Chapter 13
Regression, Inference, and Model Building
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Building a Simple Linear Regression
Model
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Simple Linear Regression Model
Definition
The simple linear regression model is given by the
linear equation
yi  b0  b1 xi  ei ,
where
b0 is the y-intercept for the population data,
b1 is the slope coefficient for the population data,
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Simple Linear Regression Model
Definition (cont.)
x i is the value of the independent (or predictor)
variable for observation i,
ei is the random error in y for observation i, and
y i is the value of the dependent (or response) variable
for observation i.
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Estimated Simple Linear Regression Equation
Definition
The estimated simple linear regression equation is
yˆi  b0  b1 xi ,
where b0 and b1 are estimates of their population
counterparts. Specifically,
b0 is an estimate of b0 , and b1 is an estimate of b1 .
yˆi is the predicted value of y for a given value of xi , and
is pronounced y-hat.
The symbol yi is reserved for the observed value of y.
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Defining a Linear Relationship
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How Do We Measure How Close a Line Is to the
Data?
Definition
The difference between the observed value of y and
the predicted value of y is called the error, estimated
error, or residual (ei). The error for each observation is
given by
Error  ei  Observed y  Predicted y  yi  yˆi .
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Sum of Squared Errors
Formula
Sum of Squared Errors (SSE)
The sum of squared errors (SSE) is given by


SSE    yi  yˆi    yi   b0  b1 xi  .
2
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Least Squares Line
Definition
The least squares line is the line that has the smallest
sum of squared errors. This is the line of best fit.
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Finding the Least Squares Line
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Slope and y-Intercept of the Least Squares Line
Formula
Slope and y-Intercept of the Least Squares Line
The equation for finding the slope is given by
b1 
SS xy
SS xx
where
SS xy    xi  x  yi  y   
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x   y 


xy 
i
i i
i
n
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Slope and y-Intercept of the Least Squares Line
Formula (cont.)
and
SS xx    xi  x    xi
2
2
x



i
n
2
.
The slope can also be calculated using
b1 
n x i y i   x i  y i
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n x i 
2
 x 
2
.
i
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Slope and y-Intercept of the Least Squares Line
Formula (cont.)
The estimate of the intercept is given by


1
b0  y  b1 x   yi  b1  xi .
n
The x i and y i referred to in the expressions are the
observed data values of x and y, respectively.
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Intercept and Slope Coefficients
Definition
The intercept coefficient, b0, is the average value of the
dependent variable, y, when the independent variable,
x, is equal to zero.
The slope coefficient, b1, is the average change in the
dependent variable, y, for a one unit change in the
independent variable, x.
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Mean Square Error
Formula
Mean Square Error
The variance of the error terms is also known as the
mean square error and is given by:
2
e
s
y



i
 yi
n2

2
SSE

.
n2
The square root of the mean square is the standard
error, or the standard deviation of the error terms.
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Evaluating the Fit of a Model
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Total Sum of Squares (TSS)
Formula
Total Sum of Squares (TSS)
The total variation in y is given by the total sum of
squares (TSS).
TSS    yi  y 
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Sum of Squares of Regression
Definition
The sum of squares of regression (SSR) denotes the
explained variation in the model.
SSR = TSS – SSE
(the explained variation, SSR, is equal to the total
variation minus the unexplained variation)
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Coefficient of Determination
Formula
Coefficient of Determination
The coefficient of determination, R2, is given by
SSR
SSE
R 
1
.
TSS
TSS
2
The coefficient of determination is a value between 0
and 1, inclusive. That is, 0  R2  1.
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Example 13.1
The SAT Reasoning Test has been used for years as a
predictor of academic success. If SAT scores are
predictors of academic success, they should be
positively related to the grade point average upon
graduation. 27 graduates of a state college were
sampled and their grade point averages (GPA) upon
graduation and SAT scores reported upon admission
are recorded. The data are given in Table 13.5.
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Example 13.1 (cont.)
Table 13.5 – SAT Scores and Graduating GPA
Student
SAT Critical
Reading
SAT
Math
SAT
Writing
SAT
Total
Graduating
GPA
1
2
3
4
5
6
7
8
9
10
11
12
440
390
410
390
490
400
450
420
370
460
370
410
550
480
360
350
590
550
430
350
390
600
400
530
495
435
385
370
540
475
440
385
380
530
385
470
1485
1305
1155
1110
1620
1425
1320
1155
1140
1590
1155
1410
2.105
2.484
2.537
2.969
3.619
2.303
2.602
2.195
2.112
3.482
2.367
2.082
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Example 13.1 (cont.)
Table 13.5 – SAT Scores and Graduating GPA
Student
SAT Critical
Reading
SAT
Math
SAT
Writing
SAT Total
Graduating
GPA
13
14
15
16
17
18
19
20
21
22
23
24
470
490
540
560
440
440
580
360
440
290
440
510
570
610
630
620
470
530
670
420
460
410
500
570
520
550
585
590
455
485
625
390
450
350
470
540
1560
1650
1755
1770
1365
1455
1875
1170
1350
1050
1410
1620
2.346
3.484
2.446
2.820
2.556
3.357
3.269
2.964
2.642
2.297
2.388
2.850
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Example 13.1 (cont.)
Table 13.5 – SAT Scores and Graduating GPA
Student
SAT Critical
Reading
SAT
Math
SAT
Writing
SAT Total
Graduating
GPA
25
26
27
320
470
550
540
580
550
430
525
550
1290
1575
1650
2.742
2.347
3.025
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Example 13.1 (cont.)
Figure 13.14
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Example 13.1 (cont.)
The scatterplot in Figure 13.14 suggests that as SAT
scores increase the GPA tends to increase, although
there is a substantial amount of variability in the
relationship. The upward sloping pattern of the data
suggests a linear model could be constructed. However,
a great deal of variation in the model’s errors should be
expected. What percent of the variation in final grade
point average can be explained by the model relating
total SAT score to graduating GPA?
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Example 13.1 (cont.)
Solution
Using the least squares method, the estimated model is
given by
Estimated Graduating GPA
 1.355  0.00094  Total SAT Score .
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Example 13.1 (cont.)
Figure 13.15
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Example 13.1 (cont.)
Figure 13.16
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Example 13.1 (cont.)
One of the differences in the production model and the
SAT/GPA model is the manner in which the data seem
to fit the model. In the production model, the data
seemed to fit closely around the line, while in the
SAT/GPA model the data are loosely clustered about
the line. While tight and loose are interesting portrayals
of the relative fit of the data, it would be desirable to
have a numerical measure to describe fit. R2 is such a
measure.
SSR 1.1862
2
R 

 0.1944
TSS 6.1013
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Example 13.1 (cont.)
Thus, approximately 19% of the variation in graduating
GPA is explained by this model.
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What is a goo R2?
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Fitting a Linear Time Trend
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Example 13.2
Many analysts believe that college tuition prices may
soon be in the same situation as housing prices were
when the housing bubble burst (causing home prices to
drop significantly). Table 13.6 contains data for the
Tuition Consumer Price Index (TCPI) from 1978 to 2009.
Use a linear time trend to model the data.
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Example 13.2 (cont.)
Table 13.6 – Tuition Consumer Price Index
Year
TCPI
Year
TCPI
1978
1979
1980
1981
1982
1983
1984
1985
1986
1987
1988
1989
60.89
65.66
71.80
80.58
91.33
100.73
110.94
112.61
130.63
140.41
125.98
137.86
1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
2000
2001
175.93
193.73
181.14
234.48
250.80
249.17
264.15
278.42
307.51
319.63
307.80
324.73
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Example 13.2 (cont.)
Table 13.6 – Tuition Consumer Price Index
Year
TCPI
Year
TCPI
2002
2003
2004
2005
348.54
371.42
343.05
476.08
2006
2007
2008
2009
507.90
456.30
531.55
607.60
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Example 13.2 (cont.)
Solution
Tuition Consumer Price Index
Figure 13.17
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Example 13.2 (cont.)
A graph of the data reveals an upward trend in the
tuition consumer price index. The data appear to be a
nonstationary time series with an upward trend. To
describe the data, we will model the trend by fitting a
line through the data with the notion of capturing how
fast (on average) the series is changing over time.
Estimating the slope of the line will provide the average
rate of change per year in the TCPI. The line is fitted
using least squares estimates in exactly the same way
as other regression models have been constructed.
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Example 13.2 (cont.)
The independent variable in a linear trend model is
always time. In this case, the dependent variable is
TCPI.
The estimated least squares equation is
Estimated TCPI  30609.4994  15.4794  Year  .
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Example 13.2 (cont.)
Tuition Consumer Price Index
Figure 13.18
The computer output for the problem is given in Figure
13.19.
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Example 13.2 (cont.)
Figure 13.19
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Example 13.2 (cont.)
The estimate of the slope, 15.4794, tells us that on
average the TCPI is increasing at a rate of 15.4794 per
year. Given how well the line fits the data (R2  0.9309),
the trend line is a good descriptor of the data.
The trend line can also be used for short-term
prediction. Suppose you wanted to estimate the TCPI
for 2010. If the data are not available, the trend model
can be used.
Estimated TCPI  30609.4994  15.4794  2010  504.0946
Prediction of the TCPI for 2010
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Confidence Interval
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The Confidence Interval for
Formula
100(1  a)% Confidence Interval for B1
The 100(1  a)% confidence interval for B1 is given by
b1  ta 2,df sb1 ,
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Example 13.3
Table 13.7 – Weekly Production
Week
Items Produced
Cost ($)
1
2
3
4
5
6
7
8
9
10
22
30
36
41
27
45
30
37
32
31
3500
3800
4500
4200
3700
4600
3600
4550
3990
3675
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Example 13.3 (cont.)
In Section 13.3, a model relating the number of items
produced to total cost was constructed.
Cost  b0  b1 Items Produced  ei
If the relationship is to be applicable for the entire
production process, then a substantial amount of data
will be required, more than we could hope to collect. If
the data given in Table 13.7 are considered a random
sample of weekly production, then a relationship can
be constructed from the sample data.
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Example 13.3 (cont.)
Specifically, the estimated least squares regression line
relating items produced to total cost is
Estimated Cost  $2227.96  $53.88 Items Produced ,
where
b0 = $2227.96 (the sample estimate of b0, the yintercept), and
b1 = $53.88 (the sample estimate of b1, the slope).
Note: Both estimates were determined using Microsoft
Excel and rounded to the nearest hundredth.
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Example 13.3 (cont.)
Figure 13.20
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Example 13.3 (cont.)
The manual calculation of sb1 is tedious. Virtually every
statistical analysis program that performs regression
analysis calculates sb1 . The summary output from
Microsoft Excel is given in Figure 13.20. Most software
packages will automatically include a confidence
interval for b1 or it will include the pieces required to
compute a confidence interval. Microsoft Excel
automatically displays the 95% confidence interval for
b1, and is capable of displaying an interval for any level
of confidence you choose.
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Example 13.3 (cont.)
95% Confidence Interval for
1
b1  ta 2,df sb1
53.88  2.306 10.9778 
53.88  25.3148
28.57 to 79.19
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Example 13.4 (cont.)
99% Confidence Interval for b1
b1  ta 2,df sb1
53.88  3.35510.9778 
53.88  36.8305
17.05 to 90.71
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Example 13.4 (cont.)
Note: The confidence interval in this example was
calculated using rounded values from the summary
output. Microsoft Excel calculates the confidence
interval using unrounded values as 17.05 to 90.72.
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Testing a Hypothesis Concerning b1
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Example 13.6
Using the data in Example 13.3, determine if there is
overwhelming evidence at the a = 0.05 level of a
relationship between the number of items produced
and the total production cost.
Solution
Step 1: State the hypotheses in plain English.
• Null Hypothesis: There is not a linear
relationship between the number of items
produced and total production cost.
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Example 13.6 (cont.)
• Alternative Hypothesis: There is a linear
relationship between the number of items
produced and total production cost.
Step 2: Select the appropriate statistical measure.
Since we are interested in determining if b1 = 0, the
sample estimate of b1, namely b1, will be used to
evaluate whether the hypothesis b1 = 0 is reasonable.
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Example 13.6 (cont.)
Step 3: Determine whether the hypothesis should be
one-sided or two-sided.
The alternative hypothesis should be two-sided since
we are interested in discovering any relationship
(positive or negative) between items produced and
total cost.
Step 4: Specify the hypotheses using the appropriate
statistical measure.
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Example 13.6 (cont.)
H0 : b1  0 (No linear relationship exists between items
produced and total cost.)
Ha : b1  0 (A linear relationship exists between items
produced and total cost.)
Step 5: Specify the level of the test.
The level of the test has been given in the problem
statement as the 0.05 level.
Step 6: Select the appropriate test statistic.
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Example 13.6 (cont.)
Formula
Test Statistic for Testing the Hypothesis b1 ≠ 0
The test statistic for testing the hypothesis b1 ≠ 0 is
given by
b1  0 b1
t
 .
sb1
sb1
The test statistic follows a t-distribution with n  2
degrees of freedom.
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Example 13.6 (cont.)
The test statistic is similar in nature to the other test
statistics developed in Chapter 10. It measures how far
b1 is from the hypothesized value of b1, which is 0. This
distance is measured in standard deviation units. If t is
close to 0, then b1 is close to 0 and H0: b1 = 0 is the
more reasonable conclusion. However, if t is far from
zero, then b1 is far from its hypothesized value and
Ha: b1 ≠ 0 would seem more reasonable. This criterion
is defined by the critical value of the test statistic.
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Example 13.6 (cont.)
Step 7: Determine the critical value.
The test is two-tailed and the level of the test is
specified to be 0.05, which implies
a  0.05 and a 2  0.05 2  0.025.
The test statistic has a t-distribution with
df  n  2  10  2  8.
The critical value corresponds to
t0.025,8  2.306.
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Example 13.6 (cont.)
t-Distribution, df = 8
Figure 13.22
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Example 13.6 (cont.)
Step 8: Compute the test statistic.
Table 13.9 – Regression Results
Predictor
Coefficient
Standard Deviation of Coefficient t-value
Intercept
2227.96
370.1488
6.019
Items Produced
53.88
10.9778
4.908
53.88
b1  0 b1
 4.908

t

sb1
sb1 10.9778
The estimated value of b1 is almost five standard
deviations above zero. This is very persuasive evidence
that b1 ≠ 0.
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Example 13.6 (cont.)
Step 9: Make the decision.
Since the value of the test statistic falls into the
rejection region, reject the null hypothesis in favor of
the alternative.
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Example 13.6 (cont.)
Step 10: State the conclusion in terms of the original
question.
There is overwhelming evidence at the 0.05 level that
b1 ≠ 0 so we reject the null hypothesis in favor of the
alternative. This implies that it is reasonable to believe
(at the 0.05 level) that there is a linear relationship
between the number of items produced and total cost.
In fact, there appears to be a positive linear
relationship between items produced and production
cost. However, our hypothesis test did not address the
issue of a positive relationship, so we cannot make this
conclusion.
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