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Regression Interval Estimates
© 2010 Pearson Prentice Hall. All rights reserved
Confidence intervals for a mean response are
intervals constructed about the predicted value of y, at a
given level of x, that are used to measure the accuracy
of the mean response of all the individuals in the
population.
Prediction intervals for an individual response are
intervals constructed about the predicted value of y that
are used to measure the accuracy of a single
individual’s predicted value.
14-2
Confidence Interval for the Mean Response of y
A (1-)100% confidence interval for yˆ , the mean response
of y for a specified value of x, is given by
yˆ  t 2  se
x  x

1

n  x i  x 2
yˆ  t 2  se
x  x

1

2
n  x i  x 
*
Lower bound:

*
Upper bound:

2
2
where x* is the given value of the explanatory variable, n is
the number of observations, and t/2 is the critical value
with n-2 degrees of freedom.
14-3

Parallel Example 1: Constructing a Confidence Interval for a
Mean Response
Construct a 95% confidence interval about the predicted
mean time to drill 5 feet for all drillings started at a
depth of 110 feet.
14-4

Solution
The least squares regression line is yˆ  0.0116x  5.5273
.
To find the predicted mean time to drill 5 feet for all
drillings started at 110 feet, let x*=110 in the regression
equation and obtain yˆ  6.8033
.

Recall:
•
se=0.5197

•
x  126.25
•
xi  x   30006.25
•
t0.025=2.228 for 10 degrees of freedom
14-5
2
Solution
Therefore,
Lower bound:
1 110 126.25
6.8033  2.228  0.5197

 6.45
12
30006.25
2
Upper bound:


1 110 126.25
6.8033  2.228  0.5197

 7.15
12
30006.25
2
14-6
Solution
We are 95% confident that the mean time to drill 5
feet for all drillings started at a depth of 110 feet is
between 6.45 and 7.15 minutes.
14-7
Prediction Interval for an Individual
Response about y
ˆ.
A (1-)100% prediction interval for the individual response of y, is
given by
Lower bound:
Upper bound:

yˆ 
t 2  se
yˆ  t 2  se
1
1 
n
x  x 
x  x 
1
1 
n
x  x 
x  x 
2
*
2
i
2
*
2
i
where x* is the given value of the explanatory variable, n is
the number of observations, and t/2 is the critical value
with n-2
degrees of freedom.

14-8
Parallel Example 2: Constructing a Prediction Interval for an
Individual Response
Construct a 95% prediction interval about the predicted
time to drill 5 feet for a single drilling started at a depth
of 110 feet.
14-9
Solution
The least squares regression line is
yˆ  0.0116x  5.5273 . To find the predicted
mean time to drill 5 feet for all drillings started at
110 feet, let x*=110 in the regression equation and
obtain yˆ  6.8033.

Recall:
• se=0.5197

• x  126.25
• x i  x 2  30006.25
• t0.025=2.228 for 10 degrees of freedom
14-10
Solution
Therefore,
Lower bound:
1 110 126.25
6.8033  2.228  0.5197 1 
 5.59
12
30006.25
2
Upper bound:


1 110 126.25
6.8033  2.228  0.5197 1 
 8.01
12
30006.25
2
14-11
Solution
We are 95% confident that the time to drill 5
feet for a random drilling started at a depth of 110
feet is between 5.59 and 8.01 minutes.
14-12
Model effectiveness
Our regression model can be considered effective
for producing confidence interval estimates if:
6 * se  Range ( y)
Our regression model can be considered effective
for producing prediction interval estimates if:
6 * se  IQR ( y)
This will ensure that the interval estimates do not
become too wide to be meaningful to us.
14-13