Transcript Correlation Coefficient

S519: Evaluation of
Information Systems
Social Statistics
Inferential Statistics
Chapter 13: correlation
coefficient
This week





Testing correlation coefficient
The interpretation
PEARSON
SPEARMAN
Using Excel and SPSS to calculate
correlation coefficient
Which test to use




Figure 13.1
The relationship between variables, and not
the difference between groups, is being
examined.
Only two variables are being used
The appropriate test statistic to use is the t
test for the correlation coefficient
Example
Quality of
Marriage
Quality of parent-child
relationship
76
81
78
76
76
78
76
78
98
88
76
66
44
67
65
59
87
77
79
85
68
76
77
98
99
98
87
67
78
43
33
23
34
31
51
56
43
44
45
32
33
28
39
31
38
21
27
43
46
41
41
48
56
55
45
68
54
33
Correlation coefficient

CORREL() and PEARSON()




Same value
There is no significant difference
Spearman’s rank correlation coefficient
Kendall's tau
T test for the significance of
the correlation coefficient

Step1: A statement of the null and research
hypotheses


Null hypothesis: there is no relationship between
the quality of the marriage and the quality of the
relationship between parents and children
Research hypothesis: (two-tailed, nondirectional)
there is a relationship between the two variables
H 0 :  xy  0
H1 : rxy  0
T test for the significance of
the correlation coefficient

Step2: setting the level of risk (or the level of
significance or Type I error) associated with
the null hypothesis


0.05 or 0.01
What does it mean?


on any test of the null hypothesis, there is a 5% (1%)
chance you will reject it when the null is true when
there is no group difference at all.
Why not 0.0001?

So rigorous in your rejection of false null hypothesis
that you may miss a true one; such stringent Type I
error rate allows for little leeway
T test for the significance of
the correlation coefficient

Step 3 and 4: select the appropriate test
statistics



The relationship between variables, and not the
difference between groups, is being examined.
Only two variables are being used
The appropriate test statistic to use is the t test for
the correlation coefficient
T test for the significance of
the correlation coefficient

Step5: determination of the value needed for rejection of
the null hypothesis using the appropriate table of critical
values for the particular statistic.





Table B4
compute the correlation coefficient (r=0.393)
Compute df=n-2 (df=27)
If obtained value>the critical value reject null hypothesis
If obtained value<the critical value accept null hypothesis
T test for the significance of
the correlation coefficient

Step6: compare the obtained value with the
critical value


obtained value: 0.393
critical value: 0.349
T test for the significance of
the correlation coefficient


Step 7 and 8: make decisions
What could be your decision? And why, how
to interpret?



obtained value: 0.393 > critical value: 0.349 (level
of significance: 0.05)
Coefficient of determination is 0.154, indicating
that 15.4% of the variance is accounted for and
84.6% of the variance is not.
There is a 5% chance that the two variables are
not related at all
A simplified solution

Using statistical software packages as SPSS,
we do not need to determine level of risk or
critical value. Instead, the exact probability
will be calculated. Moreover, numbers will be
automatically highlighted if they are bigger
than corresponding critical values.
Causes and associations

Two variables are related to each other
One causes another



having a great marriage cannot ensure that the
parent-child relationship will be of a high quality
as well;
The two variables maybe correlated because they
share some traits that might make a person a
good husband or wife and also a good parent;
It’s possible that someone can be a good
husband or wife but have a terrible relationship
with his/her children.
A critique


a correlation can
be taken as
evidence for a
possible causal
relationship, but
cannot indicate
what the causal
relationship, if any,
might be.
These examples
indicate that the
correlation
coefficient, as a
summary statistic,
cannot replace the
individual
examination of the
data.
Exercises




S-P267
1
2
3
S-P267-Q1
n
degree of
freedom
correlation
coefficient
level tail
critical value
20
18
0.567
0.01
one
0.516
80
78
-0.45
0.05
one
0.183
50
48
0.37
0.05
two
0.273
S-P267-Q2


Excel
SPSS



Pearson
Spearman
Kendal