1 Chapter 11 Comparisons Involving Proportions and A Test of Independence.
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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 _____._____.
Example: MRAExample: MRA
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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.
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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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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
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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
Example: MRAExample: MRA
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Hypothesis Tests about p1 - p2
– 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.
Example: MRAExample: MRA
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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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Example: Finger Lakes Homes (A)
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)
n Multinomial Distribution Goodness of Fit Test
• 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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n Multinomial Distribution Goodness of Fit Test
• Rejection Rule
With With = .05 = .05 andand
kk - 1 = 4 - 1 = 3 - 1 = 4 - 1 = 3 degrees ofdegrees of
freedomfreedom
Example: Finger Lakes Homes (A)
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7.815 7.815
Do Not Reject H0Do Not Reject H0 Reject H0Reject H0
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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
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.
Example: Finger Lakes Homes (A)
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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
Example: Finger Lakes Homes (B)
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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
Example: Finger Lakes Homes (B)Example: Finger Lakes Homes (B)
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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
Example: Finger Lakes Homes (B)
. .052 7 81. .052 7 81
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n Contingency Table (Independence) Test
• 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
( . ).
( ). .
( . ).
.
Example: Finger Lakes Homes (B)Example: Finger Lakes Homes (B)
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n Contingency Table (Independence) Test
• 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.
Example: Finger Lakes Homes (B)Example: Finger Lakes Homes (B)
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End of Chapter 11