SPSS Homework

1
(No Transcript)
2
SPSS Homework
3
Pearson r
4
Pearson r
Use when you want to examine the relationship
between two continuous variables
5
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

6
Point-Biserial Correlation

• The dichotomous variable must be coded with a
  number
• Gender
• 1 male
• 2 female
• Computer
• 1 IBM
• 2 Mac

7
Point-Biserial Correlation

• The dichotomous variable must be coded with a
  number (it doesnt matter what the numbers are)
• Gender
• 9 male
• -2 female
• Computer
• 0 IBM
• 98 Mac

8
Point-Biserial Correlation
Computer 1 IBM 0 Mac
Comp IQ 1.00 110.00 1.00 120.00 1.00 98.00 1.0
0 115.00 1.00 99.00 .00 90.00 .00 100.00 .00
82.00 .00 101.00 .00 91.00
9
Point-Biserial Correlation
Simply do a normal r
Computer 1 IBM 0 Mac
Comp IQ 1.00 110.00 1.00 120.00 1.00 98.00 1.0
0 115.00 1.00 99.00 .00 90.00 .00 100.00 .00
82.00 .00 101.00 .00 91.00
10
Point-Biserial Correlation
Simply do a normal r Cov 4.33 Scomp .527 SIQ
11.70
Computer 1 IBM 0 Mac
Comp IQ 1.00 110.00 1.00 120.00 1.00 98.00 1.0
0 115.00 1.00 99.00 .00 90.00 .00 100.00 .00
82.00 .00 101.00 .00 91.00
11
Point-Biserial Correlation
Simply do a normal r Cov 4.33 Scomp
.527 SIQ 11.70 r .70
Computer 1 IBM 0 Mac
Comp IQ 1.00 110.00 1.00 120.00 1.00 98.00 1.0
0 115.00 1.00 99.00 .00 90.00 .00 100.00 .00
82.00 .00 101.00 .00 91.00
12
Point-Biserial Correlation
Comp IQ -90.00 110.00 -90.00 120.00 -90.00 98.
00 -90.00 115.00 -90.00 99.00 12.20 90.00 12.20
100.00 12.20 82.00 12.20 101.00 12.20 91.00
Computer -90 IBM 12.2 Mac
13
Point-Biserial Correlation
Simply do a normal r Cov -442.867 Scomp
53.86 SIQ 11.70
Comp IQ -90.00 110.00 -90.00 120.00 -90.00 98.
00 -90.00 115.00 -90.00 99.00 12.20 90.00 12.20
100.00 12.20 82.00 12.20 101.00 12.20 91.00
Computer -90 IBM 12.2 Mac
14
Point-Biserial Correlation
Simply do a normal r Cov -442.867 Scomp
53.86 SIQ 11.70 r -.70
Comp IQ -90.00 110.00 -90.00 120.00 -90.00 98.
00 -90.00 115.00 -90.00 99.00 12.20 90.00 12.20
100.00 12.20 82.00 12.20 101.00 12.20 91.00
Computer -90 IBM 12.2 Mac
15
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

16
(No Transcript)
17
Do a t-test
Computer 1 IBM 0 Mac
Comp IQ IBM 110.00 IBM 120.00 IBM 98.00 IBM 1
15.00 IBM 99.00 MAC 90.00 MAC 100.00 MAC 82.00
MAC 101.00 MAC 91.00
18
(No Transcript)
19
Point-Biserial Correlation

• Good effect size estimate to use for independent
  t-tests
• How to compute an r from a t

20
(No Transcript)
21
t 2.793 df 8
22
t 2.793 df 8
23
(No Transcript)
24
Phi Coefficient

• Use when you have TWO dichotomous variables
• An old friend. . . .

25
Phi

• Use with 2x2 tables

N sample size
26
Remember
Ever Bullied
27
?2
28
Remember
29
Phi Coefficient

• Another way to think about this data is

Ever Bullied
30
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
31
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
32
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
33
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

34
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

35
Examples

• Ø .50 N 10
• Ø .50 N 100
• Ø .20 N 50

36
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
37

• Rate on a scale of 1 5
• I want to be
• Healthy
• Wealthy
• Wise

38

• Rate on a scale of 1 5

Britney Spears Beyonce Justin
Timberlake Christina Agulera Michael
Jackson
39

• Now rank them in order of importance
• Healthy
• Wealthy
• Wise

40

• Now rank them

Britney Spears Beyonce Justin
Timberlake Christina Agulera Michael
Jackson
41
Ranked Data

Justin Timberlake 1
Britney Spears 2
Christina Agulera 3
Beyonce 4
Michael Jackson 5
Person 1
42
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
43
Spearmans rho

• Spearmans 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

44
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
COV 2.00 SP1 1.58 SP2 1.58 rs .80
45
Spearmans rho

• Problem These data are not normally distributed
• No agreed method for significance testing

46
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.

47
Kendalls Tau

• Kendalls t
• Also used for rank ordered data
• Based on the number of inversions in the rankings

48
Inversions

Justin Timberlake 1 1
Britney Spears 2 3
Christina Agulera 3 2
Beyonce 4 5
Michael Jackson 5 4
Person 1 Person 2
49
Inversions

Justin Timberlake 1 1
Britney Spears 2 3
Christina Agulera 3 2
Beyonce 4 5
Michael Jackson 5 4
Person 1 Person 2
50
Inversions

Justin Timberlake 1 1
Britney Spears 2 3
Christina Agulera 3 2
Beyonce 4 5
Michael Jackson 5 4
Person 1 Person 2
Two inversions
51
Kendalls Tau
I number of inversions N number of pairs of
objects
52
Kendalls Tau
I number of inversions N number of pairs of
objects
53
Kendalls Tau
I number of inversions N number of pairs of
objects
54
Kendalls Tau
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.
55
Kendalls Tau

• Significance testing
• H1 t is gt than zero
• Ho t is lt or to zero
• Note you are looking for gt, thus it is a
  one-tailed test

56
Kendalls Tau

• Significance testing

t tau N number of pairs of objects
57
Kendalls Tau

• Significance testing

t tau N number of pairs of objects
58
Kendalls Tau

• Significance testing

t tau N number of pairs of objects
59
Kendalls Tau

• Significance testing

p.07
Look up Z value in table to find exact p value
(smaller area of curve) One-tailed test makes
most sense If p lt .05 reject Ho and accept H1 (t
is greater than 0) If p gt .05 fail to reject Ho
(t is not greater than 0)
60
Practice

Lord of the Rings
21 Grams
Lost in Translation
Seabiscut
In American
American Splendor
50 First Dates
Love Actually
Mystic River
61
Practice
Holly 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
62
Practice
Holly 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
63
Practice
Holly 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
64
Kendalls Tau
I number of inversions N number of pairs of
objects
65
Kendalls Tau
I number of inversions N number of pairs of
objects
66
Kendalls Tau

• Significance testing

t tau N number of pairs of objects
67
Kendalls Tau

• Significance testing

p lt .0006
68
Kendalls Tau vs. Spearmans 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.

69
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)

70
(No Transcript)