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Mata kuliah
Tahun
: A0392 - Statistik Ekonomi
: 2010
Pertemuan 09 & 10
Pengujian Hipotesis
1
Outline Materi :
• Uji hipotesis rata-rata dan proporsi
• Uji hipotesis beda dua rata-rata
• Uji hipotesis beda dua proporsi
2
Introduction
• When the sample size is small, the
estimation and testing procedures of large
samples are not appropriate.
• There are equivalent small sample test and
estimation procedures for
m, the mean of a normal population
m1-m2, the difference between two population
means
s2, the variance of a normal population
The ratio of two population variances.
3
The Sampling Distribution
of the Sample Mean
• When we take a sample from a normal
population, the sample mean x has a normal
distribution for any sample size n, and
x-m
z
s/ n
x-m
is not normal!
s/ n
• has a standard normal distribution.
• But if s is unknown, and we must use s to
estimate it, the resulting statistic is not normal.
4
Student’s t Distribution
•
Fortunately, this statistic does have a sampling
distribution that is well known to statisticians,
called the Student’s t distribution, with n-1
degrees of freedom.
x-m
t
s/ n
•We can use this distribution to create
estimation testing procedures for the population
5
mean m.
Properties of Student’s t
•Mound-shaped and
symmetric about 0.
•More variable than
z, with “heavier tails”
•
•
•
Shape depends on the sample size n or the degrees
of freedom, n-1.
As n increase , variability of t decrease because the
estimate s is based on more and more information.
As n increases the shapes of the t and z distributions
become almost identical.
Applet
6
Using the t-Table
•
•
Table 4 gives the values of t that cut off certain
critical values in the tail of the t distribution.
Index df and the appropriate tail area a to find
ta,the value of t with area a to its right.
For a random sample of size n =
10, find a value of t that cuts
off .025 in the right tail.
Row = df = n –1 = 9
Column subscript = a = .025
t.025 = 2.262
7
Small Sample Inference
for a Population Mean m
•
The basic procedures are the same as those
used for large samples. For a test of
hypothesis:
Test H 0 : m  m 0 versus H a : one or two tailed
using the test statistic
x - m0
t
s/ n
using p - values or a rejection region based on
a t - distributi on with df  n - 1.
8
Example
A sprinkler system is designed so that the average
time for the sprinklers to activate after being turned
on is no more than 15 seconds. A test of 5 systems
gave the following times:
17, 31, 12, 17, 13, 25
Is the system working as specified? Test using
a = .05.
H 0 : m  15 (working as specified)
H a : m  15 (not worki ng as specified)
9
Example
Data: 17, 31, 12, 17, 13, 25
First, calculate the sample mean and
standard deviation, using your calculator or
the formulas in Chapter 2.
 xi 115
x

 19.167
n
6
( x)
115
x 2477 n 
6  7.387
s
n -1
5
2
2
2
10
Example
Data: 17, 31, 12, 17, 13, 25
Calculate the test statistic and find the
rejection region for a =.05.
Test statistic :
Degrees of freedom :
x - m 0 19.167 - 15
t

 1.38 df  n - 1  6 - 1  5
s / n 7.387 / 6
Rejection Region: Reject H0 if
t > 2.015. If the test statistic falls
in the rejection region, its p-value
will be less than a = .05.
11
Conclusion
Data: 17, 31, 12, 17, 13, 25
Compare the observed test statistic to the
rejection region, and draw conclusions.
H 0 : m  15
H a : m  15
Test statistic : t  1.38
Rejection Region :
Reject H 0 if t  2.015.
Conclusion: For our example, t = 1.38 does not fall in the
rejection region and H0 is not rejected. There is insufficient
evidence to indicate that the average activation time is greater
than 15.
12
Approximating the
p-value
• You can only approximate the p-value
for the test using Table 4.
Since the observed value
of t = 1.38 is smaller
than t.10 = 1.476,
p-value > .10.
13
Applet
The exact p-value
• You can get the exact p-value
using some calculators or a computer.
p-value = .113 which
is greater than .10 as
we approximated
using Table 4.
One-Sample T: Times
Test of mu = 15 vs mu > 15
Variable
Times
Variable
Times
N
6
Mean
19.17
95.0% Lower Bound
13.09
StDev
7.39
T
1.38
SE Mean
3.02
P
0.113
14
Testing the Difference
between Two Means
Asin Chapter9, independent random samplesof sizen1 and n2 are drawn
from populations1 and 2 with means μ1 and μ2 and variancess 12and s 22 .
Since the samplesizesare small,the twopopulations must be normal.
•To test:
•H0: m1-m2  D0 versus Ha: one of three
where D0 is some hypothesized difference,
usually 0.
15
Testing the Difference
between Two Means
•The test statistic used in Chapter 9
x1 - x2
z
s12 s22

n1 n2
•does not have either a z or a t distribution, and
cannot be used for small-sample inference.
•We need to make one more assumption, that
the population variances, although unknown,
are equal.
16
Testing the Difference
between Two Means
•Instead of estimating each population variance
separately, we estimate the common variance
with
2
2
•And
the
resulting
(
n
1
)
s

(
n
1
)
s
1
2
2
s2  1
test statistic,
n1  n2 - 2
has a t distribution
x1 - x2 - D0
with n1+n2-2=(n1t
1)+(n2-1) degrees of
1
2 1
s   
freedom.
n
n
2 
 1
17
Example
• Two training procedures are compared by
measuring the time that it takes trainees to
assemble a device. A different group of trainees are
taught using each method. Is there a difference in the
two methods? Use a = .01.
Time to
Assemble
Method 1
Method 2
Sample size
10
12
Sample mean
35
31
Sample Std
Dev
4.9
4.5
H 0 : m1 - m 2  0
H a : m1 - m 2  0
Test statistic :
x1 - x2 - 0
t
1
2 1
s   
 n1 n2 
18
Applet
Example
• Solve this problem by approximating the pMethod 1 Method 2
value using Time to
Assemble
Table 4.
Sample size
10
12
Sample mean
35
31
Sample Std
Dev
4.9
4.5
Calculate :
2
2
(
n
1
)
s

(
n
1
)
s
1
2
2
s2  1
n1  n2 - 2
9(4.9 2 )  11(4.52 )

 21.942
20
Test statistic :
t
35 - 31
1 1
21.942  
 10 12 
 1.99
19
Example
p - value : P(t  1.99)  P(t  -1.99)
1
P(t  1.99)  ( p - value)
2
df = n1 + n2 – 2 = 10 + 12 – 2 = 20
.025 < ½( p-value) < .05
.05 < p-value < .10
Since the p-value is
greater than a = .01, H0
is not rejected. There is
insufficient evidence to
indicate a difference in
the population means.
20
Testing the Difference
between Two Means
•If the population variances cannot be assumed
equal, the test statistic
t
x1 - x2
s12 s22

n1 n2
2
s
s 
  
n1 n2 

df  2
( s1 / n1 ) 2 ( s22 / n2 ) 2

n1 - 1
n2 - 1
2
1
2
2
•has an approximate t distribution with degrees
of freedom given above. This is most easily
done by computer.
21
The Paired-Difference
Test
•Sometimes the assumption of independent
samples is intentionally violated, resulting in a
matched-pairs or paired-difference test.
•By designing the experiment in this way, we can
eliminate unwanted variability in the experiment
by analyzing only the differences,
di = x1i – x2i
•to see if there is a difference in the two
population means, m1-m2.
22
Example
Car
1
2
3
4
5
Type A
10.6
9.8
12.3
9.7
8.8
Type B
10.2
9.4
11.8
9.1
8.3
• One Type A and one Type B tire are randomly assigned
to each of the rear wheels of five cars. Compare the
average tire wear for types A and B using a test of
hypothesis.
• But the samples are not
independent. The pairs of
H 0 : m1 - m 2  0
responses are linked because
H a : m1 - m 2  0
measurements are taken on the
same car.
23
The Paired-Difference
Test
To test H 0 : m1 - m 2  0 we test H 0 : m d  0
using the test statistic
d -0
t
sd / n
where n  number of pairs, d and sd are the
mean and standard deviation of the difference s, d i .
Use the p - value or a rejection region based on
a t - distributi on with df  n - 1.
24
Example
Car
1
2
3
4
5
Type A
10.6
9.8
12.3
9.7
8.8
Type B
10.2
9.4
11.8
9.1
8.3
Difference
.4
.4
.5
.6
.5
H 0 : m1 - m 2  0
H a : m1 - m 2  0
Test statistic :
 di
Calculated 
 .48
n

 di 
-
d -0
.48 - 0
t

 12.8
sd / n .0837 / 5
2
sd 
d
2
i
n -1
n
 .0837
25
Example
Car
1
2
3
4
5
Type A
10.6
9.8
12.3
9.7
8.8
Type B
10.2
9.4
11.8
9.1
8.3
Difference
.4
.4
.5
.6
.5
Rejection region: Reject H0 if
t > 2.776 or t < -2.776.
Conclusion: Since t = 12.8,
H0 is rejected. There is a
difference in the average tire
wear for the two types of tires.
26
Key Concepts
27
Test statistic for proportions
The test statistic is a value computed from
the sample data, and it is used in making
the decision about the rejection of the null
hypothesis.

z=p-p

pq
n
Test statistic for
proportions
28
Solution: The preceding example showed that the given claim results in the
following null and alternative hypotheses: H0: p = 0.5 and H1: p > 0.5. Because we
work under the assumption that the null hypothesis is true with p = 0.5, we get the
following test statistic:
z = p – p = 0.56 - 0.5 = 3.56


pq
n

(0.5)(0.5)
880
Figure 7-3
Inferences about
Two Proportions
Assumptions
1. We have proportions from two
independent simple random samples.
2. For both samples, the conditions np  5
and nq  5 are satisfied.
31
Notation for
Two Proportions
For population 1, we let:
p1 = population proportion
n1 = size of the sample
x1 = number of successes in the sample
^
p1 = x1 (the sample proportion)
n1
q^1 = 1 – ^
p1
The corresponding meanings are attached
to p2, n2 , x2 , ^
p2. and q^2 , which come from
population 2.
32
Pooled Estimate of
p and p
1
2
 The pooled estimate of p1 and
p2
is denoted by p.
x1 + x2
p= n +n
1
2

q =1–p
33
Test Statistic for
Two Proportions
For H0: p1 = p2 , H0: p1  p2 ,
H1: p1  p2 ,
z=
H0: p1 p2
H1: p1 < p2 , H1: p 1> p2
^ )–(p –p )
( p^1 – p
2
1
2
pq
pq
n1 + n2
34
Test Statistic for
Two Proportions
For H0: p1 = p2 , H0: p1  p2 ,
H1: p1  p2 ,
where
p1 – p 2 = 0
p^
1
p=
x1 + x2
n1 + n2
x1
= n
1
H0: p1 p2
H1: p1 < p2 , H1: p 1> p2
(assumed in the null hypothesis)
and
and
p^
2
x2
=
n2
q=1–p
35
Example:
Example:
n1= 200
x1 = 24
^p1 = x1 = 24 = 0.120
n1 200
n2 = 1400
x2 = 147
^p2 = x2 = 147 = 0.105
n2 1400
H0: p1 = p2, H1: p1 > p2
p = x1 + x2 = 24 + 147 = 0.106875
n1 + n2 200+1400
q = 1 – 0.106875 = 0.893125.
Example:
n1= 200
(0.120 – 0.105) – 0
z=
x1 = 24
^p1 = x1 = 24 = 0.120
n1 200
n2 = 1400
x2 = 147
^p2 = x2 = 147 = 0.105
n2 1400
(0.106875)(0.893125) + (0.106875)(0.893125)
200
1400
z = 0.64
Example:
n1= 200
x1 = 24
^p1 = x1 = 24 = 0.120
n1 200
n2 = 1400
x2 = 147
^p2 = x2 = 147 = 0.105
n2 1400
z = 0.64
This is a right-tailed test, so the Pvalue is
the area to the right of the test statistic
z = 0.64. The P-value is 0.2611.
Because the P-value of 0.2611 is
greater than the significance level of a
= 0.05, we fail to reject the null
hypothesis.
Example: For the sample data listed in
Table 8-1, use a 0.05 significance level to test
the claim that the proportion of black drivers stopped by the police is greater than the
proportion of white drivers who are stopped.
n1= 200
x1 = 24
^p1 = x1 = 24 = 0.120
n1 200
n2 = 1400
x2 = 147
^p2 = x2 = 147 = 0.105
n2 1400
z = 0.64
Because we fail to reject the null
hypothesis, we conclude that there is not
sufficient evidence to support the claim
that the proportion of black drivers
stopped by police is greater than that for
white drivers. This does not mean that
racial profiling has been disproved. The
evidence might be strong enough with
more data.
Example:
n1= 200
x1 = 24
^p1 = x1 = 24 = 0.120
n1 200
n2 = 1400
x2 = 147
^p2 = x2 = 147 = 0.105
n2 1400
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