Confidence interval:
Confidence interval can be used for both estimation and hypothesis testing. The two
1001   % confidence intervals for  0 and  1 are
b0  t n2,1 / 2 s.e.(b0 )  [b0  t n2,1 / 2 s.e.(b0 ), b0  t n2,1 / 2 s.e.(b0 )]
and ,
b1  t n2,1 / 2 s.e.(b1 )  [b1  t n2,1 / 2 s.e.(b1 ), b1  t n2,1 / 2 s.e.(b1 )] ,
where
P(| Tn2 | t n2,1 / 2 )   .
To test H 0 :  0  c and H 0 : 1  c , we only need to examine if c falls into the
above two confidence intervals.
The interrelation between testing hypothesis and confidence interval can be explained
in the following simple case. Let X 1 , X 2 ,, X n ~ N (u,1) . Suppose we want to test
H 0 : u  c , then we reject H 0 as
random variable and
n ( X  c)  z / 2 , where Z is the standard normal
P( Z  z / 2 ) 

2
. Therefore, according the above rule,
we do not reject H 0 as
n ( X  c )  z / 2 
n ( c  X )  z / 2  X 
z / 2
n
c X 
That is we do not reject H 0 as c falls into the interval [ X 
matter of fact, [ X 
[X 
z / 2
n
,X 
z / 2
n
z / 2
n
,X 
z / 2
n
z / 2
n
z / 2
,X 
n
z / 2
n
.
] . As a
] is a 100(1   )% confidence interval for c.
]
Example
Yi   0  1 X i   i , i  1,2,,20. Use F-test, T-test, and confidence interval to test
H 0 : 1  0 given level of significance equal to 0.05 and
20
X
i 1
i
20
20
20
20
i 1
i 1
i 1
i 1
 1330,  Yi  1862.8,  X i2  90662,  Yi 2  173554.26,  X i Yi  124206.9.
1
[solution]
1.
(a) F-test:
20
20
i 1
i 1
20
 X i  Yi
i 1
20
S XY   X i Yi 
 124206.9 
1330 * 1862.8
 330.7.
20
2
 20

 Xi 
20
1330 2
S XX   X i2   i 1   90662 
 2217.
20
20
i 1
20
20
20
2
S
330.7
 b1  XY 
 0.149,  Yˆi  Y  b12 S XX  49.22,  (Yi  Y ) 2   Yi 2 
S XX
2217
i 1
i 1
i 1


2
 20 
  Yi 
2
20
 i 1   173554.26  (1862.8)  53.068  (Y  Yˆ ) 2  53.068  49.22  3.848.

i
i
20
20
i 1
Then, we have the following ANOVA table
Source
SS( b1 | b0 )
residual
Total(corrected)
As H 0 is true, F 
df
SS
MS
1
18
19
49.22
3.848
53.068
49.22
0.214
49.22
 230  f1,18,0.05  4.41 , where P( F1,18  f1,18, 0.05 )  0.05 .
0.214
 reject H 0
(b) T-test:
S 1XX/ 2 b1
t 

s
2217 * 0.149
0.214
 15.16  t18,0.975  2.10  reject H 0 .
2. 95% confidence interval:
The 95% confidence interval for  1 is
b1  t18, 0.975
s
S
1/ 2
XX
 0.149  (2.101 *
0.214
2217
)  0.149  0.021  0.128,0.179 .
Since 0 does not fall into the above interval, we reject H 0 .
2
Example:
Suppose the model is

yi   0  1 xi   i , i  1,  ,10,  i ~ N 0,  2
,
and
10
x
i 1
10
10
i 1
i 1
 320,  yi  462,  xi2  11550,
i
10
y
i 1
2
i
10
 21710,  xi yi  15390
i 1
(a) Find the least square estimate and the fitted regression equation
(b) Provide an ANOVA table and use F statistic to test H 0 : 1  0 at   0.05 .
(c) Find the 95% confidence interval for  1 .and use the confidence interval to test
H 0 : 1  0 .
(d) Find the ANOVA table for the hypothesis H 0 :  0  1  0 and use F statistic to
test H 0 :  0  1  0 at   0.01.
[solution:]
(a)
2
 320 
  X  10  X  11550  10  
  1310
10


i 1
10
 320   462 
  X iYi  10  XY  15390  10  

  606
10
10




i 1
10
S XX
S XY
2
i
2
Then, the least square estimate is
b1 
S XY 606
 462 
 320 

 0.4626, b0  Y  b1 X  
  0.4626  
  31.3968
S XX 1310
 10 
 10 
The fitted regression equation is
Yˆ  31.3968  0.4626 X .
(b)
Since
3
2
 462 
SYY   Yi  10  Y  21710  10  
  365.6
 10 
i 1
SS (b1 | b0 )  b1 S XY  0.4626  606  280.3356
10
2
2
SS ( residual )  SYY  SS b1 | b0   365.6  280.3356  85.2644
The ANOVA table is
Source
df
SS
MS
F
Regression
1
280.3356
SS (b1 | b0 )
1
 280.3356
Residual
(Error)
n-2=8
85.26444
SS (residual )
 10.6581
8
Total
(corrected)
9
365.6
F
280.3356
10.6581
 26.3026
Since F  26.3026  5.3177  f1,8, 0.05 , we reject H 0 : 1  0 .
(c)
The 95% confidence interval for  1 is
 s2
b1  t n2, 
2 S
 XX



1
 10.6581 
 0.4626  t8,0.025  

 1310 
2
1
2
 0.2546,0.6706 .
Since 0  0.2546,0.6706, we reject H 0 : 1  0 .
(d)
Since
10
10
10
i 1
i 1
i 1

 Yˆi 2   Yi 2   Yi  Yˆi

2
 21710  85.2644  21624.74 , the ANOVA table
for H 0 :  0  1  0 is
Source
df
SS
MS
Regression
2
21624.74
21624.74
 10812.37
2
Residual
(Error)
n-2=8
85.26444
SS (residual )
 10.6581
8
Total
10
21710
F
F
Since F  1014.474  8.65  f 2,8, 0.01 , we reject H 0 :  0  1  0 .
4
10812.37
10.6581
 1014.474