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Chapter 13: Linear Correlation
and Regression Analysis
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Chapter Goals
• More detailed look at linear correlation
and regression analysis.
• Develop a hypothesis test to determine the
strength of a linear relationship.
• Consider the line of best fit. Use this to
make confidence interval estimations.
13.1: Linear Correlation Analysis
• The coefficient of linear correlation, r, is a
measure of the strength of a linear
relationship.
• Consider another measure of dependence:
covariance.
• Recall: bivariate data - ordered pairs of
numerical values.
Derivation of the covariance:
Goal: a measure of the linear relationship between two
variables.
Consider the following set of bivariate data:
{(8, 22), (5, 28), (8, 18), (4, 16), (13, 27), (15, 23), (17, 17),
(12, 13)}
x  10.25
y  20.50
Consider a graph of the data:
1. The point ( x, y ) is the centroid of the data.
2. A vertical and horizontal line through the centroid divides
the graph into four sections.
Graph of the data, with centroid.
3
0
2
8
( x  x)
2
6
2
4
2
2
( y  y)
(
1
0
.
2
5
,
2
0
.
5
)
y
2
0
1
8
1
6
1
4
1
2
1
0
468
1
0
1
2
1
4
1
6
1
8
2
0
x
Note:
1. Each point (x, y) lies a certain distance from each of the
two lines.
2. ( x  x ) : the horizontal distance from (x, y) to the vertical
line passing through the centroid.
3. ( y  y ) : the vertical distance from (x, y) to the horizontal
line passing through the centroid.
4. The distances may be positive, negative, or zero.
5. Consider the product: ( x  x)( y  y )
a. If the graph has lots of points to the upper right and
lower left of the centroid (positive linear relationship),
most products will be positive.
b. If the graph has lots of points to the upper left and lower
right of the centroid (negative linear relationship), most
products will be negative.
Covariance:
The covariance of x and y is defined as the sum of the
products of the distances of all values x and y from the
centroid divided by n  1.
n
covar ( x, y ) 
 ( xi  x)( yi  y)
i 1
n 1
Note:
 ( x  x)  0
and
 ( y  y)  0
always!
Calculations for finding covar(x, y):
Points
(8, 22)
(5, 28)
(8, 18)
(4, 16)
(13, 27)
(15, 23)
(17, 17)
(12, 13)
Total
xx
-2.25
-5.25
-2.25
-6.25
2.75
4.75
6.75
1.75
0.00
 16
covar ( x, y ) 
 2.2857
7
y y
1.5
7.5
-2.5
-4.5
6.5
2.5
-3.5
-7.5
0.0
( x  x)( y  y )
-3.375
-39.375
5.625
28.125
17.875
11.875
-23.625
-13.125
-16.000
Data and Covariance:
Positive covariance
8
7
6
5
y
4
( x, y )
3
2
1
0
012345678
x
Negative covariance
9
8
7
6
y
5
4
( x, y )
3
2
1
0
012345678
x
Covariance near 0
9
8
7
6
y
5
4
( x, y )
3
2
1
0
0123456789
x
Problem:
1. The covariance does not have a standardized unit of
measure.
2. Suppose we multiply each data point in the example in this
section by 15.
The covariance of the new data set is -514.29.
3. The amount of the dependency between x and y seems
stronger. But the relationship is really the same.
4. We must find a way to eliminate the effect of the spread of
the data when we measure the strength of a linear
relationship.
Solution:
1. Standardize x and y:
xx
x' 
sx
and
y y
y' 
sy
2. Compute the covariance of x’ and y’.
3. This covariance is not affected by the spread of the data.
4. This is exactly what is accomplished by the coefficient of
linear correlation:
covar ( x, y )
r  covar ( x' , y ' ) 
sx  s y
Note:
1. The coefficient of linear correlation standardizes the
measure of dependency and allows us to compare the
relative strengths of dependency of different sets of data.
2. Also commonly called Pearson’s product moment, r.
Calculation of r (for the data presented in this section):
s x  4.71
r
and
s y  5.37
covar ( x, y )
 2.2857

 0.0904
sx  s y
(4.71)(5.37)
Alternative (Computational) Formula for r:
 ( x  x)( y  y)
covar ( x, y )
r

sx  s y
n 1
sx  s y
SS( xy )

SS( x)  SS( y )
1. This formula avoids the separate calculations of the means,
standard deviations, and the deviations from the means.
2. This formula is easier and more accurate: minimizes
round-off error.
13.2 Inferences About the Linear
Correlation Coefficient
• Use the calculated value of the coefficient
of linear correlation, r*, to make an
inference about the population correlation
coefficient, r.
• Consider a confidence interval for r and a
hypothesis test concerning r.
Assumptions for inferences about linear correlation
coefficient:
The set of (x, y) ordered pairs forms a random sample and the
y-values at each x have a normal distribution. Inferences use
the t-distribution with n  2 degrees of freedom.
Caution:
The inferences about the linear correlation coefficient are
about the pattern of behavior of the two variables involved
and the usefulness of one variable in predicting the other.
Significance of the linear correlation coefficient does not
mean there is a direct cause-and-effect relationship.
Confidence Interval Procedure:
1. A confidence interval may be used to estimate the value of
the population correlation coefficient, r.
2. Use a table showing confidence belts.
3. Table 10, Appendix B: confidence belts for 95%
confidence intervals.
4. Table 10 utilizes n, the sample size.
Example: A random sample of 25 ordered pairs of data have a
calculated value of r = 0.45. Find a 95% confidence interval
for r, the population linear correlation coefficient.
Solution:
1. Population Parameter of Concern:
The linear correlation coefficient for the population, r.
2. The Confidence Interval Criteria:
a. Assumptions: The ordered pairs form a random sample,
and for each x, the y-values have a mounded
distribution.
b. Test statistic: The calculated value of r.
c. Confidence level: 1  a = 0.95
3. Sample Evidence:
n = 25 and r = 0.45
4. The Confidence Interval:
The confidence interval is read from Table 10, Appendix B.
Find r = 0.45 at the bottom of Table 10.
Visualize a vertical line through that point.
Find the two points where the belts marked for the correct
sample size cross the vertical line.
Draw a horizontal line through each point to the vertical
scale on the left and read the confidence interval.
The values are 0.68 and 0.12
5. The Results:
0.68 to 0.12 is the 95% confidence interval for r.
Hypothesis-Testing Procedure:
1. Null hypothesis: the two variables are linearly unrelated,
r = 0.
2. Alternative hypothesis: one- or two-tailed, usually r  0
3. Test statistic: calculated value of r.
4. Probability bounds or critical values for r: Table 11,
Appendix B.
5. Number of degrees of freedom for the r-statistic: n  2.
Example: In a study of 32 randomly selected ordered pairs,
r = 0.421. Is there any evidence to suggest the linear
correlation coefficient is different from 0 at the 0.05 level of
significance?
Solution:
1. The Set-up:
a. Population parameter of concern: The linear correlation
coefficient for the population, r.
b. The null and the alternative hypothesis:
H0: r = 0
Ha: r  0
2. The Hypothesis Test Criteria:
a. Assumptions: The ordered pairs form a random sample
and we will assume that the y-values at each x have a
mounded distribution.
b. Test statistic:
r* (calculated value of r) with df = 32  2 = 30
c. Level of significance: a = 0.05
3. The Sample Evidence:
n = 32 and r = r* = 0.421
4. The Probability Distribution (Classical Approach):
a. Critical Value: The critical value is found at the
intersection of the df = 30 row and the two-tailed 0.05
column of Table 11: 0.349
b. r* is in the critical region.
4. The Probability Distribution (p-Value Approach):
a. The p-value: Use Table 11: 0.01 < P < 0.02
b. The p-value is smaller than the level of significance, a.
5. The Results:
a. Decision: Reject H0.
b. Conclusion: At the 0.05 level of significance, there is
evidence to suggest x and y are correlated.
13.3: Linear Regression Analysis
• Line of best fit results from an analysis of
two (or more) related variables.
• Try to predict the value of the dependent,
or output, variable.
• The variable we control is the independent,
or input, variable.
Method of Least Squares:
The line of best fit: yˆ  b0  b1 x
The slope: b1 
SS( xy )
SS( x)
The y-intercept: b0 
1
 y  b1  x 
n
Note:
1. A scatter plot may suggest curvilinear regression.
2. If two or more input variables are used: multiple
regression.
The Linear Model: yˆ  b 0  b1 x  e
This equation represents the linear relationship between the
two variables in a population.
b0: The y-intercept, estimated by b0.
b1: The slope, estimated by b1.
e: Experimental error, estimated by e  y  yˆ
The random variable e is called the residual.
e is the difference between the observed value of y and
the predicted value of y at a given x.
The sum of the residuals is exactly zero.
Mean value of experimental error is zero: me = 0
2

Variance of experimental error: e
Estimating the Variance of the Experimental Error:
Assumption: The distribution of y’s is approximately normal
and the variances of the distributions of y at all values of x are
the same (The standard deviation of the distribution of y
about yˆ is the same for all values of x.).
2
(
x

x
)

2
Consider the sample variance: s 
n 1
1. The variance of y involves an additional complication:
there is a different mean for y at each value of x.
2. Each “mean” is actually the predicted value, yˆ
3. Variance of the error e estimated by:
2
(
y

y
)
ˆ
se2  
n2
Degrees of freedom: n  2
2
Rewriting se
2
(
y

y
)
ˆ
se2  
n2
2
(
y

b

b
x
)

0
1

n2
2
y
 b0  y   b1  xy 


n2
SSE

n2


SSE = sum of squares for error
Example: A recent study was conducted to determine the
relation between advertising expenditures and sales of
statistics texts (for the first year in print). The data is given
below (in thousands). Find the line of best fit and the
variance of y about the line of best fit.
Adv. Costs (x ) Sales (y ) Adv. Costs (x ) Sales (y )
40
289
60
470
55
423
52
408
35
250
39
320
50
400
47
415
43
335
38
389
Solution:
2
2


x
(
459
)
  21677 
SS( x)   x 2 
 608.9
n
10
x y
(459)(3699)

SS( xy )   xy 
 174163 
 4378.9
n
b1 
10
SS( xy ) 4378.9

 7.1915
SS( x)
608.9
y  b1   x  3699  (7.1915)(459)

b0 

 39.8105
n
10
The equation for the line of best fit: yˆ  39.81  7.19 x
The variance of y about the regression line:
2


y
 b0  y   b1  xy 

2
s 
e
n2
(1410485)  (39.81)(3699)  (7.1915)(174163)

8
10734.5955

 1341.8244
8
Note: Extra decimal places are often needed for this type of
calculation.
Scatter plot, regression line, and random errors as line
segments:
5
0
0
4
7
5
4
5
0
4
2
5
4
0
0
Sales
3
7
5
3
5
0
3
2
5
3
0
0
2
7
5
2
5
0
2
2
5
3
03
54
04
55
05
56
06
5
A
d
v
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i
s
i
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C
o
s
t
s
Minitab Output:
Regression Analysis
The regression equation is
C2 = 39.8 + 7.19 C1
Predictor
Constant
C1
Coef
39.81
7.191
StDev
69.11
1.484
S = 36.63
R-Sq = 74.6%
T
0.58
4.84
P
0.580
0.001
R-Sq(adj) = 71.4%
Analysis of Variance
Source
Regression
Residual Error
Total
DF
1
8
9
SS
31491
10734
42225
MS
31491
1342
F
23.47
P
0.001
13.4: Inferences Concerning the
Slope of the Regression
Line
• Null hypothesis: the equation of the line of
best fit is of no value in predicting y given x
(b1 = 0).
• Use a t test.
Sampling Distribution of the Slope b1:
Assume: Random samples of size n are repeatedly taken from
a bivariate population.
1. b1 has a sampling distribution that is approximately
normal.
2. The mean of b1 is b1.
3. The variance of b1 is
 e2
2
 b1 
2
(
x

x
)

provided there is no lack of fit.
Estimator for  b21
2
s
e
sb21 

2
 ( x  x)
se2
2


x

2
x


se2

SS( x)
n
The standard error of regression (slope) is  b1
and is estimated by sb1
Example (continued): For the advertising costs and sales data:
2
s
1341.8244
sb21  e 
 2.2037
SS( x)
608.9
Assumptions for inferences about the slope parameter b1:
The set of (x, y) ordered pairs forms a random sample and the
y-values at each x have a normal distribution. Since the
population standard deviation is unknown and replaced with
the sample standard deviation, the t-distribution will be used
with n  2 degrees of freedom.
Confidence Interval Procedure:
The 1  a confidence interval for b1 is given by
b1  t (n  2,a / 2)  sb1
Example: Find the 95% confidence interval for the population
slope b1 for the advertising costs and sales example.
Solution:
1. Population parameter of Interest:
The slope, b1, for the line of best fit for the population.
2. The Confidence Interval Criteria:
a. Assumptions: The ordered pairs form a random sample
and we will assume the y-values (sales) at each x
(advertising costs) have a mounded distribution.
b. Test statistic: t with df = 10  2 = 8
c. Confidence level: 1  a = 0.95
3. Sample Evidence:
2
n

10
,
b

7
.
1915
,
s
Sample information:
1
b1  2.2037
4. The Confidence Interval:
a. Confidence coefficients:
t(df, a/2) = t(8, 0.025) = 2.31
b. Interval:
b1  t (n  2,a / 2)  sb1  7.1915  (2.31)  2.2037
 7.1915  1.4845
 (5.707, 8.676)
5. The Results:
The slope of the line of best fit of the population from
which the sample was drawn is between 5.707 and 8.676
with 95% confidence.
Hypothesis-Testing Procedure:
1. Null hypothesis is always H0: b1 = 0
2. Use the Students t distribution with df = n  2.
3. The test statistic:
b1  b1
t* 
sb1
Example: In the previous example, is the slope for the line of
best fit significant enough to show that advertising cost is
useful in predicting the first year sales? Use a = 0.05
Solution:
1. The Set-up:
a. Population parameter of concern: The parameter of
concern is b1, the slope of the line of best fit for the
population.
b. The null and alternative hypothesis:
H0: b1 = 0 (x is of no use in predicting y)
Ha: b1 > 0 (we expect sales to increase as costs increase)
2. The Hypothesis Test Criteria:
a. Assumptions: The ordered pairs form a random sample
and we will assume the y-values (sales) at each x
(advertising costs) have a mounded distribution.
b. Test statistic: t* with df = n  2 = 8
c. Level of significance: a = 0.05
3. The Sample Evidence:
2
a. Sample information: n  10, b1  7.1915, sb1  2.2037
b. Calculate the value of the test statistic:
b1  b1 7.1915  0.0
t* 

 4.8444
sb1
2.2037
4. The Probability Distribution (Classical Approach):
a. Critical value: t(8, 0.05) = 1.86
b. t* is in the critical region.
4. The Probability Distribution (p-Value Approach):
a. The p-value: P = P(t* > 4.8444, with df = 8) < 0.001
b. The p-value is smaller than the level of significance, a.
5. The Results:
a. Decision: Reject H0.
b. Conclusion: At the 0.05 level of significance, there is
evidence to suggest the slope of the line of best fit is
greater than zero. The evidence indicates there is a
linear relationship and that advertising cost (x) is useful
in predicting the first year sales (y).
13.5: Confidence Interval
Estimates for Regression
• Use the line of best fit to make predictions.
• Predict the population y-value at a given x.
• Predict the individual y-value selected at
random that will occur at a given value of x.
• The best point estimate, or prediction, for
both is yˆ
Notation:
1. Mean of the population y-values at a given value of x: m y|x0
2. The individual y-value selected at random for a given
value of x: y x0
Background:
1. Recall: the development of confidence intervals for the
population mean m when the variance was known and
when the variance was estimated.
2. The confidence interval for m y|x0 and the prediction interval
for y x0 are constructed in a similar fashion.
3. yˆ replaces x as the point estimate.
4. The sampling distribution of yˆ is normal.
5. The standard deviation in both cases is computed by
multiplying the square root of the variance of the error by
an appropriate correction factor.
6. The line of best fit passes through the centroid: ( x, y )
Consider a confidence interval for the slope b1.
If we draw lines with slopes equal to the extremes of that
confidence interval through the centroid, the value for y
fluctuates considerably for different values of x (See the
Figure on the next slide.).
It is reasonable to expect a wider confidence interval as we
consider values of x further from x .
We need a correction factor to adjust for the distance
between x0 and x .
This factor must also adjust for the variation of the yˆ
values about y.
Lines Representing the Confidence Interval for Slope:
5
0
0
Slope is 8.676
4
7
5
4
5
0
4
2
5
4
0
0
Sales
3
7
5
3
5
0
Slope is 5.707
( x, y )
3
2
5
3
0
0
2
7
5
2
5
0
2
2
5
3
03
54
04
55
05
56
06
5
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d
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C
o
s
t
s
Confidence interval for the mean value of y at a given value
of x, m y|x0
standard error of yˆ
1 ( x0  x) 2
yˆ  t (n  2,a / 2)  se 

n  ( x  x) 2
1 ( x0  x) 2
 yˆ  t (n  2,a / 2)  se 

n
SS( x)
Note:
1. The numerator of the second term under the radical sign is
the square of the distance of x0 from x
2. The denominator is closely related to the variance of x and
has a standardizing effect on this term.
Example: It is believed that the amount of nitrogen fertilizer
used per acre has a direct effect on the amount of wheat
produced. The data below shows the amount of nitrogen
fertilizer used per test plot and the amount of wheat harvested
per test plot.
a. Find the line of best fit.
b. Construct a 95% confidence interval for the mean amount
of wheat harvested for 45 pounds of fertilizer.
Pounds of
Fertilizer (x )
30
36
41
49
53
55
60
65
100 Pounds
of Wheat (y )
14
9
18
16
23
17
28
33
Pounds of
Fertilizer (x )
74
76
81
88
93
94
101
109
100 Pounds
of Wheat (y )
20
24
29
35
34
39
28
33
Solution:
Using Minitab, the line of best fit: yˆ  4.42  0.298 x
Confidence Interval:
1. Population Parameter of Interest:
The mean amount of wheat produced for 45 pounds of
fertilizer, m y| x 45
2. The Confidence Interval Criteria:
a. Assumptions: The ordered pairs form a random sample
and the y-values at each x have a mounded distribution.
b. Test statistic: t with df = 16  2 = 14
c. Confidence level: 1  a = 0.95
3. Sample Information:
se2  25.97
y x 45 :
se  25.97  5.096
yˆ  4.42  0.298(45)  17.83
4. The Confidence Interval:
1 ( x0  x) 2
yˆ  t (n  2, a / 2)  se 

n
SS( x)
1 (45  69.06) 2
 17.83  (2.14)(5.096)

16
8746.94
 17.83  (2.14)(5.096) 0.0625  0.0662
 17.83  (2.14)(5.096)(0.3587)
 17.83  3.91
13.92 to 21.74, 95% confidence interval for m y| x 45
Confidence interval: green vertical line.
Confidence interval belt: upper and lower boundaries of all
95% confidence intervals.
4
5
4
0
Line of best fit
3
5
Upper boundary
3
0m
for y|x0
Wheat
2
5
2
0
1
5
1
0
Lower boundary for m y|x0
5
2
0
3
0
4
0
5
0
6
0
7
0
8
0
9
0
1
0
0
1
1
0
1
2
0
F
e
r
t
i
l
i
z
e
r
Prediction interval of the value of a single randomly selected y:
1 ( x0  x) 2
yˆ  t (n  2, a / 2)  se  1  
n
SS( x)
Example: Find the 95% prediction interval for the amount of
wheat harvested for 45 pounds of fertilizer.
Solution:
1. Population Parameter of Interest:
yx=45, the amount of wheat harvested for 45 pounds of
fertilizer
2. The Confidence Interval Criteria:
a. Assumptions: The ordered pairs form a random sample
and the y-values at each x have a mounded distribution.
b. Test statistic: t with df = 16  2 = 14
c. Confidence level: 1  a = 0.95
3. Sample Information:
se2  25.97
y x 45 :
se  25.97  5.096
yˆ  4.42  0.298(45)  17.83
4. The Confidence Interval:
1 ( x0  x) 2
yˆ  t (n  2, a / 2)  se  1  
n
SS( x)
1 (45  69.06) 2
 17.83  (2.14)(5.096) 1  
16
8746.94
 17.83  (2.14)(5.096) 1  0.0625  0.0662
 17.83  (2.14)(5.096) 1.1287
 17.83  (2.14)(5.096)(1.0624)
 17.83  11.5859
6.24 to 29.41, 95% prediction interval for y x 45
Prediction belts for y x0
4
5
Line of best fit
4
0
3
5
Upper boundary on
individual y-values
3
0
Wheat
2
5
2
0
1
5
Lower boundary for 95% prediction
interval on individual y-values at any x
1
0
5
2
0
3
0
4
0
5
0
6
0
7
0
8
0
9
0
1
0
0
1
1
0
1
2
0
x0 = 45
F
e
r
t
i
l
i
z
e
r
Precautions:
1. The regression equation is meaningful only in the domain
of the x variable studied. Estimation outside this domain is
risky; it assumes the relationship between x and y is the
same outside the domain of the sample data.
2. The results of one sample should not be used to make
inferences about a population other than the one from
which the sample was drawn.
3. Correlation (or association) does not imply causation. A
significant regression does not imply x causes y to change.
Most common problem: missing, or third, variable effect.
13.6: Understanding the
Relationship Between
Correlation and Regression
• We have considered correlation and
regression analysis.
• When do we use these techniques?
• Is there any duplication of work?
Remarks:
1. The primary use of the linear correlation coefficient is in
answering the question “Are these two variables related?”
2. The linear correlation coefficient may be used to indicate
the usefulness of x as a predictor of y (if the linear model is
appropriate).
The test concerning the slope of the regression line
(H0: b1 = 0) tests the same basic concept.
3. Lack-of-fit test: Is the linear model appropriate?
Consider the scatter diagram.
Conclusions:
1. Linear correlation and regression measure different
characteristics. It is possible to have a strong linear
correlation and have the wrong model.
2. Regression analysis should be used to answer questions
about the relationship between to variables.
a. What is the relationship?
b. How are the two variables related?