SPSS Homework Pearson r Use when you want to examine the relationship between two continuous...

SPSS Homework

Pearson r

YX

XY

SS

COVr

Pearson r

YX

XY

SS

COVr

Use when you want to examine the relationship between two continuous variables

Point-Biserial Correlation

• Use when you want to examine the association between two variables and:

• One variable is continuous– Extraversion (1 – 5)– IQ (1 – 200)– Time (in seconds)

• One variable is dichotomous– Male vs. Female– Married vs. Single– IBM user vs. Mac user

Point-Biserial Correlation

• The dichotomous variable must be coded with a number

• Gender1 = male2 = female

• Computer1 = IBM2 = Mac

Point-Biserial Correlation

• The dichotomous variable must be coded with a number (it doesn’t matter what the numbers are)

• Gender9 = male-2 = female

• Computer0 = IBM98 = Mac

Point-Biserial Correlation

Comp IQ1.00 110.001.00 120.001.00 98.001.00 115.001.00 99.00.00 90.00.00 100.00.00 82.00.00 101.00.00 91.00

Computer

1 = IBM

0 = Mac

Point-Biserial Correlation

Comp IQ1.00 110.001.00 120.001.00 98.001.00 115.001.00 99.00.00 90.00.00 100.00.00 82.00.00 101.00.00 91.00

Computer

1 = IBM

0 = Mac

Simply do a normal r

Point-Biserial Correlation

Comp IQ1.00 110.001.00 120.001.00 98.001.00 115.001.00 99.00.00 90.00.00 100.00.00 82.00.00 101.00.00 91.00

Computer

1 = IBM

0 = Mac

Simply do a normal r

Cov = 4.33

Scomp = .527

SIQ = 11.70

Point-Biserial Correlation

Comp IQ1.00 110.001.00 120.001.00 98.001.00 115.001.00 99.00.00 90.00.00 100.00.00 82.00.00 101.00.00 91.00

Computer

1 = IBM

0 = Mac

Simply do a normal r

Cov = 4.33

Scomp = .527

SIQ = 11.70

r = .70

Point-Biserial Correlation

Comp IQ-90.00 110.00-90.00 120.00-90.00 98.00-90.00 115.00-90.00 99.0012.20 90.0012.20 100.0012.20 82.0012.20 101.0012.20 91.00

Computer

-90 = IBM

12.2 = Mac

Point-Biserial Correlation

Comp IQ-90.00 110.00-90.00 120.00-90.00 98.00-90.00 115.00-90.00 99.0012.20 90.0012.20 100.0012.20 82.0012.20 101.0012.20 91.00

Simply do a normal r

Cov = -442.867

Scomp = 53.86

SIQ = 11.70

Computer

-90 = IBM

12.2 = Mac

Point-Biserial Correlation

Comp IQ-90.00 110.00-90.00 120.00-90.00 98.00-90.00 115.00-90.00 99.0012.20 90.0012.20 100.0012.20 82.0012.20 101.0012.20 91.00

Simply do a normal r

Cov = -442.867

Scomp = 53.86

SIQ = 11.70

r = -.70

Computer

-90 = IBM

12.2 = Mac

Point-Biserial Correlation

• The direction of the r (+ or - ) is changes depending on how the dichotomous variable was coded

• rpb = r

• Calculate a point-biserial correlation the same way as before, you just need to remember to label it differently!

• Significance tests for rpb are exactly the same as before

Do a t-test

Comp IQIBM 110.00IBM 120.00IBM 98.00IBM 115.00IBM 99.00MAC 90.00MAC 100.00MAC 82.00MAC 101.00MAC 91.00

Computer

1 = IBM

0 = Mac

5 108.4000 9.7108 4.3428

5 92.8000 7.8549 3.5128

COMP1.00

.00

IQN Mean

Std.Deviation

Std. ErrorMean

Group Statistics

.514 .494 2.793 8 .023 15.6000 5.5857 2.7194 28.4806

2.793 7.665 .024 15.6000 5.5857 2.6208 28.5792

Equalvariancesassumed

Equalvariancesnotassumed

IQF Sig.

Levene's Test forEquality of Variances

t dfSig.

(2-tailed)Mean

DifferenceStd. ErrorDifference Lower Upper

95% ConfidenceInterval of the Mean

t-test for Equality of Means

Independent Samples Test

Point-Biserial Correlation

• Good effect size estimate to use for independent t-tests

• How to compute an r from a t

dft

trpb

2

2

dft

trpb

2

2

t = 2.793

df = 8

880.7

80.770.

t = 2.793df = 8

Phi Coefficient

• Use when you have TWO dichotomous variables

• An old friend. . . .

Phi

• Use with 2x2 tables

N

2 N = sample size

Remember

Height Yes No Total

Short 42 50 92

Not short 30 87 117

Total 72 137 209

Ever Bullied

2

O E O - E (O - E)2 (O - E)2

E42 31.69 10.31 106.30 3.35

50 60.30 -10.30 106.09 1.76

30 40.30 -10.30 106.09 2.63

87 76.69 10.31 106.30 1.392 = 9.13

Remember

21.209

13.9

Phi Coefficient

• Another way to think about this data is

Height Yes No Total

Short 42 50 92

Not short 30 87 117

Total 72 137 209

Ever Bullied

Bullied Height

1 1

1 0

0 1

1 1

0 0

0 0

0 0

1 1

0 0

Bullied

1 = Yes

0 = No

Height

1 = Tall

0 = Short

Bullied Height

1 1

1 0

0 1

1 1

0 0

0 0

0 0

1 1

0 0

Bullied

1 = Yes

0 = No

Height

1 = Tall

0 = Short

Can simply do a normal correlation!

r = .21

Bullied Height

-1 1

-1 0

0 1

-1 1

0 0

0 0

0 0

-1 1

0 0

Bullied

-1 = Yes

0 = No

Height

1 = Tall

0 = Short

As before, the sign of the correlation will change depending on how it is coded

r = -.21

Phi Coefficient

• The direction of the r (+ or - ) is changes depending on how the dichotomous variable was coded

• Ø = r

• To calculate a Ø by hand it is probably easier to simply use the X2 method

Phi Coefficient

• You can also test Ø for significance by using the X2

df = 1

• If you are given a Ø you can compute X2 for this test

• X2 = NØ2

df = 1

Examples

• Ø = .50; N = 10

• Ø = .50; N = 100

• Ø = .20; N = 50

Examples

• Ø = .50; N = 10• X2 = 10(.502) = 2.5

• Ø = .50; N = 100• X2 = 100(.502) = 25

• Ø = .20; N = 50• X2 = 50(.202) = 2

X2 crit = 3.84

• Rate on a scale of 1 – 5

• I want to be:

• Healthy

• Wealthy

• Wise

• Rate on a scale of 1 – 5

Britney Spears Beyonce Justin Timberlake

Christina Agulera Michael Jackson

• Now rank them in order of importance

• Healthy

• Wealthy

• Wise

Now rank them

Britney Spears Beyonce Justin Timberlake

Christina Agulera Michael Jackson

Ranked Data

Justin Timberlake 1

Britney Spears 2

Christina Agulera 3

Beyonce 4

Michael Jackson 5

Person 1

Ranked Data

Justin Timberlake 1 1

Britney Spears 2 3

Christina Agulera 3 2

Beyonce 4 5

Michael Jackson 5 4

Person 1 Person 2

Spearman’s rho

• Spearman’s correlation coefficient for ranked data (rs)

• Used to determine relations between two sets of rankings – Typically used for reliability issues

• Exact same formula as a Pearson r!rs = r

Ranked Data

Justin Timberlake 1 1

Britney Spears 2 3

Christina Agulera 3 2

Beyonce 4 5

Michael Jackson 5 4

COV = 2.00

SP1 = 1.58

SP2 = 1.58

rs = .80

Person 1 Person 2

Spearman’s rho

• Problem: These data are not normally distributed

• No agreed method for significance testing

Why?

• Different correlations exist primarily because in the past formulas were created to make calculations easier – The r is a pain to calculate by hand if N is big!

• You may run into trouble with people who do not understand that rpb, Ø, rs, are all equivalent to the Pearson r.

Kendall’s Tau

• Kendall’s τ

• Also used for rank ordered data

• Based on the number of inversions in the rankings

Inversions

Justin Timberlake 1 1

Britney Spears 2 3

Christina Agulera 3 2

Beyonce 4 5

Michael Jackson 5 4

Person 1 Person 2

Inversions

Justin Timberlake 1 1

Britney Spears 2 3

Christina Agulera 3 2

Beyonce 4 5

Michael Jackson 5 4

Person 1 Person 2

Inversions

Justin Timberlake 1 1

Britney Spears 2 3

Christina Agulera 3 2

Beyonce 4 5

Michael Jackson 5 4

Two inversions

Person 1 Person 2

Kendall’s Tau

2/)1(

)(21

NN

I

I = number of inversions

N = number of pairs of objects

Kendall’s Tau

2/)4(5

)2(21

I = number of inversions

N = number of pairs of objects

Kendall’s Tau

10

)2(2160.

I = number of inversions

N = number of pairs of objects

Kendall’s Tau

10

)2(2160.

If a pair of objects is sampled at random, the probability that two judges will rank these objects in the same order is .60 higher than the probability that they will rank them in the reverse order.

Kendall’s Tau

• Significance testing

• H1: t is > than zero

• Ho: t is < or = to zero

• Note: you are looking for >, thus it is a one-tailed test

Kendall’s Tau

• Significance testing

)1(9)52(2

NNN

Z

t = tau

N = number of pairs of objects

Kendall’s Tau

• Significance testing

)15)(5(9)5)5(2(2

60.

Z

t = tau

N = number of pairs of objects

Kendall’s Tau

• Significance testing

)15)(5(9)5)5(2(2

60.47.1

t = tau

N = number of pairs of objects

Kendall’s Tau

• Significance testing

Look up Z value in table to find exact p value (smaller area of curve)

One-tailed test makes most sense

If p < .05 reject Ho and accept H1 (t is greater than 0)

If p > .05 fail to reject Ho (t is not greater than 0)

)15)(5(9)5)5(2(2

60.47.1

p=.07

Practice

Lord of the Rings

21 Grams

Lost in Translation

Seabiscut

In American

American Splendor

50 First Dates

Love Actually

Mystic River

PracticeHolly George

Lord of the Rings 4 3

21 Grams 2 2

Lost in Translation 5 4

Seabiscut 3 5

In American 6 6

American Splendor 1 1

50 First Dates 7 7

Love Actually 8 8

Mystic River 9 9

PracticeHolly George

American Splendor 1 1

21 Grams 2 2

Seabiscut 3 5

Lord of the Rings 4 3

Lost in Translation 5 4

In American 6 6

50 First Dates 7 7

Love Actually 8 8

Mystic River 9 9

PracticeHolly George

American Splendor 1 1

21 Grams 2 2

Seabiscut 3 5

Lord of the Rings 4 3

Lost in Translation 5 4

In American 6 6

50 First Dates 7 7

Love Actually 8 8

Mystic River 9 9

2 inversions

Kendall’s Tau

I = number of inversions

N = number of pairs of objects

2/)1(

)(21

NN

I

Kendall’s Tau

2/)8(9

)2(2189.

I = number of inversions

N = number of pairs of objects

Kendall’s Tau

• Significance testing

)1(9)52(2

NNN

Z

t = tau

N = number of pairs of objects

Kendall’s Tau

• Significance testing

)19)(9(9)5)9(2(2

89.34.3

p < .0006

Kendall’s Tau vs. Spearman’s rho

• Although you can get a p value from a Tau. . . . Simply calculating an r value on the raw data is still what is most often done.

Practice Questions• Page 314

– 10.1• Stime = .489• Sperf = 11.743• COV = -3.105

• Page 316– 10.11

• Sadd = .471• Salchol=.457• COV = .135

– 10.14– 10.15 (test for significance)