parametric Statistical tests

Used fornominal data (categories)ordinal (ordered categories)non-normal interval/ratio data

Goodness of fit χ2 Used with nominal dataTests a DISTRIBUTION (not a mean)Sees if observed data FITS an expected distribution

H0=true frequency distribution is expected

H1=true frequency distribution has some other form

VEGAS BABY!!!

Rolling dice at the MirageLots of Snake Eyes coming up Are the dice fixed?Test with goodness of fitDoes our distribution FIT the expected distribution

VEGAS BABY!!!

Expected distribution for 120 rolls if fair:

Each die(dice) has 1/6 chance

1/6 X 120 = 20 of each type

Expected Distribution =[20,20,20,20,20,20]

VEGAS BABY!!!

Actual distribution for 120 rolls is:

[28,16,23,23,17,13]

Are these dice fair?

Use Goodness of fit χ2

VEGAS BABY!!!

Determine critical χ2 value:

df = number of categories – 1= 6-1 = 5

χ2 critical for df=5 is 11.07 from table

Cat Oi Ei (Oi-Ei) (Oi-Ei) 2 (Oi-Ei) 2 / Ei

1 28 20 8 64 3.2

2 16 20 -4 16 0.8

3 23 20 3 9 0.45

4 23 20 3 9 0.45

5 17 20 -3 9 0.45

6 13 20 -7 49 2.45

Σ 120 120 0 7.8

FAIR!!!

Cat Oi Ei (Oi-Ei) (Oi-Ei) 2 (Oi-Ei) 2 / Ei

1 56 40 16 256 6.4

2 32 40 -8 64 1.6

3 46 40 6 36 0.9

4 46 40 6 36 0.9

5 34 40 -6 36 0.9

6 26 40 -14 196 4.9

Σ 240 240 0 15.6

CHEAT!!!

Test of independence χ2

Used with nominal dataTests whether DISTRIBUTION 1 is dependent upon DISTRIBUTION 2

H0= Distribution 1 is independent of Distribution 2

H1= Distribution 1 is related to Distribution 2

Example: Are Men more likely to have supported was in IRAQ

100 Subjects (50 male, 50 female)Asked yes or no question about supporting war

in Iraq

H0= Gender does not affect likelihood of supporting war

H1= Gender does affect likelihood of supporting war

Determine critical Value

Df = (R-1) (C-1)

Df = (Category 1 Size -1) size X Category 2 Size -1)

= (2-1) X (2-1) = 1 X 1 = 1Critical Value from A-3 is 3.84

Set up Data

Males Females TotalSupport war 32 21 53Not support war 18 29 47

Total 50 50 100

Set up Data

Males Females TotalSupport war 32 (26.5) 21(26.5) 53 Not support war 18 (23.5) 29(23.5) 47

Total 50 50 100

Category Oi Ei (Oi-Ei) (Oi-Ei) 2 (Oi-Ei) 2 / Ei

M/S 32 26.5 5.5 30.3 1.14

M/N 18 23.5 -5.5 30.3 1.29

F/S 21 26.5 -5.5 30.3 1.14

F/N 29 23.5 5.5 30.3 1.29

Σ 100 100 0 4.86

Calculate observed χ2

Test observed against critical

observed χ2 = 4.86 critical χ2 = 3.84

So we reject the idea that gender does not affect support of war and conclude

Gender DOES affect support of war

McNemar test for significance of change

Used with nominal dataTests whether DISTRIBUTION 1 is dependent upon DISTRIBUTION 2

Same as test of dependence but uses SAME person to test nominal data before and after some event

Example: Are Men more likely to have supported was in IRAQ

100 Subjects Do you favor the pledge allegiance?Before and After terrorist attacks

H0= proportion of individuals supporting pledge before attacks is same as after attacks

H1= proportion of individuals supporting pledge before attacks is different after attacks

Determine critical Value

Df = 1 for all McNemar testsCritical Value is 3.84

Set up Data Before AttacksYes No Total

After Attacks Yes 33 20 53No 9 38 47

Total 42 58 100

Set up Data Before AttacksYes No Total

After Attacks Yes 33 20 (14.5) 53

No 9 (14.5) 38 47

Total 42 58

Category Oi Ei (Oi-Ei) (Oi-Ei) 2 (Oi-Ei) 2 / Ei

1 9 14.5 -5.5 30.3 2.09

2 20 14.5 5.5 30.3 2.09

Σ 29 29 0 4.17

Calculate observed χ2

Test observed against critical

observed χ2 = 4.71 critical χ2 = 3.84

So we reject the idea that the proportions are the same

Conclusion: Attacks did change the proportion who support pledge of allegiance