# Chapter 11a: Comparisons Involving Proportions and a Test of Independence Inference about the Difference between the Proportions of Two Populations Hypothesis.

## Presentation on theme: "Chapter 11a: Comparisons Involving Proportions and a Test of Independence Inference about the Difference between the Proportions of Two Populations Hypothesis."— Presentation transcript:

Chapter 11a: Comparisons Involving Proportions and a Test of Independence Inference about the Difference between the Proportions of Two Populations Hypothesis Test for Proportions of a Multinomial Population H o : p 1 - p 2 = 0 H a : p 1 - p 2 = 0

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

Expected Value Standard Deviation Sampling Distribution of where: n 1 = size of sample taken from population 1 n 2 = size of sample taken from population 2 n 2 = size of sample taken from population 2

If the sample sizes are large, the sampling distribution If the sample sizes are large, the sampling distribution of can be approximated by a normal probability of can be approximated by a normal probability distribution. distribution. If the sample sizes are large, the sampling distribution If the sample sizes are large, the sampling distribution of can be approximated by a normal probability of can be approximated by a normal probability distribution. distribution. The sample sizes are sufficiently large if all of these The sample sizes are sufficiently large if all of these conditions are met: conditions are met: The sample sizes are sufficiently large if all of these The sample sizes are sufficiently large if all of these conditions are met: conditions are met: n1p1 > 5n1p1 > 5n1p1 > 5n1p1 > 5 n 1 (1 - p 1 ) > 5 n2p2 > 5n2p2 > 5n2p2 > 5n2p2 > 5 n 2 (1 - p 2 ) > 5 Sampling Distribution of

p 1 – p 2

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

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.

Example: MRA A survey conducted immediately after the new 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 Does the data support the position that the advertising campaign has provided an increased awareness of the client’s product?

Point Estimator of the Difference Between the Proportions of Two Populations = sample proportion of households “aware” of the = sample proportion of households “aware” of the product after the new campaign product after the new campaign = sample proportion of households “aware” of the = sample proportion of households “aware” of the product before the new campaign product before the new campaign p 1 = proportion of the population of households p 1 = proportion of the population of households “aware” of the product after the new campaign “aware” of the product after the new campaign p 2 = proportion of the population of households p 2 = proportion of the population of households “aware” of the product before the new campaign “aware” of the product before the new campaign

.08 + 1.96(.0510).08 +.10 Interval Estimate of p 1 - p 2 : Large-Sample Case Hence, the 95% confidence interval for the difference Hence, the 95% confidence interval for the difference in before and after awareness of the product is -.02 to +.18. For  =.05, z.025 = 1.96:

Using Excel to Develop an Interval Estimate of p 1 – p 2 n Formula Worksheet Note: Rows 16-251 are not shown.

n Value Worksheet Using Excel to Develop an Interval Estimate of p 1 – p 2 Note: Rows 16-251 are not shown.

Hypothesis Tests about p 1 - p 2 Hypotheses Test Statistic H 0 : p 1 - p 2 < 0 H a : p 1 - p 2 > 0

Hypothesis Tests about p 1 - p 2 n Point Estimator of where p 1 = p 2 where:

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? Hypothesis Tests about p 1 - p 2

p 1 = proportion of the population of households p 1 = proportion of the population of households “aware” of the product after the new campaign “aware” of the product after the new campaign p 2 = proportion of the population of households p 2 = proportion of the population of households “aware” of the product before the new campaign “aware” of the product before the new campaign n Hypotheses H 0 : p 1 - p 2 < 0 H a : p 1 - p 2 > 0

Rejection Rule Test Statistic Hypothesis Tests about p 1 - p 2 Reject H 0 if z > 1.645

n Conclusion Hypothesis Tests about p 1 - p 2 Using  =.05, we cannot conclude that the Using  =.05, we cannot conclude that the proportion of households aware of the client’s proportion of households aware of the client’s product increased after the new campaign. product increased after the new campaign. z = 1.56 < 1.645. Do not reject H 0.

Using Excel to Conduct a Hypothesis Test about p 1 – p 2 n Formula Worksheet Note: Rows 17-251 are not shown.

n Value Worksheet Using Excel to Conduct a Hypothesis Test about p 1 – p 2 Note: Rows 17-251 are not shown.

Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population 1. Set up the null and alternative hypotheses. 2. Select a random sample and record the observed frequency, f i, for each of the k categories. frequency, f i, for each of the k categories. 3. Assuming H 0 is true, compute the expected frequency, e i, in each category by multiplying the frequency, e i, in each category by multiplying the category probability by the sample size. category probability by the sample size.

Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population 5. Reject H 0 if (where  is the significance level and there are k - 1 degrees of freedom). and there are k - 1 degrees of freedom). 4. Compute the value of the test statistic.

Example: Finger Lakes Homes (A) Multinomial Distribution Goodness of Fit Test Finger Lakes Homes manufactures four models of prefabricated homes, a two-story colonial, a log cabin, a split-level, and an A-frame. To help in production planning, management would like to determine if previous customer purchases indicate that there is a preference in the style selected.

Example: Finger Lakes Homes (A) n Multinomial Distribution Goodness of Fit Test The number of homes sold of each model for 100 sales over the past two model for 100 sales over the past two years is shown below. Split- A- Split- A- Model Colonial Log Level Frame # Sold 30 20 35 15

n Hypotheses Multinomial Distribution Goodness of Fit Test where: p C = population proportion that purchase a colonial p C = population proportion that purchase a colonial p L = population proportion that purchase a log cabin p L = population proportion that purchase a log cabin p S = population proportion that purchase a split-level p S = population proportion that purchase a split-level p A = population proportion that purchase an A-frame p A = population proportion that purchase an A-frame H 0 : p C = p L = p S = p A =.25 H a : The population proportions are not p C =.25, p L =.25, p S =.25, and p A =.25 p C =.25, p L =.25, p S =.25, and p A =.25

n Rejection Rule 22 22 7.815 Do Not Reject H 0 Reject H 0 Multinomial Distribution Goodness of Fit Test With  =.05 and k - 1 = 4 - 1 = 3 k - 1 = 4 - 1 = 3 degrees of freedom degrees of freedom if  2 > 7.815. Reject H 0 if  2 > 7.815.

Expected Frequencies Test Statistic Multinomial Distribution Goodness of Fit Test e 1 =.25(100) = 25 e 2 =.25(100) = 25 e 3 =.25(100) = 25 e 4 =.25(100) = 25 e 3 =.25(100) = 25 e 4 =.25(100) = 25 = 1 + 1 + 4 + 4 = 10

Conclusion Multinomial Distribution Goodness of Fit Test We reject, at the.05 level of significance, We reject, at the.05 level of significance, the assumption that there is no home style preference.  2 = 10 > 7.815

n Worksheet (showing data) Using Excel to Conduct a Goodness of Fit Test Note: Rows 13-101 are not shown.

Using Excel to Conduct a Goodness of Fit Test n Formula Worksheet Note: Columns A-B and rows 13-101 are not shown.

n Value Worksheet Using Excel to Conduct a Goodness of Fit Test Note: Columns A-B and rows 13-101 are not shown.

Download ppt "Chapter 11a: Comparisons Involving Proportions and a Test of Independence Inference about the Difference between the Proportions of Two Populations Hypothesis."

Similar presentations