Pengujian Hipotesis Proporsi dan Beda Proporsi Pertemuan 21

Matakuliah : I0134/Metode StatistikaTahun : 2007

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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 H0: p < p0 H0: p = p0

Ha: p < p0 Ha: p > p0 Ha: p p0

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Tests about a Population Proportion:Large-Sample Case (np > 5 and n(1 - p) > 5)

• Test Statistic

where:

• Rejection Rule One-Tailed Two-Tailed

H0: pp Reject H0 if z > z

H0: pp Reject H0 if z < -z

H0: pp Reject H0 if |z| > z

zp p

p

0

z

p p

p

0

pp p

n

0 01( ) pp p

n

0 01( )

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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.

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Example: NSC• Two-Tailed Test about a Population Proportion:

Large n– Hypothesis

H0: p = .5

Ha: p .5

– Test Statistic

0 (67/ 120) .51.278

.045644p

p pz

0 (67/ 120) .51.278

.045644p

p pz

0 0(1 ) .5(1 .5).045644

120p

p p

n

0 0(1 ) .5(1 .5)

.045644120p

p p

n

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Example: NSC• Two-Tailed Test about a Population Proportion:

Large n– Rejection Rule

Reject H0 if z < -1.96 or z > 1.96

– ConclusionDo not reject H0.

For z = 1.278, the p-value is .201. If we rejectH0, we exceed the maximum allowed risk of committing a Type I error (p-value > .050).

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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.

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Calculating the Probability of a Type II Error

in Hypothesis Tests about a Population Mean1. 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.

5. Using the sampling distribution of for any value of from the alternative hypothesis, and the acceptance region from step 4, compute the probability that the sample mean will be in the acceptance region.

xx

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Inferences About the Difference Between the Proportions of Two

Populations• Sampling Distribution of

• Interval Estimation of p1 - p2

• Hypothesis Tests about p1 - p2

p p1 2p p1 2

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• Expected Value

• Standard Deviation

• Distribution FormIf the sample sizes are large (n1p1, n1(1 - p1),

n2p2,

and n2(1 - p2) are all greater than or equal to 5), thesampling distribution of can be approximatedby a normal probability distribution.

Sampling Distribution of p p1 2p p1 2

E p p p p( )1 2 1 2 E p p p p( )1 2 1 2

p pp pn

p pn1 2

1 1

1

2 2

2

1 1 ( ) ( ) p p

p pn

p pn1 2

1 1

1

2 2

2

1 1 ( ) ( )

p p1 2p p1 2

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Interval Estimation of p1 - p2• Interval Estimate

• Point Estimator of

p p z p p1 2 2 1 2 /p p z p p1 2 2 1 2 /

p p1 2 p p1 2

sp pn

p pnp p1 2

1 1

1

2 2

2

1 1 ( ) ( )

sp pn

p pnp p1 2

1 1

1

2 2

2

1 1 ( ) ( )

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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?

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Example: MRA

• Point Estimator of the Difference Between the Proportions of Two Populations

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 = sample proportion of households “aware” of the

product after the new campaign = sample proportion of households “aware” of the

product before the new campaign

p p p p1 2 1 2120250

60150

48 40 08 . . .p p p p1 2 1 2120250

60150

48 40 08 . . .

p1p1

p2p2

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Example: MRA

• Interval Estimate of p1 - p2: Large-Sample Case

For = .05, z.025 = 1.96:

.08 + 1.96(.0510) .08 + .10or -.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.

. . .. (. ) . (. )

48 40 1 9648 52250

40 60150

. . .. (. ) . (. )

48 40 1 9648 52250

40 60150

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Hypothesis Tests about p1 - p2

• Hypotheses H0: p1 - p2 < 0

Ha: p1 - p2 > 0

• Test statistic

• Point Estimator of where p1 = p2

where:

zp p p p

p p

( ) ( )1 2 1 2

1 2

zp p p p

p p

( ) ( )1 2 1 2

1 2

s p p n np p1 21 1 11 2 ( )( )s p p n np p1 21 1 11 2 ( )( )

pn p n pn n

1 1 2 2

1 2

pn p n pn n

1 1 2 2

1 2

p p1 2 p p1 2

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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 Reject H0 if z > 1.645

– Test Statistic

– Conclusion Do not reject H0.

p

250 48 150 40250 150

180400

45(. ) (. )

.p

250 48 150 40250 150

180400

45(. ) (. )

.

sp p1 245 55 1

2501150 0514 . (. )( ) .sp p1 2

45 55 1250

1150 0514 . (. )( ) .

z

(. . ).

..

.48 40 00514

080514

1 56z

(. . ).

..

.48 40 00514

080514

1 56