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12.2 Correlation 299
This test statistic follows Student’s t distribution with n–2 degrees of freedom.
Example 12.11 As part of Covid-19 project, the number of hospitalization days
for 28 patients by gender is listed in Table
12.8. Find a point-biserial correlation
coefficient between sex and duration of hospitalization, %95 confidence interval,
and test statistic.
Solution: Standard deviation of the number of hospitalization days for 28 patients
was
S
Y
= 6.130, the first sample size n
1
= 13, the second sample size n
2
= 15,the
total sample size
n = 28, the proportion of the first sample size to the total sample
size was
p =
13
28
= 0.4643 and
(
1 − p
)
= 1 − 0.4643 = 0.5357, the total mean is
Y = 11.2143, the mean of the first group Y
1
= 7.846, and the mean of the second
group is
Y
2
= 14.133. By substituting these values in the formula, the point-biserial
correlation coefficient is obtained (Table
12.11).
r
pb
=
Y
1
− Y
2
S
Y
n
1
× n
2
n
(
n − 1
)
=
14.133 − 7.846
6.130
13 × 15
28
(
28 − 1
)
= 0.5209
The 95% confidence interval is:
r
pb
± Z
1−
α
2
×
1 − r
2
pb
2
n
1 + r
2
pb
1 − 6p
(
1 − p
)
4p
(
1 − p
)
= 0.5209 ± 1.96 ×
1 − 0.5209
2
2
28
1
+0.5209
2
1 − 6 × 0.4643
(
1 − 0.4643
)
4 × 0.4643
(
1 − 0.4643
)
= 0.5209 ± 1.96 ×
0.01896
1 + 0.271
−0.4924
0.9949
= 0.5209 ± 1.96 × 0.1281
=
(
0.2698 to 0.7720
)(
0.183 to 0.748
)
And test statistic is:
Table 12.11 Number of hospitalization days for Covid-19 patients according to gender
Male 17 17 8 20 22 8 6 13 11 8 16 13 18 28 7
Female 10 12 7 9 10 5 7 4 15 4 12 3 4 − −
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300 12 Correlation and Regression
t
pb
= r
pb
×
√
n − 2
1 − r
2
pb
=
0.521
√
28 − 2
√
1 − 0.52
2
=
2.6515
0.8542
= 3.104
which is greater than the critical value with a degree of freedom of 28–2 =
28 (
3.104 > t
(
0.975,26
)
= 2.056) and the null hypothesis is rejected. That is, t he
point-biserial correlation coefficient cannot be considered equal to zero.
If we plot the scatter diagram of two variables with SPSS and then fit the regression
line, it can be observed that the mean difference between the two groups is equal
to the slope of the regression line and the p-value obtained for the t-test and for the
point-biserial correlation coefficient is exactly identical.
Correlations
Gender Covid-19 Days hospitalization
Gender Covid-19 Pearson correlation 1 0.521
**
Sig. (two-tailed) 0.004
N 28 28
Days hospitalization Pearson correlation 0.521
**
1
Sig. (two-tailed) 0.004
N 28 28
**. Correlation is significant at the 0.01 level (two-tailed)
12.2.10 Running Point-Biserial Correlation on SPSS
To calculate the point-biserial correlation coefficient in SPSS, first select the
Analyze\Correlate\Bivariate path like the Pearson correlation coefficient to open a
window.
https://t.me/med1917
12.2 Correlation 301
Select the Dichotomous and the quantitative variable from the box on the left
and move it to the box on the right. Then mark the Pearson option to calculate the
point-biserial correlation coefficient.
As mentioned, the point-biserial correlation coefficient is a special case of Pearson
correlation coefficient, one of the variables of which is dichotomous. Therefore, we
need to record the quantities of both groups together in one column and assign
another column to the dichotomous variable to determine which value belongs to
group one and which value belongs to the other group (figure on the top left). Then, by
confirming OK, the program is executed and the value of the point-biserial correlation
coefficient is obtained.
Note that the dichotomous variable in SPSS must be entered by nominal scale so
that the digits entered in this column are considered code. Therefore, code (0 and 1)
or code (1, 2) or code (4 and 5) will not change the calculations.
12.2.11 Partial Correlation Coefficient
Partial correlation is a linear correlation between two quantitative variables (x and y)
while controls or eliminates the effect of another quantitative confounding variable
(z). Therefore, when the correlation coefficient between the two variables (x and y)
is calculated, another confounding variable such as z may influence the relationship
between the two variables (x and y). In such cases, the effect of the third variable (z)
can be adjusted by calculating the partial correlation coefficient. For this purpose,
we need to calculate the correlation between the mutual variables separately (
r
zy
, r
xz
and r
xy
) and then calculate the partial correlation using the following formula:
r
xy.z
= r
xy|z
=
r
xy
− r
xz
× r
zy
1 − r
2
xz
×
1 − r
2
zy
The test statistic for the partial correlation coefficient is the same as for the Pearson
correlation coefficient, except that df = n−3.
Example 12.12
For the data in Table 12.1, obtain the correlation coefficient between
the final grade of students and the number of student study hours after adjusting the
number of absence sessions.
Solution: In this example, we consider the exam score as x, the number of study
hours y, and the number of absence sessions as z. Like calculating Pearson correlation
coefficient, their mutual correlation coefficient is equal to:
r
xy
= 0.912, r
xz
=−0.964, and r
zy
=−0.930
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302 12 Correlation and Regression
By substituting above values in the relevant formula, the adjusted correlation
coefficient of the final grade and study hours for the number of student absence
sessions is obtained.
r
xy.z
= r
xy|z
=
r
xy
− r
xz
× r
zy
1 − r
2
xz
×
1 − r
2
zy
=
0.912 −
(
−0.964
)
×
(
−0.930
)
1 −
(
−0.964
)
2
×
1 −
(
−0.930
)
2
=
0.01548
0.09973
= 0.158
It can be seen that after adjusting and controlling the number of student absence
sessions, the correlation coefficient of the final grade and study hours, which was
previously 0.912, decreased to 0.158.
12.2.12 Comparing Two Correlation Coefficients
How can two correlation coefficients be compared whether their differences are
significant? These coefficients may be dependent, i.e., between one variable and two
other variables (e.g.,
r
xy
and r
xz
in the same sample, or independent coefficients, i.e.,
between variable X and Y in two different samples (
r
1
xy
and r
2
xy
).
Dependent Correlations
If correlation coefficients are dependent, the hypothesis test of equality of two
dependent correlation coefficients is used. The null and alternative hypotheses for
this test are:
H
0
: ρ
xy
= ρ
xz
⇒ ρ
xy
− ρ
xz
= 0
H
0
: ρ
xy
/= ρ
xz
⇒ ρ
xy
− ρ
xz
/= 0
The test statistic is defined as below:
t
r
xy
− r
xz
=
(r
xy
− r
xz
) ×
(
n − 3
)
1 + r
yz
2
1 − r
2
xz
− r
2
yz
− r
2
xy
+ 2r
xz
.r
yz
.r
yz
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12.2 Correlation 303
where r
xy
is the correlation coefficient between the two variables x and y, r
xz
is the
correlation coefficient between the two variables
x and z, r
yz
is the correlation
coefficient between the two variables
y and z and n is the sample size. This test
statistic should be compared with the critical value of table t with degree of
freedom (n−3).
Example 12.13 The correlation coefficients between stiffness (x), pain (y), and
activity level (z) in 24 elderly are
r
xy
= 0.861, r
yz
= 0.657, and r
xz
= 0.824,
respectively. Is there a significant difference between the correlation coefficients of
stiffness and pain (
ρ
xy
) with the correlation between pain and activity level (ρ
xy
)?
Solution: The critical value of t from t student table with degree of freedom df =
24–3 = 21 is 2.080 and test statistic is as below:
t
r
xy
− r
xz
=
(r
xy
− r
xz
) ×
(
n − 3
)
1 + r
yz
2
1 − r
2
xz
− r
2
yz
− r
2
xy
+ 2r
xz
.r
yz
.r
yz
=
(
0.861 − 0.657
)
×
√
(
24 − 3
)(
1 + 0.824
)
2
1 − 0.824
2
− 0.657
2
− 0.861
2
+ 2 × 0.824 × 0.657 × 0.861
=
1.2625
0.1606
= 7.862
Since the test statistics is greater than critical value (7.862 > 2.080), the null
hypothesis is rejected and the two correlation coefficients are not equal.
Example 12.14
Inasampleof40(n = 40), the correlation coefficients between
height of child (C) with father (F)is
r
CF
= 0.604, child with mother is r
CM
= 0.262,
and mother with father is
r
MF
= 0.151. Is there a significant difference between the
correlation coefficients of child and father (
ρ
CF
) with the correlation between child
and mother (
ρ
CM
)?
Solution: The critical value of t from t student table with degree of freedom df =
40–3 = 37 is 2.026 and test statistic is as below:
t
(
r
CM
− r
CF
)
=
(r
CM
− r
CF
) ×
√
(
n − 3
)(
1 + r
MF
)
2
1 − r
2
CM
− r
2
CF
− r
2
MF
+ 2r
CM
.r
CF
.r
MF
=
(
0.604 − 0.262
)
×
√
(
40 − 3
)(
1 + 0.151
)
2
1 − 0.604
2
− 0.262
2
− 0.151
2
+ 2 × 0.604 × 0.262 × 0.151
=
2.2318
1.1831
= 1.886
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304 12 Correlation and Regression
Since the test statistics is greater than critical value (1.886 < 2.026), the null
hypothesis is not rejected and the two correlation coefficients are equal.
Independent Correlations
If correlation coefficients are independent, the hypothesis test of equality of two
independent correlation coefficients is used. The null and alternative hypotheses for
this test are:
H
0
: ρ
1
= ρ
2
⇒ ρ
1
− ρ
2
= 0
H
0
: ρ
1
/= ρ
2
⇒ ρ
1
− ρ
2
/= 0
Since it is a little difficult to calculate the combined distribution of two correlation
coefficients, to test this hypothesis, we first transform both correlation coefficients
using the z Fisher score as follows:
Z
1
=
1
2
ln
1 + r
1
1 − r
1
, Z
2
=
1
2
ln
1 + r
2
1 − r
2
Note that the reverse transformation of Z
1
and Z
2
to r
1
of r
2
is obtained by:
r
1
=
e
z
1
− e
−z
1
e
z
1
+ e
−z
1
, r
2
=
e
z
2
− e
−z
2
e
z
2
+ e
−z
2
Because the combined distribution of Z
1
and Z
2
is approximately normal, the test
statistic of z is used for it:
Z
(
r
1
− r
2
)
=
Z
1
− Z
2
1
n
1
−3)
+
1
n
2
−3
And it should be compared with the critical value of z score in the standard normal
distribution.
Example 12.15
The correlation coefficients between stiffness (x) and activity (z)
in 16 male elderly are
r
1
= 0.709 and in 8 female elderly are r
2
= 0.849.Are the
correlation coefficients different between the two groups?
Solution: The transformed values of
r
1
and r
2
are equal to:
Z
1
=
1
2
ln
1 + r
1
1 − r
1
=
1
2
ln
1 + 0.709
1 − 0.709
= 0.8852
Z
2
=
1
2
ln
1 + r
2
1 − r
2
=
1
2
ln
1 + 0.849
1 − 0.849
= 1.2526
The test statistic for comparing the two correlation coefficients is equal to:
https://t.me/med1917
12.2 Correlation 305
Z
(
r
1
− r
2
)
=
Z
1
− Z
2
1
n
1
−3)
+
1
n
2
−3
=
1.2526 − 0.8852
1
16−3)
+
1
8−3
=
0.3674
0.526
= 0.6984
Because the test statistic is less than the critical value of 1.96, the null
hypothesis is not rejected, i.e., the difference in correlation coefficient between the
two communities is not significant.
Example 12.16 The correlation coefficients between height of son (S) and father
(F) in 20 adult males are
r
1
= r
SF
= 0.241, and correlation coefficients between
height of daughter (D) and father (M ) in 20 adult females are
r
2
= r
DM
= 0.306.
Are the correlation coefficients different between the two groups?
Solution: The transformed values of r
1
and r
2
are equal to:
Z
1
=
1
2
ln
1 + r
1
1 − r
1
=
1
2
ln
1 + 0.241
1 − 0.241
= 0.246
Z
2
=
1
2
ln
1 + r
2
1 − r
2
=
1
2
ln
1 + 0.306
1 − 0.306
= 0.316
The test statistic for comparing the two correlation coefficients is equal to:
Z
(
r
1
− r
2
)
=
Z
1
− Z
2
1
n
1
−3)
+
1
n
2
−3
=
0.246 − 0.316
1
20−3)
+
1
20−3
=
−0.070
0.343
=−0.204
Because the test statistic is less than the critical value of 1.96 (1.88 < 1.96), the
null hypothesis is not rejected, i.e., the difference in correlation coefficient between
the two communities is not significant.
12.2.13 Precautions
Before calculating the Pearson correlation coefficient, first draw the scatter diagram
of the two variables so that if there is a linear relationship between the two variables,
the Pearson correlation coefficient is calculated.
When there is an outlier in the dataset, be cautious in applying the Pearson
correlation coefficient. Because in these cases the value of the correlation
coefficient changes by eliminating an outlier that may have been wrongly measured
or taken from another population.
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306 12 Correlation and Regression
When the data source is several populations or sub-populations, it is necessary to
be cautious in applying the coefficient of correlation. Because each sub-population
may have a different correlation coefficient than the correlation coefficient of the
whole population.
When one of the variables is pre-specified, it is not appropriate to use a correlation
coefficient. For example, in measuring the response to different doses of a drug, the
correlation coefficient cannot be clear, because the choice of a specific dose level of
the drug is determined by the experimenter and is not random.
12.3 Regression
In medicine and health science, there are inherent relationships between two or more
variables. Such relationships are occurring every day in the practice of health and
medical fields. For example, blood pressure (BP) levels have a relationship with the
body mass index (BMI), and the osteoporosis is associated with physical activity and
dairy consumption, or tobacco consumption is related to many of cardiac and lung
disorders.
The medical researchers have to depict and describe the relationship between
these variables to be able to predict with some degree of precision, the value of one
variable for an individual with the help of the other variables. That is, how much a unit
increases or decreases in one variable, how much change it makes in another variable
(Fig.
12.2). It is a question that correlation coefficient is clearly unable to answer
and left the answer to the regression models. Because the correlation coefficient just
indicates the strength and direction of the relationship between two variables with a
single number between −1 and + 1.
Fig. 12.2 Variations in variable Y per unit change in variable X
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12.3 Regression 307
A way to describe the relationship between these two variables, and to predict the
value of one variable for each individual from the known value of the other variable,
is called regression, which was proposed by Francis Galton in the late-nineteenth
century.
Therefore, the purposes of regression models can be summarized as follows:
To be able to predict the value of one variable from other variables that are
associated to it.
Help to understand the mechanism of such relationships, particularly by studying
the effect of a changing in the value of one variable if other conditions do not change.
For this purpose, a line should be fitted to the set of observed values that is closest
to them, i.e., the vertical distance of each point with the fitted line should be minimize.
The general equation of this line ( regression) is as follows:
Y = a + bx + e
where Y is dependent variable, outcome or response variable, X is named
independent or predictor variable, b is slope of regression line or regression
coefficient, a is intercept and e is called error term or residual.
Typically, the above line minimizes the vertical distance between the observed
data and the fitted line. There are several approaches to estimate a and b, but the
“least squares method” is their most common one. Therefore, using this method to
fit a regression line leads to minimal errors, which is the sum of the squares of the
vertical distances of the observations from the line. Figure
12.3 shows difference
between observed values from fitted regression line for systolic blood pressure and
age in a sample of 11 girl school students (Table
12.12).
The values of b and a can be estimated from sampled data by the following
equations:
b =
(
x −
x
)(
y − y
)
(
x −
x
)
2
=
n.
xy −
x
y
n.
x
2
−
x
2
And
a =
(
y)
x
2
−
x
xy
n.
x
2
−
x
2
Or
a = y − b.x
The calculations of the above equation may be a bit complicated and difficult; to
simplify the calculations, it is recommended to assign five columns for
x, y, x
2
, y
2
, xy as in Table 12.2. Then, substitute the sum of each column directly
in formula of the regression line.
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308 12 Correlation and Regression
Fig. 12.3 Difference between observed values from the fitted regression line
Table 12.12 Age and systolic and diastolic blood pressures in millimeters of mercury, for 11
children girl students
Age 5 6 7 8 9 10 11 12 13 14 15
Systolic 99.8 102.6 103.0 105.4 106.8 107.2 110.6 111.5 114.0 114.3 115.0
Diastolic 66.9 67.4 67.8 69.6 69.9 69.8 71.5 71.6 72.7 74.1 74.8
12.3.1 Assumptions for Linear Regression
Like all parametric models, before using regression analysis, it is important to
consider and evaluate four assumptions in order to validate the regression method.
The dependent variable (Y ) should have a normal distribution for each value of
the independent variable X (Fig.
12.4);
(The independent variable of X (predictor) doesn’t need to be a random variable or
have an approximately normal distribution (Unlike for correlation). The regression
model is still valid even if the researcher chose specific values of X.)
The variance (or standard deviation) of Y should be the same for each value of X.
The relation between the two variables should be linear.
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