Matakuliah
Tahun
: I0272 - STATISTIK PROBABILITAS
: 2009
PENGUJIAN HIPOTESIS 1
Pertemuan 9
Materi
• Uji hipotesis nilai tengah
• Uji hipotesis beda dua nilai tengah
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Hypothesis Testing
• Developing Null and Alternative Hypotheses
• Type I and Type II Errors
• One-Tailed Tests About a Population Mean:
Large-Sample Case
• Two-Tailed Tests About a Population Mean:
Large-Sample Case
• Tests About a Population Mean:
Small-Sample Case
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Developing Null and Alternative Hypotheses
• Hypothesis testing can be used to determine whether a
statement about the value of a population parameter
should or should not be rejected.
• The null hypothesis, denoted by H0 , is a tentative
assumption about a population parameter.
• The alternative hypothesis, denoted by Ha, is the
opposite of what is stated in the null hypothesis.
• Hypothesis testing is similar to a criminal trial. The
hypotheses are:
H0: The defendant is innocent
Ha: The defendant is guilty
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• Testing Research Hypotheses
– The research hypothesis should be expressed as
the alternative hypothesis.
– The conclusion that the research hypothesis is
true comes from sample data that contradict the
null hypothesis.
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A Summary of Forms for Null and Alternative
Hypotheses about a Population Mean
• The equality part of the hypotheses always appears in
the null hypothesis.
• In general, a hypothesis test about the value of a
population mean  must take one of the following three
forms (where 0 is the hypothesized value of the
population mean).
H0:  > 0
Ha:  < 0
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H0:  < 0
Ha:  > 0
H0:  = 0
Ha:  ≠ 0
77
Type I and Type II Errors
Contoh Soal: Metro EMS
Conclusion
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Population Condition
H0 True
Ha True
(  )
( )
Accept H0
(Conclude  
Correct
Conclusion
Reject H0
(Conclude 
Type I
rror
Type II
Error
Correct
Conclusion
88
The Steps of Hypothesis Testing






Determine the appropriate hypotheses.
Select the test statistic for deciding whether or not to
reject the null hypothesis.
Specify the level of significance for the test.
Use
to develop the rule for rejecting H0.
Collect the sample data and compute the value of the
test statistic.
a) Compare the test statistic to the critical value(s) in
the rejection rule, or
b) Compute the p-value based on the test statistic
and compare it to
to determine whether or not to
reject H0.
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One-Tailed Tests about a Population Mean: LargeSample Case (n > 30)

Hypotheses
H0:   
Ha: 


 Known
x  0
z
/ n
Test Statistic
or
H0:   
Ha: 
 Unknown
x  0
z
s/ n
Rejection Rule
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Reject H0 if z > zReject H0 if z < -z
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Two-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
• Hypotheses
• Test Statistic
H0:   
Ha:   
 Known
x  0
z
/ n
 Unknown
z
x  0
s/ n
• Rejection Rule
Reject H0 if |z| > z
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Example: One Tail Test
Q. Does an average box of cereal
contain more than 368 grams of
cereal? A random sample of 25
boxes showed X = 372.5. The
company has specified s to be
15 grams. Test at the a = 0.05
level.
368 gm.
H0: 368
H1: > 368
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Finding Critical Value: One Tail
What is Z given  = 0.05?
Z 1
Z
.95
 = .05
0 1.645
Critical Value
= 1.645
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Standardized Cumulative Normal
Distribution Table (Portion)
Z
.04
.05
.06
1.6 .9495 .9505 .9515
1.7 .9591 .9599 .9608
1.8 .9671 .9678 .9686
1.9 .9738 .9744 .9750
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Example Solution: One Tail Test
H0: 368
H1:  > 368
Test Statistic:
Z
 = 0.5
n = 25
Critical Value: 1.645
Decision:
1.645
1.50
Conclusion:
Do Not Reject at  = .05
.05
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
 1.50
n
Reject
0
X 
Z
No evidence that true
mean is more than 368
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p -Value Solution
p-Value is P(Z 1.50) = 0.0668
Use the
alternative
hypothesis to
find the
direction of
the rejection
region.
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P-Value =.0668
1.0000
- .9332
.0668
0
From Z Table:
Lookup 1.50 to
Obtain .9332
1.50
Z
Z Value of Sample
Statistic
15 15
(continued)
p -Value Solution
(p-Value = 0.0668)  ( = 0.05)
Do Not Reject.
p Value = 0.0668
Reject
 = 0.05
0
1.50
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1.645
Test Statistic 1.50 is in the Do Not Reject Region
Z
16 16
Example: Two-Tail Test
Q. Does an average box of cereal
contain 368 grams of cereal?
A random sample of 25 boxes
showed X = 372.5. The
company has specified  to
be 15 grams. Test at the 
0.05 level.
368 gm.
H0:  368
H1:   368
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Example Solution: Two-Tail Test
H0: 368
H1:  368
Test Statistic:
X   372.5  368
Z

 1.50

15
n
25
 = 0.05
n = 25
Critical Value: ±1.96
Reject
.025
-1.96
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.025
0 1.96
1.50
Z
Decision:
Do Not Reject at  = .05
Conclusion:
No Evidence that True
Mean is Not 368 18 18
p-Value Solution
(p Value = 0.1336)  ( = 0.05)
Do Not Reject.
p Value = 2 x 0.0668
Test
Statistic
1.50 is in
the Do Not
Reject
Region
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Reject
Reject
 = 0.05
0
1.50
1.96
Z
19 19
Tests about a Population Mean:
Small-Sample Case (n < 30)
x  0
x  Unknown
0
t
t
s/ n
/ n
•
Test Statistic  Known
•
This test statistic has a t distribution with n - 1 degrees of freedom.
Rejection Rule
One-Tailed
Two-Tailed
Ha: 
Ha: 
Ha:  
Reject H0 if t > t
Reject H0 if t < -t
Reject H0 if |t| > t

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Example: One-Tail t Test
Does an average box of
cereal contain more than
368 grams of cereal? A
random sample of 36
boxes showed X = 372.5,
and s  15. Test at the 
0.01 level.
 is not given
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368 gm.
H0:  368
H1:  368
21 21
Example Solution: One-Tail
H0: 368
H1:  368
Test Statistic:
t
 = 0.01
n = 36, df = 35
Critical Value: 2.4377
X 
372.5  368

 1.80
S
15
n
36
Decision:
Reject
.01
0 2.437
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1.80
Do Not Reject at  = .01
Conclusion:
No evidence that true mean is
more than 368
t35
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p -Value Solution
(p Value is between .025 and .05)  ( = 0.01).
Do Not Reject.
p Value = [.025, .05]
Reject
 = 0.01
0
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1.80
2.4377
t35
Test Statistic 1.80 is in the Do Not Reject Region
23 23
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
H0: 1 - 2 = 0
Ha: 1 - 2  0
Test Statistic
Large-Sample
z
Small-Sample
( x1  x2 )  ( 1   2 )
12
n1   22
n2
where,
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t
( x1  x2 )  ( 1   2 )
s2 (1 n1  1 n2 )
2
2
(
n

1)
s

(
n

1)
s
1
2
2
s2  1
n1  n2  2
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Critical Value untuk statistik uji z:
Ha
wilayah kritik (tolak H )
•
1 - 2 >
d0
z > z
•
1 - 2 <
d0
z < -z
•
1 - 2 
d0
z   z  atau z   z 
0
2
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Critical Value untuk statistik uji t:
Ha
wilayah kritik
•
1 - 2 >
d0
t > t
•
1 - 2 <
d0
t < -t
•
1 - 2 
d0
t  t  atau t  t 
2
2
v = n1 – n2 - 2 derajat bebas
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Contoh Soal: Specific Motors
• Hypothesis Tests About the Difference Between the
Means of Two Populations: Small-Sample Case
– Rejection Rule
Reject H0 if t > 1.734
( = 0.05, d.f. = 18)
– Test Statistic
t
where:
( x1  x2 )  ( 1   2 )
s2 (1 n1  1 n2 )
(n1  1)s12  (n2  1)s22
s 
n1  n2  2
2
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