Matakuliah
Tahun
Versi
: I0134 – Metoda Statistika
: 2005
: Revisi
Pertemuan 18
Pengujian Hipotesis Lanjutan
1
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Mahasiswa dapat menghasilkan simpulan
dari uji hipotesis beda rataan, proporsi dan
varians.
2
Outline Materi
• Uji hipotesis beda rataan
• Uji hipotesis beda proporsi
• Uji hipotesis homogenitas varians
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Hypothesis Tests About the Difference
Between the Means of Two
Populations: Independent Samples
• Hypotheses
H0: 1 - 2 < 0
Ha: 1 - 2 > 0
H0: 1 - 2 > 0
Ha: 1 - 2 < 0
• Test Statistic
Large-Sample
z
( x1  x2 )  ( 1   2 )
12 n1   22 n2
H0: 1 - 2 = 0
Ha: 1 - 2  0
Small-Sample
t
( x1  x2 )  ( 1   2 )
s2 (1 n1  1 n2 )
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Inference About the Difference Between the
Means of Two Populations: Matched
Samples
• With a matched-sample design each
sampled item provides a pair of data
values.
• The matched-sample design can be
referred to as blocking.
• This design often leads to a smaller
sampling error than the independentsample design because variation
between sampled items is eliminated as
a source of sampling error.
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Example: Express Deliveries
• Inference About the Difference Between the Means of
Two Populations: Matched Samples
Let d = the mean of the difference values for the
two delivery services for the population of
district offices
– Hypotheses
H0: d = 0, Ha: d 
– Rejection Rule
Assuming the population of difference values is
approximately normally distributed, the t distribution
with n - 1 degrees of freedom applies. With  = .05,
t.025 = 2.262 (9 degrees of freedom).
Reject H0 if t < -2.262 or if t > 2.262
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Example: Express Deliveries
• Inference About the Difference Between
the Means of Two Populations: Matched
Samples
 di ( 7  6... 5)
d 
n

10
 2. 7
2
76.1
 ( di  d )
sd 

 2. 9
n 1
9
t
d  d
2. 7  0

 2. 94
sd n 2. 9 10
– Conclusion
Reject H0.
There is a significant difference between the
mean delivery times for the two services.
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Hypothesis Tests about p1 - p2
• Hypotheses
H0: p1 - p2 < 0
Ha: p1 - p2 > 0
• Test statistic
z
( p1  p2 )  ( p1  p2 )
• Point Estimator of
where:
n1 p1  n2 p2
p
n1  n2
 p1  p2
 p1  p2
where p1 = p2
s p1  p2  p (1  p )(1 n1  1 n2 )
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Example: MRA
• Hypothesis Tests about p1 - p2
Can we conclude, using a .05 level of
significance, that the proportion of households aware
of the client’s product increased after the new
advertising campaign?
p1 = proportion of the population of households
“aware” of the product after the new campaign
p2 = proportion of the population of households
“aware” of the product before the new campaign
– Hypotheses
H0: p1 - p2 < 0
Ha: p1 - p2 > 0
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Example: MRA
• Hypothesis Tests about p1 - p2
– Rejection Rule
– Test Statistic
Reject H0 if z > 1.645
250(. 48)  150(. 40) 180
p

. 45
250  150
400
s p1  p2  . 45(.55)( 1
 1 ) . 0514
250 150
(. 48. 40)  0
. 08
z

 1.56
. 0514
. 0514
– Conclusion
Do not reject H0.
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Hypothesis Testing About the
Variances of Two Populations

One-Tailed Test
• Hypotheses
H 0 :  12   22
H a :  12   22
• Test Statistic
s12
F 2
s2
• Rejection Rule
Reject H0 if F > F where the value of F is based
on an F distribution with n1 - 1 (numerator) and
n2 - 1 (denominator) d.f.
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Hypothesis Testing About the
Variances of Two Populations

Two-Tailed Test
• Hypotheses
H 0 :  12   22
Ha : 12   22
• Test Statistic
s12
F 2
s2
• Rejection Rule
Reject H0 if F > F/2 where the value of F/2 is
based on an F distribution with n1 - 1
(numerator) and n2 - 1 (denominator) d.f.
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Example: Buyer’s Digest
Buyer’s Digest has conducted the same test, as was
described earlier, on another 10 thermostats, this time
manufactured by TempKing. The temperature readings
of the ten thermostats are listed below.
We will conduct a hypothesis test with  = .10 to see
if the variances are equal for ThermoRite’s thermostats
and TempKing’s thermostats.
Therm. 1
2
3
4
5
6
7
8
9
10
Temp. 66.4 67.8 68.2 70.3 69.5 68.0 68.1 68.6 67.9 66.2
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Example: Buyer’s Digest
• Hypothesis Testing About the Variances of Two
Populations
– Hypotheses
H 0 :  12   22 (ThermoRite and TempKing thermostats have same temperature variance)
H a :  12   22 (Their variances are not equal)
– Rejection Rule
The F distribution table shows that with  = .10,
9 d.f. (numerator), and 9 d.f. (denominator),
F.05 = 3.18.
Reject H0 if F > 3.18
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Example: Buyer’s Digest
• Hypothesis Testing About the Variances
of Two Populations
–
–
Test Statistic
ThermoRite’s sample variance is .70.
TempKing’s sample variance is 1.52.
F = 1.52/.70 = 2.17
Conclusion
We cannot reject H0. There is insufficient
evidence to conclude that the population
variances differ for the two thermostat
brands.
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• Selamat Belajar Semoga Sukses.
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