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Chapter 7
Hypothesis testing
§7.1 The basic concepts of
hypothesis testing

1 An example

Example 7.1 We selected 20 newborns
randomly from a region in 2002, the average
weight of them is 3160g, the sample standard
deviation of the weight is 300g, and based on
past statistics, the average weight of newborn
is 3140g. If the weight of X obeys normal
distribution. Is there any significant difference
about the weight between the newborn in 2002
and the old ones?


Let X denote the weight of newborn, then based on
the hypothesis, we have X ~ N ( ,  2 ) .The problem is that
whether the mean of population is equal to 3140g or not,
which can be expressed as
,
H 0 :   3140
This hypothesis is called zero hypotheses or the
original hypothesis.
If H 0 :   3140 is not correct, so H1 :   3140 is correct.
This hypothesis could be called the alternative
hypothesis. The above hypothesis testing problem is
often expressed as
.
H 0 :   3140  H1 :   3140
2, significant test

According to statistics, in the past, the
average weight of newborns is 3140g,
while in 2002, the average weight of
newborn samples is 3160g, a difference in
20g, this difference may arise in two
situations. One is that there is no
essential difference in them, the
difference in 20g is only caused by the
randomness of the sample; another is
caused by the essential difference in them.
So the point is whether the difference can
be explained by the randomness of
sample or not.

The sample mean is a good
estimation of the population mean ,
if   3140 , | X  3140 | should be
relatively small, we should establish
a reasonable limit C ,
When| X  3140 | C ,we accept the zero
hypothesis ;Otherwise ,we will
accept alternative hypothesis.


We know that
X 
T
~ t (n  1)
S/ n
,
Setting  =0.01, P(| T | t0.005 (n  1))    0.01 , if the
observation of | T | satisfies | t |  t0.005 (n  1) ,that
is to say the small probability event {| T | t0.005 (n  1)}
occur. Generally speaking, small probability
events will not occur in one experiment, so we
believe H 0 is unreasonable, we call {| t | t0.005 (n  1)}
critical region. Otherwise, we have not enough
H0
evidence to reject
,so we
accept H 0 .Which is called significant test.
3,Two types of errors


1. The original hypothesis is actually correct, but the
test result wrongly reject it, which commit "abandoning
true" errors, often referred to as type I error.
2. The original hypothesis is not right, but the test
result wrongly accept it, which commit "maintaining
false " errors, often referred to as type II error. As the
sample is random, so we are committing two types of
errors on certain probability. In statistics, we call the
probability of committing type I error the significance
level, abbreviated as the level. Naturally, people
desire the probability of committing two types of errors
as small as possible, but for a given sample size, We
can not reduce the probability of committing two types
of errors simultaneously, Commonly, we often fix the
upper bound of the probability of committing type I
error, and then select a test with smaller probability of
committing type II error.
§7.2 single normal population
hypothesis testing


Let ( X1 , , X n ) denote a random sample from a normal
distribution X ~ N ( ,  2 ) ,
is significance level.
1 Test of mean


1.1 Variance known
The problem as follows
H 0 :   0  H1 :   0


When H 0 is true, then
U

thus
X 
/ n
~ N (0,1)
P(| U | u /2 )  
So the critical region is
,
{| u | u /2 }
.

Example 7.2
Let us assume X N (4.55, 0.1082 ) ,we have nine
random samples,and the mean is
4.484,suppose the variance have no change,
testing the following problem with the
significance level   0.05,
H 0 :   4.55  H1 :   4.55 .
u0.025=1.96,  0.108, x  4.484, n  9 ，we get
4.484  4.55
| u ||
| 1.83  1.96

,
0.108 / 3
Then H can be accepted.

0
1.2 Variance unknown

The problem as follows
H 0 :   0  H1 :   0
When H is true, then
0
X 
T

Thus
S/ n
~ t (n  1)
P(| T | t /2 (n  1))   ,
,
,
So the critical region is {| t | t /2 (n  1)} .


Example 7.3
Let us assume X have normal
distribution,we have 25 random
samples,and the mean is 66.5,and the
standard deviation of sample is 15,
testing the following problem with the
significance level   0.05,
H 0 :   70  H1 :   70

.
n  25，t0.025 (24) 2.06, x  66.5 ,we get
66.5  70
| t ||
| 1.167  2.06
15 / 5

,
Then H 0 can be accepted.

There are still two other kinds of
problem of testing mean as
follows
H 0 :   0  H1 :   0
H 0 :   0  H1 :   0
2 Test of variance


The problem as follows
H 0 :     H1 :   
2
2
0
2
2
0
When H 0 is true, then

2 
(n  1) S 2
2
~  2 (n  1)
thus
2
2
2
2
P(   1 /2 (n  1)或   /2 (n  1))  
So the critical region is
{ 2  12 /2 (n  1)} { 2  2 /2 (n  1)}

Example 7.4



Let us assume X have normal
distribution,we have 5 random samples,
1.32, 1.55, 1.36, 1.40, 1.44,
and the standard deviation is   0.048,
testing the following problem with the
significance level   0.05,
H 0 :  2  0.0482  H1 :  2  0.0482
.
2
2
(4)  11.14,  0.975
(4)  0.484 ,From the
n  5, 0.025
given sample,we have S 2  0.00778 ,hence
2 
4  0.00778
 13.51  11.14
2
,
0.048
So we reject H 0 .

There are still two other kinds of
problem of testing variance as
follows
2
2
2
2 ;
H :    H :  
0

0
1
0
H 0 :     H1 :    .
2
2
0
2
2
0
§7.3 double normal population
hypothesis testing


Let ( X 1 , X 2 ,..., X n ) denote a random sample
of size n1 from a distribution that is
N ( 1 ,  12 ), (Y1 , Y2 ,..., Yn ) denote a random
sample of size n from a distribution
2
2
2
2
that is N ( 2 ,  2 ),where 1 ,  1 ,  2 ,  2 are
unknown parameters, the two random
samples are independent
1
2
2
2



2
1. 1




, testing of 1  2
The problem as follows
H 0 : 1  2  a  H1 : 1  2  a
When H 0 is true, then
T
(X Y)  a
S
1 1

n1 n2
~ t (n1  n2  2)
Thus P(| T | t (n  n  2))  
 /2 1
2
So the critical region is

,
{| T | t /2 (n1  n2  2)} .
,

Example 7.4
Let us assume X and Y both have
normal distribution with equal
variances,we have the following random
samples,
X 24.3, 20.8, 23.7, 21.3, 17.4;
Y 18.2, 16.9, 20.2, 16.7.
testing the following problem with the
significance level   0.05,

H 0 : 1  2  H1 : 1  2

Using similar methods above to
discuss the following problems
H 0 : 1  2  a  H1 : 1  2  a


H 0 : 1  2  a  H1 : 1  2  a
2 testing of  / 
2
1


2
2
The problem as follows
H 0 :     H1 :   
2
1
2
2
2
1
2
2
When H 0 is true, then

F S
,
hence
2
1
S
2
2
F (n1  1, n2  1)
P( F  F1 2 (n1  1, n2  1)或F  F 2 (n1  1, n2  1))  

So the critical region is .
{F  F1 2 (n1  1, n2  1)或F  F 2 (n1  1, n2  1)}

Using similar methods above to
discuss the following problems,
H 0 :     H1 :   
2
1
2
2
2
1
2
2

Example 7.5

There are two team A and B participate a
paper contest, A team have 9 people,and B
team have 8 people,the score as follows:
A team 85, 59, 66, 81, 35, 57, 76, 63,
78,
B team 65, 72, 69, 65, 58, 68, 52, 64,
Can we believe the variance of B team is
significantly greater than A team’s concerning
the significance level 0.05?