spss homework pearson r use when you want to examine the relationship between two continuous...
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SPSS Homework
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Pearson r
YX
XY
SS
COVr
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Pearson r
YX
XY
SS
COVr
Use when you want to examine the relationship between two continuous variables
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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
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Point-Biserial Correlation
• The dichotomous variable must be coded with a number
• Gender1 = male2 = female
• Computer1 = IBM2 = Mac
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Point-Biserial Correlation
• Good effect size estimate to use for independent t-tests
• How to compute an r from a t
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dft
trpb
2
2
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dft
trpb
2
2
t = 2.793
df = 8
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880.7
80.770.
t = 2.793df = 8
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Phi Coefficient
• Use when you have TWO dichotomous variables
• An old friend. . . .
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Phi
• Use with 2x2 tables
N
2 N = sample size
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Remember
Height Yes No Total
Short 42 50 92
Not short 30 87 117
Total 72 137 209
Ever Bullied
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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
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Remember
21.209
13.9
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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
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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
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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
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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
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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
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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
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Examples
• Ø = .50; N = 10
• Ø = .50; N = 100
• Ø = .20; N = 50
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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
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• Rate on a scale of 1 – 5
• I want to be:
• Healthy
• Wealthy
• Wise
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• Rate on a scale of 1 – 5
Britney Spears Beyonce Justin Timberlake
Christina Agulera Michael Jackson
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• Now rank them in order of importance
• Healthy
• Wealthy
• Wise
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Now rank them
Britney Spears Beyonce Justin Timberlake
Christina Agulera Michael Jackson
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Ranked Data
Justin Timberlake 1
Britney Spears 2
Christina Agulera 3
Beyonce 4
Michael Jackson 5
Person 1
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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
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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
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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
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Spearman’s rho
• Problem: These data are not normally distributed
• No agreed method for significance testing
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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.
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Kendall’s Tau
• Kendall’s τ
• Also used for rank ordered data
• Based on the number of inversions in the rankings
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Inversions
Justin Timberlake 1 1
Britney Spears 2 3
Christina Agulera 3 2
Beyonce 4 5
Michael Jackson 5 4
Person 1 Person 2
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Inversions
Justin Timberlake 1 1
Britney Spears 2 3
Christina Agulera 3 2
Beyonce 4 5
Michael Jackson 5 4
Person 1 Person 2
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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
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Kendall’s Tau
2/)1(
)(21
NN
I
I = number of inversions
N = number of pairs of objects
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Kendall’s Tau
2/)4(5
)2(21
I = number of inversions
N = number of pairs of objects
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Kendall’s Tau
10
)2(2160.
I = number of inversions
N = number of pairs of objects
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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.
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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
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Kendall’s Tau
• Significance testing
)1(9)52(2
NNN
Z
t = tau
N = number of pairs of objects
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Kendall’s Tau
• Significance testing
)15)(5(9)5)5(2(2
60.
Z
t = tau
N = number of pairs of objects
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Kendall’s Tau
• Significance testing
)15)(5(9)5)5(2(2
60.47.1
t = tau
N = number of pairs of objects
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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
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Practice
Lord of the Rings
21 Grams
Lost in Translation
Seabiscut
In American
American Splendor
50 First Dates
Love Actually
Mystic River
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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
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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
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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
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Kendall’s Tau
I = number of inversions
N = number of pairs of objects
2/)1(
)(21
NN
I
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Kendall’s Tau
2/)8(9
)2(2189.
I = number of inversions
N = number of pairs of objects
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Kendall’s Tau
• Significance testing
)1(9)52(2
NNN
Z
t = tau
N = number of pairs of objects
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Kendall’s Tau
• Significance testing
)19)(9(9)5)9(2(2
89.34.3
p < .0006
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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.
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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)
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