Chapter 11 Comparisons Involving Proportions and a Test of Independence

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

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Ho: p1 - p2 = 0

Ha: p1 - p2 = 0

• Inference about the Difference Between the

Proportions of Two Populations

• A Hypothesis Test for Proportions of a

Multinomial Population

• Test of Independence: Contingency Tables
Inferences About the Difference between the Proportions of Two Populations
• Sampling Distribution of
• Interval Estimation of p1 - p2
• Hypothesis Tests about p1 - p2
Sampling Distribution of
• Expected Value
• Standard Deviation

where: n1 = size of sample taken from population 1

n2 = size of sample taken from population 2

Sampling Distribution of

• 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 can be approximated

by a normal probability distribution.

Interval Estimation of p1 - p2
• (Confidence) Interval Estimate
• 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 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?

Example: MRA
• Point Estimator of the Difference Between the Proportions of Two Populations is

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

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

For = .05, z.025 = _______:

.08 + 1.96(.0510)

.08 + .10

-.02 to +.18

Example: MRA

• Interval Estimate of p1 - p2: Large-Sample Case
• 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 ___ to _____.

Hypothesis Tests about p1 - p2
• Hypotheses

H0: p1 - p2< 0

Ha: p1 - p2 > 0

• Test statistic

where takes the _______ part of the value in H0.

Hypothesis Tests about p1 - p2

• Point Estimator of where p1 = p2

where:

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?

Example: MRA

• Hypothesis Tests about p1 - p2
• Hypotheses
• H0: p1 - p2< 0

Ha: p1 - p2 > 0

p1 = proportion of the population of households

“aware” of the product ______ the new campaign

p2 = proportion of the population of households

“aware” of the product ______ the new campaign

Example: MRA
• Hypothesis Tests about p1 - p2
• Rejection RuleReject H0 if z > 1.645
• Test Statistic

Example: MRA

• Hypothesis Tests about p1 - p2
• Conclusion
• z = ______ < 1.645.Do not reject H0. We cannot conclude, with at least 95 % confidence, that the proportion of households aware of the client’s product increased after the new advertising campaign.
Hypothesis (Goodness of Fit) Testfor Proportions of a Multinomial Population

1. Set up the null and alternative hypotheses.

2. Select a random sample and record the observed

frequency, fi , for each of the k categories.

3. Assuming H0 is true, compute the expected frequency,ei , in each category by multiplying the category probability by the sample size.

Hypothesis (Goodness of Fit) Testfor Proportions of a Multinomial Population

4. Compute the value of the test statistic.

5. Reject H0 if (where  is the significance level and there are k - 1 degrees of freedom).

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 ranch, 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)

• Multinomial Distribution Goodness of Fit Test

The number of homes sold of each model for ______

sales over the past two years is shown below.

Model Colonial Ranch Split-Level A-Frame

# Sold 30 20 35 15

Example: Finger Lakes Homes (A)
• Multinomial Distribution Goodness of Fit Test

Let:

pC = population proportion that purchase a colonial

pR = population proportion that purchase a ranch

pS = population proportion that purchase a split-level

pA = population proportion that purchase an A-frame

Example: Finger Lakes Homes (A)

• Multinomial Distribution Goodness of Fit Test
• Hypotheses

H0: pC = pR = pS = pA = .25

Ha: The population proportions are not pC = .25,

pR = .25, pS = .25, and pA = .25

Example: Finger Lakes Homes (A)

• Multinomial Distribution Goodness of Fit Test
• Rejection Rule

With  = .05 and

k - 1 = 4 - 1 = 3 degrees of

freedom

Do Not Reject H0

Reject H0

2

7.815

Example: Finger Lakes Homes (A)
• Multinomial Distribution Goodness of Fit Test
• Expected Frequencies

e1 = .25(100) = 25 e2 = .25(100) = 25

e3 = .25(100) = 25 e4 = .25(100) = 25

• Test Statistic

= 1 + 1 + 4 + 4

= 10

Example: Finger Lakes Homes (A)
• Multinomial Distribution Goodness of Fit Test
• Conclusion

c2 = 10 > 7.815. We reject the assumption there is no home style preference, at the .05 level of significance.

Test of Independence: Contingency Tables

1. Set up the null and alternative hypotheses.

2. Select a random sample and record the observed

frequency, fij , for each cell of the contingency table.

3. Compute the expected frequency, eij , for each cell.

Test of Independence: Contingency Tables

4.Compute the test statistic.

5. Reject H0 if (where  is the significance level and with n rows and m columns there are

(n - 1)(m - 1) degrees of freedom).

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.

Example: Finger Lakes Homes (B)

• Contingency Table (Independence) Test

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 \$99,000 or less or more than \$99,000.

Price Colonial Ranch Split-Level A-Frame

< \$99,000 18 6 19 12

> \$99,000 12 14 16 3

Example: Finger Lakes Homes (B)
• Contingency Table (Independence) Test
• Hypotheses

H0: Price of the home is independent of the style of the home that is purchased

Ha: Price of the home is not independent of the

style of the home that is purchased

Example: Finger Lakes Homes (B)

• Contingency Table (Independence) Test
• Expected Frequencies

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

Example: Finger Lakes Homes (B)
• Contingency Table (Independence) Test
• Rejection Rule

With  = .05 and (2 - 1)(4 - 1) = 3 d.f.,

Reject H0 if 2 > 7.81

Example: Finger Lakes Homes (B)

• Contingency Table (Independence) Test
• Test Statistic

= .1364 + 2.2727 + . . . + 2.0833 = ______

Example: Finger Lakes Homes (B)

• Contingency Table (Independence) Test
• Conclusion

2 = ____ > 7.81, so we reject H0, the assumption that the price of the home is independent of the style of the home that is purchased.