Stat 301 – Lecture 9
Regression model
Y   y| x  
•Y
represents a value of the response
variable.
• y| x represents the population mean
response for a given value of the
explanatory variable, x.
• represents the random error
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Linear Regression Model
Y   y| x     0  1 x  
0
The Y-intercept parameter.
1
The slope parameter.
2
 y | x   0  1 x
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Stat 301 – Lecture 9
Estimated Mean
Estimated Mean Response
ˆ y| x  ˆ0  ˆ1 x
2007: CO2=382.43 ppmv
ˆ y| x  9.8815  0.012584382.43
ˆ y| x  14.694 o C
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Confidence Interval
We have estimated a
population mean response
for a given value of the
explanatory variable. We
can expand this into a
confidence interval.
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Confidence Interval
ˆ y| x  t * seˆ y| x 
 1  x0  x 2 

seˆ y| x   MS Error  
 n   x  x 2 


t * from t - table with df  n  2
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Stat 301 – Lecture 9
Standard Error Calculation
 1  x0  x 2 

seˆ y| x   MS Error  
 n   x  x 2 


MS Error  0.010725,   x  x   5061.2
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n  20, x0  382.43, x  341.23
seˆ y| x   0.0643
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Confidence Interval
ˆ y| x  t * seˆ y| x 
seˆ y| x   0.0643
t *  2.101
14.694  2.1010.0643
14.559 o C to 14.829 o C
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Interpretation – Part 1
The population mean
temperature when the
CO2=382.43 ppmv can be
any value between 14.56 oC
and 14.83 oC
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Stat 301 – Lecture 9
Interpretation – Part 2
We are 95% confident that
intervals based on random
samples from the population
with capture the actual
population mean value.
This is confidence in the
process.
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14.829
14.694
14.559
382.43
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 y | x   0  1 x
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Stat 301 – Lecture 9
Predicted Individual
Predicted Individual Response
Yˆ  ˆ0  ˆ1 x
2007: CO2=382.43 ppmv
yˆ  9.8815  0.012584382.43
yˆ  14.694 o C
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Prediction Interval
We have predicted an
individual response for a
given value of the
explanatory variable. We
can expand this into a
prediction interval.
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Prediction Interval
yˆ  t * se yˆ 
 1  x0  x 2 

se yˆ   MS Error 1  
 n  x  x 2 


t * from t - table with df  n  2
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Stat 301 – Lecture 9
Standard Error Calculation
 1  x  x 2 

se yˆ   MS Error 1   0
 n   x  x 2 


MS Error  0.010725,   x  x   5061.2
2
n  20, x0  382.43, x  341.23
se yˆ   0.1219
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Standard Error Calculation
se yˆ   MS Error  seˆ y| x 
2
MS Error  0.010725, seˆ y| x   0.0643
se yˆ   0.010725  0.0643
2
se yˆ   0.1219
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Prediction Interval
yˆ  t * se yˆ 
se yˆ   0.1219
t *  2.101
14.694  2.1010.1219 
14.438 o C to 14.950 o C
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Stat 301 – Lecture 9
Interpretation
We are 95% confident that
the annual global
temperature when the
CO2=382.43 ppmv can be
any value between 14.44 oC
and 14.95 oC
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14.950
14.694
Temp
14.438
382.43
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 y | x   0  1 x
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