pptx

```STA291
Statistical Methods
Lecture 29
Standard Errors for Mean Values
Confidence Interval for the Mean Response
Last time, we said we were modeling our line to infer
the “line of means”—the expected value of our
response variable for each given value of the
explanatory variable.
The confidence interval for the mean response, v, at a
value xv, is:
ˆyv  tn*2  SE ˆ v
 
where:
2
e
s
SEˆ v   SE b1  xv  x  
n
2
2
SE, and the confidence interval, becomes smaller
with increasing n.
SE, and the confidence interval, are larger for
samples with more spread around the line (when se
is larger).
Standard Errors for Mean Values
Confidence Interval for the Mean Response
The confidence interval for the mean response, v, at a
value xv, is: y
ˆ  t *  SE ˆ
v
 v
n 2
where: SEˆ  
v
2
e
s
SE b1  xv  x  
n
2
SE becomes larger the further
xν gets from x . That is, the
as you move away from x .
(See figure at right.)
2
Standard Errors for Predicted Values
Prediction Interval for an Individual Response
Now, we tackle the more difficult (as far as additional
variability) of predicting a single value at a value xv.
When conditions are met, that interval is:
where:
*
yˆv ± tn-2
´ SE ( yˆv )
2
s
2
2
e
ˆ
SE ( yv ) = SE ( b1 ) ´ ( xv - x ) + + se
n
2
2
s
Because of the extra term e, the
confidence interval for individual
values is broader that those for the
predicted mean value.
Difference Between Confidence and
Prediction Intervals
Confidence interval for a mean:
yˆ v  t
*
n2
2
e
s
SE b1  xv  x  
n
2
2
The result ˆ 10.1  4.55  0.15 at 95% means:
“We are 95% confident that the mean value
of y is between 4.40 and 4.70 when x =
10.1.”
Difference Between Confidence and
Prediction Intervals
Prediction interval for an individual value:
2
e
s
yˆ v  t
SE b1  xv  x    se2
n
The result yˆ 10.1  4.55  0.60 at 95% means:
*
n2
2
2
“We are 95% confident that a single
measurement of y will be between 3.95
and 5.15 when x = 10.1.”
Using Confidence and Prediction
Intervals
Example : External Hard Disks
A study of external disk drives reveals a linear relationship
between the Capacity (in GB) and the Price (in \$). Regression
resulted in the following:
Price = 18.64 + 0.104Capacity
se = 17.95, and SE(b1) = 0.0051
Find the predicted Price of a 1000 GB hard drive.
Find the 95% confidence interval for the mean Price of all 1000
GB hard drives.
Find the 95% prediction interval for the Price of one 1000 GB
hard drive.
Using Confidence and Prediction
Intervals
Example : External Hard Disks
A study of external disk drives reveals a linear
relationship between the Capacity (in GB) and the Price
(in \$). Regression resulted in the following:
Price = 18.64 + 0.104Capacity
se = 17.95, and SE(b1) = 0.0051
Find the predicted Price of a 1000 GB hard drive.
Price = 18.64 + 0.104(1000) = 122.64
Using Confidence and Prediction Intervals
Example : External Hard Disks
A study of external disk drives reveals a linear
relationship between the Capacity (in GB) and the Price
(in \$). Regression resulted in the following:
Price = 18.64 + 0.104Capacity
se = 17.95, and SE(b1) = 0.0051
Find the 95% confidence interval for the mean Price of
all 1000 GB hard drives.
yˆ v  t n* 2
2
s
SE 2 b1  xv  x   e
n
2
2
17
.
95
2
 \$122.64  2.571 0.00512  1000 1110 
7
 \$122.64  \$17.50  \$105.14,\$140.14
Using Confidence and Prediction Intervals
Example : External Hard Disks
A study of external disk drives reveals a linear
relationship between the Capacity (in GB) and the Price
(in \$). Regression resulted in the following:
Price = 18.64 + 0.104Capacity
se = 17.95, and SE(b1) = 0.0051
Find the 95% prediction interval for the price of one
1000 GB hard drive.
*
yˆv ± tn-2
2
s
SE 2 ( b1 ) ´ ( xv - x ) + e + se2
n
2
2
17.95
= \$122.64 ± 2.571 0.00512 ´ (1000 -1110 ) +
+17.952
7
= \$122.64 ± \$49.36 = [ 73.28,172.00 ]. CI: [ \$105.14, \$140.14]
2
Looking back
o Construct
and interpret a confidence
interval for the mean value
o Construct and interpret a prediction
interval for an individual value
```