1 chapter 11 comparisons involving proportions and a test of independence inference about the...

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Chapter 11 Comparisons Involving Proportions

and a Test of Independence

Inference about the Difference Between the Proportions of Two PopulationsA Hypothesis Test for Proportions of a Multinomial PopulationTest of Independence: Contingency Tables

HH oo: : pp 11 - - pp 22 = 0 = 0

HH aa: : pp 11 - - pp 22 = 0 = 0

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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

where: n1 = size of sample taken from population 1

n2 = size of sample taken from population 2

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 ( ) ( )

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Sampling Distribution of Sampling Distribution of

n 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.

p p1 2p p1 2

p p1 2p p1 2

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Sampling Distribution of Sampling Distribution of p p1 2p p1 2

pp11 – – pp22pp11 – – pp22

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

(Confidence) 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.

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Example: MRAExample: MRA

A survey conducted immediately after the A survey conducted immediately after the new campaign showed 120 of 250 households new campaign showed 120 of 250 households “aware” of the client’s product.“aware” of the client’s product.

Does the data support the position that Does the data support the position that the advertising campaign has provided an the advertising campaign has provided an increased awareness of the client’s product?increased awareness of the client’s product?

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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

08.40.48.150

60

250

12021 pp 08.40.48.

150

60

250

12021 pp

p1p1

p2p2

Example: MRA

21 pp

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Interval Estimate of p1 - p2: Large-Sample Case

For = .05, z.025 = _______:

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

-.02 to +.18

. . .. (. ) . (. )

48 40 1 9648 52

25040 60

150 . . .

. (. ) . (. )48 40 1 96

48 52250

40 60150

Example: MRA

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

• Conclusion

At a 95% confidence level, the interval At a 95% confidence level, the interval estimate estimate of the difference between the of the difference between the proportion of proportion of households aware of the households aware of the client’s product before client’s product before and after the new and after the new advertising campaign is ___ to advertising campaign is ___ to _____._____.


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Hypotheses

H0: p1 - p2 < 0

Ha: p1 - p2 > 0

Test statistic

where takes the _______ part of the value in H0.

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)()( 2121

pp

ppppz

)( 21 pp

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Hypothesis Tests about Hypothesis Tests about pp11 - - pp22

n Point Estimator of where Point Estimator of where pp11 = = pp22

where:where:

)11)(1( 2121nnpps pp

21 pp

21

2211

nn

pnpnp

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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

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n Hypothesis Tests about p1 - p2

• Hypotheses

HH00: : pp1 1 - - pp22 << 0 0

HHaa: : pp1 1 - - pp22 > 0 > 0

pp11 = proportion of the population of households = proportion of the population of households

“ “aware” of the product ______ the new aware” of the product ______ the new campaigncampaign

pp22 = proportion of the population of households = proportion of the population of households

“ “aware” of the product ______ the new aware” of the product ______ the new campaigncampaign


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– Rejection Rule Reject H0 if z > 1.645

– Test Statistic

p

250 48 150 40250 150

180400

45(. ) (. )

.p

250 48 150 40250 150

180400

45(. ) (. )

.

_____)1501

2501)(55(.45.

21 pps _____)150

1250

1)(55(.45.21

pps

_____0514.

08.

0514.

0)40.48(.

z _____

0514.

08.

0514.

0)40.48(.

z

Example: MRA

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n Hypothesis Tests about Hypothesis Tests about pp11 - - pp22

• ConclusionConclusion

zz = ______ < 1.645. = ______ < 1.645. Do not reject Do not reject HH00. We . We cannotcannot conclude, with at least 95 % conclude, with at least 95 % confidence, that the proportion of confidence, that the proportion of households aware of the client’s product households aware of the client’s product increased after the new advertising increased after the new advertising campaign.campaign.


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Hypothesis (Goodness of Fit) Testfor Proportions of a Multinomial

Population1. Set up the null and alternative hypotheses.2. Select a random sample and record the

observedfrequency, 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.

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4.4. Compute the value of the test statistic. Compute the value of the test statistic.

5.5. Reject Reject HH00 if if (where (where is the is the significance level and there are significance level and there are kk - 1 degrees - 1 degrees of freedom). of freedom).

Hypothesis (Goodness of Fit) TestHypothesis (Goodness of Fit) Testfor Proportions of a Multinomial for Proportions of a Multinomial

PopulationPopulation

2 2 2 2

k

i i

ii

e

ef

1

22 )(

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Example: Finger Lakes Homes (A)

Multinomial Distribution Goodness of Fit TestFinger Lakes Homes manufactures four models of

prefabricated homes, a two-story colonial, a ranch, a

split-level, and an A-frame. To help in productionplanning, management would like to determine ifprevious customer purchases indicate that there

is apreference in the style selected.

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Example: Finger Lakes Homes (A)Example: Finger Lakes Homes (A)

n Multinomial Distribution Goodness of Fit Test

The number of homes sold of each model for The number of homes sold of each model for ____________

sales over the past two years is shown below.sales over the past two years is shown below.

Model Colonial Ranch Split-Level A-Model Colonial Ranch Split-Level A-FrameFrame

# Sold# Sold 30 20 35 30 20 35 15 15

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Multinomial Distribution Goodness of Fit TestLet:pC = population proportion that purchase a colonialpR = population proportion that purchase a ranchpS = population proportion that purchase a split-levelpA = population proportion that purchase an A-frame

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Example: Finger Lakes Homes (A)Example: Finger Lakes Homes (A)


• Hypotheses

HH00: : ppCC = = ppRR = = ppSS = = ppAA = .25 = .25

HHaa: The population proportions are not : The population proportions are not ppCC = .25,= .25,

ppRR = .25, = .25, ppSS = .25, and = .25, and ppAA = .25 = .25

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• Rejection Rule

With With = .05 = .05 andand

kk - 1 = 4 - 1 = 3 - 1 = 4 - 1 = 3 degrees ofdegrees of

freedomfreedom


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7.815 7.815

Do Not Reject H0Do Not Reject H0 Reject H0Reject H0

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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

22 2 2 230 25

2520 25

2535 25

2515 25

25 ( ) ( ) ( ) ( )2

2 2 2 230 2525

20 2525

35 2525

15 2525

( ) ( ) ( ) ( )

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Multinomial Distribution Goodness of Fit Test– Conclusion

2 = 10 > 7.815. We reject the assumption there is no home style preference, at the .05 level of significance.


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Test of Independence: Contingency Tables

1. Set up the null and alternative hypotheses.2. Select a random sample and record the

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

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

Total) umn Total)(Col Row( jieij

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4.4. 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).

Test of Independence: Contingency Tables

2 2 2 2

i j ij

ijij

e

ef 22 )(

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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.

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n Contingency Table (Independence) TestThe 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)Example: Finger Lakes Homes (B)

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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


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n 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


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Contingency Table (Independence) Test– Rejection Rule

With = .05 and (2 - 1)(4 - 1) = 3 d.f., Reject H0 if 2 > 7.81


. .052 7 81. .052 7 81

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• Test Statistic

= .1364 + 2.2727 + . . . + 2.0833 = = .1364 + 2.2727 + . . . + 2.0833 = ____________

22 2 218 16 5

16 56 11

113 6 75

6 75 ( . )

.( )

. .( . )

. . 2

2 2 218 16 516 5

6 1111

3 6 756 75

( . ).

( ). .

( . ).

.


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• Conclusion

22 = ____ > 7.81, so we reject = ____ > 7.81, so we reject HH00, the , the assumption assumption that the price of the home is that the price of the home is independent of the independent of the style of the home that style of the home that is purchased.is purchased.


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End of Chapter 11

1 chapter 11 comparisons involving proportions and a test of independence inference about the...

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