 # 1 Pertemuan 09 Pengujian Hipotesis Proporsi dan Data Katagorik Matakuliah: A0392 – Statistik Ekonomi Tahun: 2006.

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1 Pertemuan 09 Pengujian Hipotesis Proporsi dan Data Katagorik Matakuliah: A0392 – Statistik Ekonomi Tahun: 2006

2 Outline Materi : Uji hipotesis proporsi Uji hipotesis beda proporsi Uji kebebasan data katagorik

3 Summary of Test Statistics to be Used in a Hypothesis Test about a Population Mean n > 30 ? s known ? Popul. approx.normal ? s known ? Use s to estimate s Use s to estimate s Increase n to > 30 Yes Yes Yes Yes No No No No

4 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 p 0 is the hypothesized value of the population proportion). H 0 : p > p 0 H 0 : p < p 0 H 0 : p = p 0 H a : p p 0 H a : p p 0

5 Tests about a Population Proportion: Large-Sample Case (np > 5 and n(1 - p) > 5) Test Statistic where: Rejection Rule One-Tailed Two-Tailed H 0 : p  p  Reject H 0 if z > z  H 0 : p  p  Reject H 0 if z < -z  H 0 : p  p  Reject H 0 if |z| > z 

6 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.

7 Example: NSC Two-Tailed Test about a Population Proportion: Large n –Hypothesis H 0 : p =.5 H a : p.5 –Test Statistic

8 Example: NSC Two-Tailed Test about a Population Proportion: Large n –Rejection Rule Reject H 0 if z 1.96 –Conclusion Do not reject H 0. For z = 1.278, the p-value is.201. If we reject H 0, we exceed the maximum allowed risk of committing a Type I error (p-value >.050).

9 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 H 0 and the conclusion reject H 0. In such situations, it is recommended that the hypothesis-testing procedure be extended to include consideration of making a Type II error.

10 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 H 0 ; 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.

11 Example: Metro EMS (revisited) Calculating the Probability of a Type II Error 1. Hypotheses are: H 0 :  and H a :  2. Rejection rule is: Reject H 0 if z > 1.645 3. Value of the sample mean that identifies the rejection region: 4. We will accept H 0 when x < 12.8323

12 Example: Metro EMS (revisited) Calculating the Probability of a Type II Error 5. Probabilities that the sample mean will be in the acceptance region: Values of   1-  14.0 -2.31.0104.9896 13.6 -1.52.0643.9357 13.2 -0.73.2327.7673 12.83 0.00.5000.5000 12.8 0.06.5239.4761 12.4 0.85.8023.1977 12.0001 1.645.9500.0500

13 Example: Metro EMS (revisited) Calculating the Probability of a Type II Error Observations about the preceding table: –When the true population mean  is close to the null hypothesis value of 12, there is a high probability that we will make a Type II error. –When the true population mean  is far above the null hypothesis value of 12, there is a low probability that we will make a Type II error.

14 Power of the Test The probability of correctly rejecting H 0 when it is false is called the power of the test. For any particular value of , the power is 1 – . We can show graphically the power associated with each value of  ; such a graph is called a power curve.

15 Determining the Sample Size for a Hypothesis Test About a Population Mean where z  = z value providing an area of  in the tail z  = z value providing an area of  in the tail  = population standard deviation  0 = value of the population mean in H 0  a = value of the population mean used for the Type II error Note: In a two-tailed hypothesis test, use z  /2 not z 

16 Relationship among , , and n Once two of the three values are known, the other can be computed. For a given level of significance , increasing the sample size n will reduce . For a given sample size n, decreasing  will increase , whereas increasing  will decrease .

17 Inferences About the Difference Between the Proportions of Two Populations Sampling Distribution of Interval Estimation of p 1 - p 2 Hypothesis Tests about p 1 - p 2

18 Expected Value Standard Deviation Distribution Form If the sample sizes are large (n 1 p 1, n 1 (1 - p 1 ), n 2 p 2, and n 2 (1 - p 2 ) are all greater than or equal to 5), the sampling distribution of can be approximated by a normal probability distribution. Sampling Distribution of

19 Interval Estimation of p 1 - p 2 Interval Estimate Point Estimator of

20 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?

21 Example: MRA Point Estimator of the Difference Between the Proportions of Two Populations p 1 = proportion of the population of households “aware” of the product after the new campaign p 2 = 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

22 Example: MRA Interval Estimate of p 1 - p 2 : Large-Sample Case For  =.05, z.025 = 1.96:.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.

23 Hypothesis Tests about p 1 - p 2 Hypotheses H 0 : p 1 - p 2 < 0 H a : p 1 - p 2 > 0 Test statistic Point Estimator of where p 1 = p 2 where:

24 Example: MRA Hypothesis Tests about p 1 - p 2 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? p 1 = proportion of the population of households “aware” of the product after the new campaign p 2 = proportion of the population of households “aware” of the product before the new campaign –Hypotheses H 0 : p 1 - p 2 < 0 H a : p 1 - p 2 > 0

25 Example: MRA Hypothesis Tests about p 1 - p 2 –Rejection Rule Reject H 0 if z > 1.645 –Test Statistic –Conclusion Do not reject H 0.

26 Test of Independence: Contingency Tables 1. Set up the null and alternative hypotheses. 2. Select a random sample and record the observed frequency, f ij, for each cell of the contingency table. 3. Compute the expected frequency, e ij, for each cell.

27 Test of Independence: Contingency Tables 4. Compute the test statistic. 5. Reject H 0 if (where  is the significance level and with n rows and m columns there are (n - 1)(m - 1) degrees of freedom).

28 Example: Finger Lakes Homes (B) Contingency Table (Independence) Test Each home sold can be classified according to price and to style. Finger Lakes Homes’ manager would like to determine if the price of the home and the style of the home are independent variables. The number of homes sold for each model and price for the past two years is shown below. For convenience, the price of the home is listed as either \$65,000 or less or more than \$65,000. Price Colonial Ranch Split-Level A- Frame < \$65,000 18 6 19 12 > \$65,000 12 14 16 3

29 n Contingency Table (Independence) Test Hypotheses Hypotheses H 0 : Price of the home is independent of the style of the home that is purchased H 0 : Price of the home is independent of the style of the home that is purchased H a : Price of the home is not independent of the H a : Price of the home is not independent of the style of the home that is purchased style of the home that is purchased Expected Frequencies Expected Frequencies Price Colonial Ranch Split-Level A-Frame Total Price Colonial Ranch Split-Level A-Frame Total < \$99K 18 6 19 12 55 > \$99K 12 14 16 3 45 Total 30 20 35 15 100 Total 30 20 35 15 100 Example: Finger Lakes Homes (B)

30 Contingency Table (Independence) Test –Test Statistic =.1364 + 2.2727 +... + 2.0833 = 9.1486 –Rejection Rule With  =.05 and (2 - 1)(4 - 1) = 3 d.f., Reject H 0 if  2 > 7.81 –Conclusion We reject H 0, the assumption that the price of the home is independent of the style of the home that is purchased. Example: Finger Lakes Homes (B)

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