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12.3 Regression 309
Fig. 12.4 Distribution of Y for each value of independent value of X
Tips
The normality assumption i s evaluated based on the residuals and can be
evaluated using a QQ-plot (plot 2). In the “ideal” normal distribution,
observations lie well along the 45-degree line in the QQ-plot. The farther the
observations are from this line, the farther the distribution is from normal.
To investigate the homoscedasticity assumption, we need to ensure that
there are no patterns in the residuals and that they are equally spread around
the y = 0 line.
To assess the linearity assumption, we make ensure that the residuals are
not too far away from 0 (standardized values less than −2 or greater than 2 are
deemed problematic).
Outliers have a strong effect on the regression equation, so the presence of outliers
should be investigated.
Fortunately assess the first three assumptions is easy by plotting residuals against
the x-values, because the residuals are informative and include the useful relevant
information. If the residuals have a normal distribution (with a mean of zero), means
that the three above assumptions are met. Then plot the residuals against the x-values.
If the above plot shows that all the residuals are scattered around the horizontal line
of mean (zero), it indicates that all the first three assumptions are met.
Example 12.17
Intraocular pressure (IOP), accommodation (ACC), and age of 11
patients are listed in Table
12.13. Find the regression line equation of IOP by age.
Table 12.13 Age, intraocular pressure (IOP), and accommodation (ACC) in a sample of 11 patients
in optometry clinic
Age 20 22 28 34 34 35 37 38 40 43 43
IOP 10 13 12 12 14 13 15 18 18 17 16
ACC 9.5 11 10.5 6.5 7.5 8 5 6 5 2.5 3.5
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310 12 Correlation and Regression
Solution: Like correlation coefficient, we perform the calculations required for
regression coefficients in Table
12.14 and substitute the sum of columns into the
regression formula.
Now, we substitute values from the last row in the formula and estimate the
regression coefficients of band a.
b =
n.
xy −
x
y
n.
x
2
−
x
2
=
11 × 5539 − 374 × 158
11 × 13316 −
(
374
)
2
=
1837
6600
= 0.278
x =
374
11
= 34,
y =
158
11
= 14.3636
a = y − b.x = 14.3636 − 0.278 × 34 = 4.900
Therefore, the equation of intraocular pressure (IOP) regression line in terms of
age is as follows:
IOP = 4.900 + 0.278 × Age
In order to predict the intraocular pressure (IOP) of a 30-year-old person, in the
above equation we replace the value of 30 instead of age.
IOP = 4.900 + 0.278 × Age
= 4.900 + 0.278 × 30
= 13.24
Table 12.14 Necessary calculations for estimating regression coefficient for data on Table 12.11
Patient Age (x) IOP (y) x
2
y
2
x.y
1 20 10 400 100 200
2 22 13 484 169 286
3 28 12 784 144 336
4 34 12 1156 144 408
5 34 14 1156 196 476
6 35 13 1225 169 455
7 37 15 1369 225 555
8 38 18 1444 324 684
9 40 18 1600 324 720
10 43 17 1849 289 731
11 43 16 1849 256 688
Sum
x = 374
y = 158
x
2
= 13316
y
2
= 2340
x.y = 5539
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12.3 Regression 311
12.3.2 Confidence Interval and Significant Test
for Regression Coefficient
Regression coefficient b is an estimate of the population β, which is obtained from the
sample data and is influenced by the sample size and sampling variations. Therefore,
like other parameters, it can be tested with the value specified in null hypothesis (
β
0
).
When we have only one independent variable in the regression model, the null and
alternative hypotheses for regression coefficient test are as follows:
Tip
The values of the independent variable x, which is substituted in this equation
to predict the dependent variable, must be in the range of the dataset (between
minimum and maximum). For instance, in the example above, age between
20 and 43 should be chosen to predict intraocular pressure (IOP). Because
the regression equation is obtained based on such data and if the data r ange
changes, the regression line equation will also change.
H
0
: β = β
0
H
A
: β /= β
0
The test statistic is:
t =
ˆ
β − β
0
Se
(
b
)
where the standard error of the regression coefficient is also equal to:
Se
(
b
)
=
S
res
S
x
S
res
=
y
2
−
(
y
)
2
n
− b
2
x
2
−
(
x
)
2
n
(
n − 2
)
S
x
=
x
2
−
x
2
n
This test statistic should be compared with the critical value of the two tailed of
the t distribution table and with the degree of freedom (n−2).
To find a 95% confidence interval for b by taking t standard errors on either side
of the estimate.
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312 12 Correlation and Regression
b ± t
(
1−
α
2
,n−2
)
× Se
(
b
)
Example 12.18 In Example 12.11 [intraocular pressure (IOP) and age of 11 patients]
test whether the regression coefficient can be equal to 0? And find a 95% confidence
interval for it.
Solution: In this example, degree of freedom is equal to df = n−2 = 11–2 = 9,
β
0
= 0 and
H
0
: β = β
0
H
A
: β /= β
0
. To find the standard error of residuals, we first calculate
S
res
and S
x
.
S
res
=
y
2
−
(
y
)
2
n
− b
2
x
2
−
(
x
)
2
n
(
n − 2
)
=
2340 −
(
158
)
2
11
− 0.278
2
13316 −
(
374
)
2
11
(
11 − 2
)
= 1.639
S
x
=
x
2
−
x
2
n
=
13316 −
(
374
)
2
11
= 24.495
And
Se
(
b
)
=
S
res
S
x
=
1.639
24.495
= 0.067
Subsisting the above value gives us the test statistic.
t =
ˆ
β − β
0
Se
(
b
)
=
0.278 − 0
0.067
= 4.149
The above statistic is greater than the critical value of t with a degree of freedom
of 9 and the null hypothesis is rejected.
4.149 > t
(
1−
α
2
,n−2
)
= t
(
0.975,9
)
=2.262
The 95% confidence interval for b is obtained as follows:
b ± t
(
1−
α
2
,n−2
)
× Se
(
b
)
= 0.278 ± 2.262 × 0.067 = 0.278 ± 0.151
=
(
0.127 to 0.429
)
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12.3 Regression 313
Because the above interval does not include the value specified in hypothesis zero
(0), so there is a significant difference between the population β with zero.
12.3.3 Running Simple Linear Regression in SPSS
To calculate the simple linear regression in SPSS, we first select the
Analyze\Regression\Linear path to open a window like the one below.
Select the independent variable of age from the left box and move to the
Independent(s) box in the middle of the window. We also select the IOP-dependent
variable from the same list and move to the Dependent box at the top of the middle
of the window. To have the confidence interval of the regression coefficient, select
the Statistics button at the top right to open another window. In this sub-window,
we mark the Confidence interval option and close it by pressing the Continue
button, and then we confirm the OK button to run the regression model and we will
have the following output.
Unstandardized
coefficients
Standardized
coefficients
95.0% confidence interval for B
B Std. error Beta t Sig. Lower bound Upper bound
(Constant) 4.900 2.323 2.110 0.064 −0.354 10.154
Age 0.278 0.067 0.812 4.169 0.002 0.127 0.429
As can be seen, the test statistic, regression coefficient, and confidence interval
are equal to manual calculations.
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314 12 Correlation and Regression
Unstandardized
coefficients
Standardized
coefficients
t Sig 95.0% confidence interval
for B
B Std. error Beta Lower
bound
Upper bound
(Constant) 4.900 2.323 2.110 0.064 −0.354 10.154
AGE 0.278 0.067 0.812 4.169 0.002 0.127 0.429
Example 12.19
Freezing of gait (FOG) and Mini-Mental State Examination
(MMSE) score in Parkinson’s patients is in Table
12.15. Find the regression
equation for FOG in terms of MMSE and calculate the test statistics and its
confidence interval.
Solution: We summarize the required calculations in Table
12.16.
b =
n.
xy −
x
y
n.
x
2
−
x
2
=
9 × 1676 − 245 × 63
9 × 6731 −
(
245
)
2
=
−351
554
=−0.634
x =
245
9
= 27.22,
y =
63
9
= 7.0
Table 12.15 Freezing of gait (FOG) and mini-mental state examination (MMSE) score in
Parkinson’s patients
MMSE 26 22 24 30 30 29 29 28 27
FOG 4 9 13 8 2 9 3 10 4
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12.3 Regression 315
Table 12.16 Necessary calculations for estimating regression equation for data on Table 12.13
Patient MMSE (x) FOG (y) x
2
y
2
x.y
1 26 5 676 25 130
2 22 9 484 81 198
3 24 13 576 169 312
4 30 8 900 64 240
5 30 2 900 4 60
6 29 9 841 81 261
7 29 3 841 9 87
8 28 10 784 100 280
9 27 4 729 16 108
Sum
x = 245
y = 63
x
2
= 6731
y
2
= 549
x.y = 1676
a = y − b.x = 7.0 −
(
−0.634
)
× 27.22 = 24.246
S
res
=
y
2
−
(
y
)
2
n
− b
2
x
2
−
(
x
)
2
n
(
n − 2
)
=
549 −
(
63
)
2
9
−
(
−0.634
)
2
6731 −
(
245
)
2
9
(
9 − 2
)
= 3.449
S
x
=
x
2
−
x
2
n
=
6731 −
(
245
)
2
9
= 7.846
and
Se
(
b
)
=
S
res
S
x
=
3.449
7.846
= 0.4396
Subsisting the above value gives us the test statistic.
t =
ˆ
β − β
0
Se
(
b
)
=
−0.634 − 0
0.4396
= 1.442
The above statistic is less than the critical value of t with a degree of freedom of
7 and the null hypothesis is not rejected.
1.442 < t
(
1−
α
2
,n−2
)
= t
(
0.975,9
)
=2.365
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316 12 Correlation and Regression
The 95% confidence interval for b is obtained as follow:
b ± t
(
1−
α
2
,n−2
)
× Se
(
b
)
=−0.634 ± 2.365 × 0.4396
=−0.634 ± 1.040
=
(
−1.674 to 0.406
)
Because the above interval includes the value specified in hypothesis zero (0), so
there is not a significant difference between the population β with zero.
12.4 Coefficient of Determination
As you’re aware, the correlation coefficient assesses the linear relationship between
two sets of data, indicating both similarity and dissimilarity. When squared (i.e., R
2
= r
2
), it yields the coefficient of determination. This measure reveals the proportion
of shared variance between two variables, often referred to as the common variance.
The coefficient of determination is valuable as it quantifies that how much of the
variance in one variable (e.g., blood pressure) can be attributed to another variable
(e.g., age). In essence, it represents the portion of the variance in the dependent
variable (Y) that can be explained by a linear regression model involving the predictor
variable (X). In simpler terms, it signifies the extent of shared variance between the
two variables X and Y.
In general, a high R
2
value indicates that the model is a good fit for the data.
The coefficient shows only the magnitude of linear association, not whether that
association is statistically significant and it can never determine whether one variable
is the cause of another variable.
The coefficient of determination takes values between 0 and 1. Since the coefficient
of determination is a squared value, the direction of variations is meaningless. As
shown in Fig.
12.4, a value of 0 indicates that the two variables have nothing in
common to explain the variance of each other. A value of 1 says that the two variables
are able to explain the entire variances for each other. An R
2
= 0.36 indicates that
%36 of the variation in the outcome has been explained just by predictor included in
the model.
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12.5 Exercises 317
Subtracting the coefficient of determination from one, (1 − R
2
), gives the
percentage of variance not shared between two variables, a quantity also called the
coefficient of alienation.
12.5 Exercises
1. Body weight by kilograms and blood plasma volume (liters) of 8 healthy men
were measured and are shown in Table
12.17.
A. Is there a correlation between these two variables?
B. If we consider the plasma volume as a dependent variable, find the regression
line equation?
C. From the regression line equation, predict plasma volume for people
weighing 60, 65, and 70 kg.
Table 12.17 Body weight and blood plasma (liters) of 8 healthy men
Weight 58.0 70.0 74.0 63.5 62.0 70.5 71.0 66.3
Plasma volume 2.75 2.86 3.37 2.76 2.62 3.48 3.05 3.12
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318 12 Correlation and Regression
Hgb PCV% Age Menopause
11.1 35 20 0
10.7 45 22 0
12.4 47 25 0
14.0 50 28 0
13.1 31 28 0
10.5 30 31 0
9.6 25 32 0
12.5 33 35 0
13.5 35 38 0
13.9 40 40 0
15.1 45 45 1
13.9 47 49 0
16.2 49 54 1
16.3 42 55 1
16,8 40 57 1
17.1 50 60 1
16.6 46 62 1
16.9 55 63 1
15.7 42 65 1
16.5 66 67 1
D. Can an individual plasma volume weighing 90 kg or 50 kg be estimated?
Why?
2. Find the correlation between weight and arm circumference for the data in Table 9.
2. Then assume arm circumference as a dependent variable and obtain the arm
circumference regression equation in terms of body weight. Then predict the
value of the arm circumference for those who weigh 75 and 85 kg.
3. Is there a significant correlation between the two methods for the data in
Table
11.15 (blood pressure of 16 patients)?
4.
Draw a scatter diagram for data in Table 2.11
. In which group is there a linear
relationship between age and calcium value?
5.
Calculate the correlation coefficient between age and calcium value for the data
in Table 2.11 and for each group separately and test their significance.
6.
The table below shows the level of hemoglobin (Hb), compressed cell volume
(PCV), age and menopausal status (0 = not menopause and 1 = menopause).
A. Draw the scattergram of age against PCV and hemoglobin separately.
B. Estimate the correlation coefficient of age with PCV.
C. Is the estimated correlation coefficient significant?
D. Obtain hemoglobin level regression equation by age.
E. Is it permissible to consider the regression line coefficient as equal to zero?
F. Estimate point-biserial correlation coefficient between menopausal status
and hemoglobin level.
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