Download 3.2 The C.I. and Hypothesis Testing

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3.2. The C.I. and Hypothesis Testing: F and
2
(I) 1   100% confidence intervals:
One sample  2 :

2
2
 n  1s , n  1s
  n21,
 n21,1
2
2

Two samples

.


 12
:
 22

s12
s12

s22
s22
,
 2
f
f n2 1, n 1,1
 n1 1, n2 1, 2
1
2
2








(II) Hypothesis testing:
One sample  2 :

n  1s 2
 
2
2
0
2
Two samples  1
 22
:
s12
f 
s22
2
 12
Summary table:
One sample
Point estimate
Classical approach
(critical values)
s
Two samples
2
s12
 n21, ,  n21,1 ,

,
2
n 1,
2
2
n 1,1
f n1 1,n2 1, , f n1 1,n2 1,1 ,
2
f n 1,n 1, , f n 1,n 1,1
1
Null distribution
(p-value)
 n21
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s 22
2
2
1
Fn1 1,n2 1
2
2
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ANOVA for testing the equality of k population means.
n x
k
j 1
f=
 n
k
j 1
2
j
j
j
 x
2
k  1 ~ Fk 1, n  k
T
 1 s 2j
nT  k
test for proportions of a multinomial population and for the
independence (contingency table).
k

i 1
p
m

i 1 j 1

f

f i  ei
2
~
ei
 eij 
 k21 ,
2
ij
eij
~  p 1m 1
2
Example 1:
A sample size of 12 provides a sample variance of 0.042. Test H 0 :  2  0.03
with   0.05 .
[solution:]
n  1s 2  11  0.042  15,4   2   2  19.68  not reject H .
2 
0
n 1,
11, 0.05
0.03
 02
Example 2:
Given the following talbe,
Sample Size
Sample 1
n1  15
Sample 2
n2  11
s1  0.8
s2  0.6
Sample Standard
Deviation
test H 0 :  12   22 with
  0.05 .
[solution:]
s12
2
2
s
1
1

 f14,10,0.975  f  2 
3.15 f10,14,0.025
 12
0.8 2
 not reject H 0 .
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0.6 2  1.77  f
 f14,10,0.025  3.55
n1 1, n2 1,
2
1
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Example 3:
The following data are from 4 different populations.
Population
1
8.2
8.7
9.4
9.2
Population
2
Population
3
7.7
8.4
8.6
8.1
8.0
6.9
5.8
7.2
6.8
7.4
Population
4
6.8
7.3
6.3
6.9
7.1
 1 ,  2 ,  3 and  4
Let
6.1
be the mean number of products of the 4
production lines. (a) Provide the ANOVA table. (b) Please test the hypothesis
H 0 : 1   2  3   4
with
  0.05 .
[solution:]
4
nj
x1  8.875, x2  8.16, x3  6.7, x4  6.88, x  7.545,   x 2j ,i  1157.89
j 1 i 1
n1  4, n2  5, n3  6, n4  5, nT  20
 x
nj
4
(a)
j 1 i 1
 n x
j
4
2
j ,i
nj
j 1 i 1
2
4
j 1
 x    x 2j ,i  nT x 2  19.35 ,
j
 x
 48.875  7.545  58.16  7.545  66.7  7.545  56.88  7.545
 15.462
2
 n
4
and
j 1
j
2
2
2
 1s 2j  19.35  15.462  3.888 .
The ANOVA table is
Source
SS
df
MS
Between
15.462
k 1  3
15.462
 5.154
3
Within
3.888
nT  k  16
3.888
 0.243
16
19.35
nT  1 19
Total
(b)Since
f  21.21  3.24  f 3,16, 0.05  f k 1, nT  k ,
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F
5.154
0.243
 21.21
f 
,we reject
H0 .
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Example 4:
Five observations were selected from each of three populations. The data
obtained follow.
Observation
Sample 1
Sample 2
Sample 3
1
32
30
30
26
32
30
6
44
43
44
46
48
45
4
33
36
35
36
40
36
6.5
2
3
4
5
Sample mean
Sample variance
(i) Set up the ANOVA table for this problem.
(ii) At the   0.01 level of significance, test the null hypothesis that the three
population means are equal?
[solution:]
 n x
3
j 1
j
 n
3
j 1
j
 x   530  37   545  37   536  37   570,
2
j
2
2
2
 1s 2j  4  6  4  4  4  6.5  66
Therefore,
Source
SS
DF
MS
F
Between
570
2
285
51.82
Within
66
12
5.5
Total
636
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(ii) f  51.82  6.93  f 2,12, 0.01  reject H 0 .
Example 5:
The following are the number of wrong answers for the number of the students.
Number of
wrong
answers
0
1
2
3
Number of
the
students
21
31
12
0
Suppose X is the random variable representing the number of wrong answers.
Please test X is distributed as Binomial(3,0.25) with
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  0.05 .
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(Note: the distribution function for Binomial(3,0.25) is
 3
x
3 x
f x    0.25 0.75 , x  0,1,2,3. .
 x
[solutions:]
As
H0
is true, the distribution for the number of wrong answers is
 3
27
0
3
p1  P X  0  
 0
0.25 0.75  64 ,
 
 3
27
1
2
p2  P X  1  
 1
0.25 0.75  64 ,
 
 3
9
2
1
p3  P X  2   
 2
0.25 0.75  64 ,
 
 3
1
3
01
p4  P X  3  
 3
0.25 0.75  64 ,
 
n  21  31  12  0  64 , the expected numbers
Since the sample size
under
H 0 : p1 
27
27
9
1
, p2 
, p3 
, p4 
64
64
64
64
are
27
27
 27, e2  np2  64 
 27,
64
64
.
9
1
e3  np3  64 
 9, e4  np4  64 
 1,
64
64
e1  np1  64 
Therefore,
4
2  
 fk
k 1
2

21  27 

27
 3.92
 ek 
ek
2
2

31  27 

27
2

12  9

9
2

 1

1
Since
 2  3.92  7.81  32,0.05   k21, ,
we do not reject
H0 .
Example 6:
Some diet pills are supposed to cause significant weight loss. The following table
shows the results of a recent study where some individuals took the diet pills and
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some did not.
Diet Pills
No Diet Pills
Total
No Weight Loss
80
20
100
Weight Loss
100
100
200
Total
180
120
300
With a 5% level of significance, test to see if losing weight is dependent on taking
the diet pills.
[solutions:]
eij
The table for the expected numbers
is
Diet Pills
No Diet Pills
Total
180  100
 60
300
120  100
 40
300
100
Weight Loss
180  200
 120
300
120  200
 80
300
200
Total
180
120
300
No Weight Loss
Thus,
p
m
  
2
i 1

f
j 1
 eij 
2
ij
eij
80  602
60

2
2

i 1
j 1
2

20  40 

40
f
 eij 
2
ij
eij
2

100  120

120
2

100  80 

80
 25
Since
 2  25  3.84  12,0.05  22121,0.05  2p1m1, , we reject
H0 .
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