Interpreting Correlation
Aims: This note addresses the following:
 What does the coefficient of correlation (r) mean?
 Is the value of r statistically meaningful?
 Worked Examples.
 How should I report my results?
What does the coefficient of correlation (r) mean?
Correlation does not means causality. That is, if we find that size of feet and IQ are positively related then this does not mean that size of feet causes higher IQ. \[ r = \large \frac{\sum{xy}  \frac{\sum{x}\sum{y}}{n} }{ \sqrt{ \left[ \sum{x^2}  \frac{(\sum{x})^2}{n} \right] \left[\sum{y^2}  \frac{(\sum{y} )^2}{n} \right] } } \]
The numerator (the upper part of the ratio) measures how much x and y vary together (using the deviations of each from their respective means). The denominator (the lower part of the ratio) is a measure of x and y vary separately.
Productmoment correlation coefficient is another name for r. Another way to express r is
\( r = \large \frac{S_{xy}}{S_x S_y} \) where
\( S_{xy} = \sum{xy}  \large \frac{\sum{x}\sum{y}}{n} \),
\( S_x = \sqrt{ \left[ \sum{x^2}  \large \frac{(\sum{x})^2}{n} \right] } \) and
\( S_y = \sqrt{ \left[ \sum{y^2}  \large \frac{(\sum{y})^2}{n} \right] } \).Note that
\( S_{xy} = ns_{xy} \) or \( n \)Covariance of \(x\) and \(y\). Distinguish the use of capital \(S\) from small \(s\).
The covariance of \(x\) and \(y\), \(s_{xy} = \frac{\sum{xy}  \large \frac{\sum{x}\sum{y}}{n} }{n}\)
Similarly \( S_x = ns_x \) or \( n \)standard deviation of \(x\).
Standard deviation of \(x\), \( s_x = \sqrt{ \frac{\left[ \sum{x^2}  \large \frac{(\sum{x})^2}{n} \right]}{n} } \) and
\( S_y = ns_y \) or \( n \)standard deviation of \(y\).
Standard deviation of \(y\), \( s_y = \sqrt{ \frac{\left[ \sum{y^2}  \large \frac{(\sum{y})^2}{n} \right]}{n} } \)We need to beware of spurious correlation in a scatter diagram. Although there may appear a relationship on the diagram and mathematically, this does not imply a relationship in reality. For example in a particular city, we find an increase in vandalism of public utilities (like public toilets and telephones) and also an increase in the number of stores supplying organic vegetables over time. In this case, it will be unreasonable to suggest that more organic vegetable stores and vandalism are related.
A rule of thumb for interpreting the \(r\) value is by using this table.
\(r\) value Strength of (linear) Correlation \( 0.75 < r \le 1 \) Strong \( 0.5 < r \le 0.75 \) Moderate \( 0.25 < r \le 0.50 \) Weak \( 0 < r \le 0.25 \) Very Weak
Is the value of my coefficient of correlation (r) statistically meaningful?

In the Excel example with weight and height, we obtained an r = 0.94 (to 2 significant figures). Well, is this value of r = 0.94 statistically meaningful? What is the probability that this value is product of random chance? Or in a set of data we obtained an r = 0.500. Does this means a positive correlation or no correlation? How do we decide?
To decide we need to consult the following table for r.v = n2 degrees of freedom
p=0.05
p=0.025
p=0.01
p=0.005
1
0.9877
0.9969
0.9995
0.9999
2
0.9000
0.9500
0.9800
0.9900
3
0.8053
0.8783
0.9343
0.9587
4
0.7292
0.8113
0.8822
0.9172
5
0.6694
0.7544
0.8329
0.8745
6
0.6215
0.7067
0.7887
0.8343
7
0.5822
0.6664
0.7497
0.7977
8
0.5493
0.6319
0.7154
0.7646
9
0.5214
0.6020
0.6850
0.7348
10
0.4972
0.5759
0.6581
0.7079
11
0.4761
0.5529
0.6338
0.6835
12
0.4575
0.5323
0.6120
0.6613
13
0.4408
0.5139
0.5922
0.6411
14
0.4258
0.4973
0.5742
0.6226
15
0.4123
0.4821
0.5577
0.6054
16
0.4000
0.4683
0.5425
0.5897
17
0.3887
0.4555
0.5285
0.5750
18
0.3783
0.4437
0.5154
0.5614
19
0.3687
0.4328
0.5033
0.5487
20
0.3597
0.4226
0.4920
0.5367
25
0.3232
0.3808
0.4450
0.4869
30
0.2959
0.3494
0.4092
0.4487
35
0.2746
0.3246
0.3809
0.4182
40
0.2572
0.3044
0.3578
0.3931
45
0.2428
0.2875
0.3383
0.3721
50
0.2306
0.2732
0.3218
0.3541
60
0.2108
0.2500
0.2948
0.3248
70
0.1954
0.2318
0.2736
0.3017
80
0.1829
0.2172
0.2565
0.2829
90
0.1725
0.2049
0.2422
0.2673
100
0.1638
0.1946
0.2300
0.2540

Actually, we will learn how to do a significance test for correlation coefficient r in this note.
Assumptions
To carry out this test, we assume that X and Y variables are jointly normally distributed with correlation coefficient ρ.
ρ is read as “rho” and we can think of this as the (true) population correlation coefficient. We also assume that the observed data consist of a random sample from all possible values (populations) of X and Y. 
Forming Null Hypothesis
The null hypothesis H_{o} is always that the correlation coefficient r = 0, that is there is no correlation between our variables.
This null hypothesis is expressed as
H_{o}: ρ = 0. 
The number of degrees of freedom is total number of observations (n) minus two. If we have ten observations (n=10) then the degrees of freedom is thus 8.

Forming Alternative Hypothesis (I)
If we have reason to believe that r is positive say from observing a scatter plot of the data (upward trend) then our alternative hypothesis is
H_{1}: ρ > 0. 
Carrying out the test
The critical value at k% significance level and v degrees of freedom is c. For a 5% significance level, we usually write " ? = 0.05 " . The value ? is the probability of wrongly rejecting the null hypothesis when it is actually true. If we take 5% significance level and 8 degrees of freedom then from the table above our critical value is approximately 0.5493. The critical region is the region in which all r is larger than 0.5493 as illustrated above. For this particular example, the pvalue (or the probability to the right of c) for the shaded region is 0.05.If our calculated r ,say r = 0.981, is in the critical region then we will reject the null hypothesis H_{o} of no correlation and accept the alternative that the correlation is positive. We write, “Since r = 0.981 is more than the critical value of 0.5493 at 5% significance level and 8 degrees of freedom then we will reject the null hypothesis and accept the alternative. We conclude that there is statistical evidence to suggest that there is a positive correlation between these data.”
If our calculated r is NOT in the critical region, say r = 0.305, then we write, “Since r = 0.305 is less than the critical value of 0.5493 at 5% significance level and 8 degrees of freedom then we failed to reject the null hypothesis and reject the alternative. We conclude that there is not enough evidence to suggest that there is any correlation between these data.”I suggest that you use the above wordings in bold face with appropriate adjustments for your conclusion. Make sure you always express your conclusion in words and in the context of your task.

Forming Alternative Hypothesis (II)
If we have reason to believe that r is negative say from observing a scatter plot of the data (downward trend) then our alternative hypothesis is
H_{1}: ρ < 0. 
Carrying out the test
The critical value at k% significance level and v degrees of freedom is c. For instance if we take 1% significance level and 35 degrees of freedom then from the table above the value is 0.3809. Since our alternative is negative correlation then our critical value is 0.3809.
The critical region is the region in which all r is less than 0.3809 as illustrated below. For this particular example, the pvalue for the shaded region is 0.01.
If our calculated r ,say r = 0.686, is in the critical region then we will reject the null hypothesis H_{o} of no correlation and accept the alternative that the correlation is positive. We write, “Since r = 0.686 is less than the critical value of 0.3809 at 1% significance level and 35 degrees of freedom then we will reject the null hypothesis and accept the alternative. We conclude that there is statistical evidence to suggest that there is a negative correlation between these data.”
If our calculated r is NOT in the critical region, say r = 0.200, then we write, “Since r = 0.200 is larger than the critical value of 0.3809 at 1% significance level and 35 degrees of freedom then we failed to reject the null hypothesis and reject the alternative. We conclude that there is not enough evidence to suggest that there is any correlation between these data.”Both alternatives above are examples of onetaileded test. For most practical investigations, we will like to establish whether or not there is a positive or negative correlation then we will be carrying a onetaileded test as above.

Forming Alternative Hypothesis (III)
If we are only interested in knowing whether or not there is a correlation regardless of sign then we will carry out a twotaileded test. In this case, the alternative is
H_{1}: ρ ≠ 0. 
Carrying out the test
For instance if we take 5% significance level and 20 degrees of freedom then the critical value at the upper tail is only p=0.025. We need to look at the column with p=0.025 and 20 degrees of freedom in the table. From the table above our critical value is 0.4226. The critical value at the lower tail is thus 0.4226. Each critical region above has a pvalue of p/2 or 0.025 in this example.
The critical region is the region in which all r is either larger than 0.4226 or less than 0.4226 as illustrated above. For this particular example, the pvalue for the shaded region is 0.025 for a combined pvalue of 0.05.If our calculated r ,say r = 0.500 or r = 0.456, is in the critical region then we will reject the null hypothesis H_{o} of no correlation and accept the alternative that there is some correlation between the two variables. We write, “Since r = 0.456 is less than the critical value of 0.456 at 5% significance level and 20 degrees of freedom then we will reject the null hypothesis and accept the alternative. We conclude that there is statistical evidence to suggest that there is some correlation between these data.”
If our calculated r is NOT in the critical region, say r = 0.305, then we write, “Since r = 0.305 is less than the critical value of 0.456 at 5% significance level and 8 degrees of freedom then we failed to reject the null hypothesis and reject the alternative. We conclude that there is not enough evidence to suggest that there is any correlation between these data.” 
Let say, for a particular sample we have 47 as our degrees of freedom. This value is not in our table and we will use the tabulated values in row v = 45. That is, we will impose a more stringent test.

Notice that the critical value for 100 degrees of freedom (or n = 102) and 0.5 % significance level is only 0.2540 which is less than 0.30. If you have a scatter diagram with total observations more than 102 with r > 0.30 then according to the table above, we have strong statistical significance for this coefficient of correlation.
Worked Examples
Example1
Let us look at our previous example
(Excel file with Macro).
X height 
Y weight 
178 
53 
174 
50 
180 
58 
182 
60 
190 
70 
195 
85 
165 
45 
168 
48 
173 
51 
175 
60 
∑x=1780 
∑y=580 
(a) Calculate
the correlation coefficient r of the data given above.
A
researcher believes that people who are taller tend to be heavier in
weight.
(b) Assuming that these data are jointly normally
distributed with correlation coefficient ρ and the data are
collected from a random sample, carry out a significance test at 5%
significance level to verify the researcher's belief.
(c) Repeat
the test at 1% significance level.
Solution
(a)\( r = \large \frac{S_{xy}}{S_xS_y} \) To calculate r manually, we need the values for ∑ x , ∑ y, ∑ xy, n, ∑ x^{2}, and ∑ y^{2}. Here, we are going to use GDC to help us find the above values. From GDC as shown in the second column, ∑ x = 1780, ∑ y = 580, ∑ x^{2} = 317612, ∑ y^{2 }= 34928, ∑ xy =104181, n = 10 S_{xx}
= [317612 (1780)^{2}/10]
= 772 The value of r is positive which indicates that the researcher's belief may be right. . 
With the help of GDC.

(b) State
H_{o}
and
H_{1}.
H_{o}
: ρ = 0 ; there is no correlation between
height (X) and weight (Y).
H_{1} :
ρ > 0 ; there is a positive correlation between height (X) and weight
(Y)
(i.e. people who are tall tend to be heavier in weight).
State
the type of test and its significance level.
We will perform an
upper tail significance test at the 5% level.
State
the rejection criterion.
Since
the sample size is 10 there are 8 degrees of freedom in these data
and at 5% significance level we will be looking at p = 0.05 in the
table above. The critical value associated with p = 0.05 at 8 degrees
of freedom is 0.5493.
We will reject H_{o}
in favour of H_{1}
if r > 0.5493.
Calculate
r.
This
calculation has been done in (a) above. r
≈ 0.944 (3 s.f.).
Conclusion
of the test.
Since r = 0.944 is more than the critical
value of 0.5493 at 5% significance level and 8 degrees of freedom. We
will reject the null hypothesis and accept the alternative.
We
conclude that there is statistical evidence to suggest that there is
a positive correlation between these data.
(c)
At 1% significance level, the critical value is 0.7154.
Verify
this value using the table above.
Since r = 0.944 is more than
the critical value of 0.7154 at 1% significance level and 8 degrees
of freedom. We will reject the null hypothesis and accept the
alternative.
We conclude that there is statistical evidence to
suggest that there is a positive correlation between these data, i.e.
people who are tall tend to be heavier in weight.
Example 2
Nine students took part in a singing competition in which their scores
from an experienced judge and a trainee judge were recorded.
Experienced judge (X) 
5.70 
6.02 
5.10 
7.01 
8.52 
4.52 
8.02 
5.08 
6.67 
Trainee judge (Y) 
6.33 
6.28 
7.82 
7.34 
7.85 
5.52 
7.08 
4.79 
6.86 
(a) Calculate the
productmoment correlation coefficient of the data above.
(b) Use
a 5% significance level to test whether or not there is any
correlation between the scores given by these judges.
(c) Repeat
the above test for 10% significance level.
Solution
(a)\( r = \large \frac{S_{xy}}{S_xS_y} \)
To calculate r manually, we need the values for ∑ x , ∑ y, ∑ xy, n, ∑ x^{2}, and ∑ y^{2}.
Here, we are going to use GDC to help us find the above values.
S_{xx} = [371.517
(56.64)^{2}/9] = 15.0626 
(b) State
H_{o}
and
H_{1}.
H_{o}
: ρ = 0 ; there is no
correlation between the scores of these two judges.
H_{1}
: ρ ≠ 0 ; there is some
correlation between the scores of these two judges.
State
the type of test and its significance level.
We
will perform a twotailed significance test at the 5% level.
State
the rejection criterion.
Since
the sample size is 9 there are 7 degrees of freedom in these data and
at 5% significance level with two tails, we will be looking at p =
0.025 in the table above. We use p =0.025 because each tail only has
2.5%. The critical value associated with p = 0.025 at 7 degrees of
freedom is 0.6664.
We will reject H_{o}
in favour of H_{1}
if r > 0.6664.
Calculate
r.
This
calculation has been done in (a) above. r
≈ 0.632 (3 s.f.).
Conclusion
of the test.
Since r = 0.632 is less than the critical
value of 0.6664 at 5% significance level and 7 degrees of freedom. We
failed to reject the null hypothesis.
We conclude that there
is not enough statistical evidence to suggest that there is a
correlation between the scores given by these judges.
(c)
At 10% significance level, the critical value is 0.5822 since each
tail now has 5% or p = 0.05.
Verify this value using the table
above.
Since r = 0.632 is more than the critical value of 0.5822
at 10% significance level and 8 degrees of freedom. We will reject
the null hypothesis and accept the alternative.
We conclude
that there is statistical evidence to suggest that there is some
correlation between the scores given by these two judges.
Here is an example where we make different conclusions given different significance level. If we are conservative and not willing to take more than 5% chance of rejecting the null hypothesis when it is actually true then we conclude that the scores given by the trainee judge is not correlated with those given by the experienced judge. If we are willing to take up to 10% chance of rejecting the null hypothesis when it is actually true then we conclude that their scores are correlated in some way.
How should I report my results?
Report the value of r.
Report the degrees of freedom (v) and compare it with relevant tabulated value in the above table.
Example: r = 0.94
Degree of freedom is 8. The significance level is less than 0.5%.
This report implicitly suggests a onetailed (upper tail) significance test since the calculated value of r is positive, and there is less than 0.5% chance of rejecting the null hypothesis of no correlation when it is actually true.