Matakuliah : I0134/Metode Statistika
Tahun
: 2007
Pengujian Hipotesis Proporsi dan Beda Proporsi
Pertemuan 21
A Summary of Forms for Null and Alternative
Hypotheses about a Population Proportion
• The equality part of the hypotheses always appears in the null
hypothesis.
• In general, a hypothesis test about the value of a population
proportion p must take one of the following three forms (where p0 is
the hypothesized value of the population proportion).
H0: p > p0
Ha: p < p0
Bina Nusantara
H0: p < p0
Ha: p > p0
H0: p = p0
Ha: p  p0
Tests about a Population Proportion:
Large-Sample Case (np > 5 and n(1 - p) > 5)
• Test Statistic
where:
z
p 
p  p0
p
p0 (1  p0 )
n
• Rejection Rule
One-Tailed
H0: p  p
H0: p  p
H0: pp
Bina Nusantara
Two-Tailed
Reject H0 if z > z
Reject H0 if z < -z
Reject H0 if |z| > z
Example: NSC
• Two-Tailed Test about a Population Proportion: Large n
For a Christmas and New Year’s week, the National Safety
Council estimated that 500 people would be killed and 25,000
injured on the nation’s roads. The NSC claimed that 50% of the
accidents would be caused by drunk driving.
A sample of 120 accidents showed that 67 were caused by
drunk driving. Use these data to test the NSC’s claim with  = 0.05.
Bina Nusantara
Example: NSC
• Two-Tailed Test about a Population Proportion: Large n
– Hypothesis
H0: p = .5 
Ha: p .5
– Test Statistic
p0 (1  p0 )
.5(1  .5)
p 

 .045644
n
120
z
Bina Nusantara
p  p0
p
(67 /120)  .5

 1.278
.045644
Example: NSC
• Two-Tailed Test about a Population Proportion: Large n
– Rejection Rule
Reject H0 if z < -1.96 or z > 1.96
– Conclusion
Do not reject H0.
For z = 1.278, the p-value is .201. If we reject
H0, we exceed the maximum allowed risk of
(p-value > .050).
Bina Nusantara
committing a Type I error
Hypothesis Testing and Decision Making
• In many decision-making situations the decision maker may want,
and in some cases may be forced, to take action with both the
conclusion do not reject H0 and the conclusion reject H0.
• In such situations, it is recommended that the hypothesis-testing
procedure be extended to include consideration of making a Type II
error.
Bina Nusantara
Calculating the Probability of a Type II Error
in Hypothesis Tests about a Population Mean
1. Formulate the null and alternative hypotheses.
2. Use the level of significance  to establish a rejection rule
based on the test statistic.
3. Using the rejection rule, solve for the value of the sample mean
that identifies the rejection region.
4. Use the results from step 3 to state the values of the sample
mean that lead to the acceptance of H0; this defines the
acceptance region.
x any value of  from the
5. Using the sampling distribution of for
alternative hypothesis, and the acceptance region from step 4,
compute the probability that the sample mean will be in the
acceptance region.
Bina Nusantara
Inferences About the Difference
Between the Proportions of Two Populations
• Sampling Distribution of p1  p2
• Interval Estimation of p1 - p2
• Hypothesis Tests about p1 - p2
Bina Nusantara
Sampling Distribution of p1  p2
• Expected Value
E ( p1  p2 )  p1  p2
• Standard Deviation
 p1  p2 
p1 (1  p1 ) p2 (1  p2 )

n1
n2
• Distribution Form
If the sample sizes are large (n1p1, n1(1 - p1), n2p2,
and n2(1 - p2) are all greater than or equal to 5), the
sampling distribution of p1  p2 can be approximated
by a normal probability distribution.
Bina Nusantara
Interval Estimation of p1 - p2
• Interval Estimate
p1  p2  z / 2  p1  p2
• Point Estimator of
s p1  p2 
Bina Nusantara
 p1  p2
p1 (1  p1 ) p2 (1  p2 )

n1
n2
Example: MRA
MRA (Market Research Associates) is conducting research
to evaluate the effectiveness of a client’s new advertising
campaign. Before the new campaign began, a telephone survey
of 150 households in the test market area showed 60
households “aware” of the client’s product. The new campaign
has been initiated with TV and newspaper advertisements
running for three weeks. A survey conducted immediately after
the new campaign showed 120 of 250 households “aware” of
the client’s product.
Does the data support the position that the advertising
campaign has provided an increased awareness of the client’s
product?
Bina Nusantara
Example: MRA
• Point Estimator of the Difference Between the Proportions of
Two Populations
120 60
p1  p2  p1  p2 

. 48. 40 . 08
250 150
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
p1 = sample proportion of households “aware” of the
product after the new campaign
p2= sample proportion of households “aware” of the
product before the new campaign
Bina Nusantara
Example: MRA
• Interval Estimate of p1 - p2: Large-Sample Case
For = .05, z.025 = 1.96:
. 48(.52) . 40(. 60)
. 48. 40  1. 96

250
150
.08 + 1.96(.0510)
.08 + .10
or -.02 to +.18
– Conclusion
At a 95% confidence level, the interval estimate of the
difference between the proportion of households aware of the
client’s product before and after the new advertising campaign
is -.02 to +.18.
Bina Nusantara
Hypothesis Tests about p1 - p2
H0: p1 - p2 < 0
Ha: p1 - p2 > 0
• Hypotheses
• Test statistic
z
• Point Estimator of
( p1  p2 )  ( p1  p2 )
 p1  p2
where p = p
 p1  p2 1 2
s p1  p2  p (1  p )(1 n1  1 n2 )
where:
Bina Nusantara
n1 p1  n2 p2
p
n1  n2
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
Bina Nusantara
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
Bina Nusantara
Do not reject H0.