Chapter 10 Statistical Inference About Means and ...dscsss/teaching/mgs9920/slides/ch10.pdf · 1...
Transcript of Chapter 10 Statistical Inference About Means and ...dscsss/teaching/mgs9920/slides/ch10.pdf · 1...
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1Slide
Chapter 10Statistical Inference About Means and
Proportions With Two Populations
2Slide
Learning objectives
1. Understand inferences About the Difference Between Two Population Means: σ 1 and σ 2 Known
4. Understand Inferences About the Difference BetweenTwo Population Proportions
3. Understand Inferences About the Difference BetweenTwo Population Means: Matched Samples
2. Understand Inferences About the Difference BetweenTwo Population Means: σ 1 and σ 2 Unknown
2
3Slide
Overview
Tests for two population
Two means(Point estimator = )
Two proportions(Point estimator = )
σ1 and σ2 are known
σ1 and σ2 are unknown
Matched (or paired) sample
1 2x x− 1 2p p−
Standard error will be different
for each case.
4Slide
Comparison between one population case and two population case.
nProcedures of getting confidence interval and of testing for two population case is very similar as those of one population case
nConfidence interval:
nTest Statistics
nPoint estimators and standard errors of two population case are different from those of one population case.
Point estimator Significance coefficient * Standard Error±
Point estimator - Hypothesized value of parameterStandard error
3
5Slide
Estimating the Difference BetweenTwo Population Means
n Let µ1 equal the mean of population 1 and µ2 equalthe mean of population 2.
n The difference between the two population means isµ1 - µ2.
n Let equal the mean of sample 1 and equal themean of sample 2.
x1x1 x2x2
n The point estimator of the difference between themeans of the populations 1 and 2 is .x x1 2−x x1 2−
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
6Slide
n Expected Value
Sampling Distribution of : σ knownx x1 2−x x1 2−
E x x( )1 2 1 2− = −µ µE x x( )1 2 1 2− = −µ µ
n Standard Deviation (Standard Error)
σ σ σx x n n1 2
12
1
22
2− = +σ σ σ
x x n n1 212
1
22
2− = +
where: σ1 = standard deviation of population 1σ2 = standard deviation of population 2n1 = sample size from population 1n2 = sample size from population 2
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
4
7Slide
n Interval Estimate
Interval Estimation of µ1 - µ2:σ 1 and σ 2 Known
2 21 2
1 2 /21 2
x x zn nασ σ
− ± +2 21 2
1 2 /21 2
x x zn nασ σ
− ± +
where:1 - α is the confidence coefficient
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
8Slide
n Example: Par, Inc.
Interval Estimation of µ1 - µ2:σ 1 and σ 2 Known
In a test of driving distance using a mechanicaldriving device, a sample of Par golf balls wascompared with a sample of golf balls made by Rap,Ltd., a competitor. The sample statistics appear on thenext slide.
Par, Inc. is a manufacturerof golf equipment and hasdeveloped a new golf ballthat has been designed toprovide “extra distance.”
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
5
9Slide
n Example: Par, Inc.
Interval Estimation of µ1 - µ2:σ 1 and σ 2 Known
Sample SizeSample Mean
Sample #1Par, Inc.
Sample #2Rap, Ltd.
120 balls 80 balls275 yards 258 yards
Based on data from previous driving distancetests, the two population standard deviations areknown with σ 1 = 15 yards and σ 2 = 20 yards.
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
10Slide
Interval Estimation of µ1 - µ2:σ 1 and σ 2 Known
Let us develop a 95% confidence interval estimateof the difference between the mean driving distances ofthe two brands of golf ball.
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
Point estimate of µ1 − µ2 = x x1 2−x x1 2−
where:µ1 = mean distance for the population
of Par, Inc. golf ballsµ2 = mean distance for the population
of Rap, Ltd. golf balls
= 275 − 258
= 17 yards
6
11Slide
x x zn n1 2 2
12
1
22
2
2 217 1 96 15
1202080
− ± + = ± +ασ σ
/ . ( ) ( )x x zn n1 2 2
12
1
22
2
2 217 1 96 15
1202080
− ± + = ± +ασ σ
/ . ( ) ( )
Interval Estimation of µ1 - µ2:σ 1 and σ 2 Known
We are 95% confident that the difference betweenthe mean driving distances of Par, Inc. balls and Rap,Ltd. balls is 11.86 to 22.14 yards.
17 + 5.14 or 11.86 yards to 22.14 yards
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
12Slide
Hypothesis Tests About µ 1 − µ 2:σ 1 and σ 2 Known
n Hypotheses
1 2 02 21 2
1 2
( )x x Dz
n nσ σ
− −=
+
1 2 02 21 2
1 2
( )x x Dz
n nσ σ
− −=
+
µ µ− ≠1 2 0: aH Dµ µ− ≠1 2 0: aH Dµ µ− =0 1 2 0: H Dµ µ− =0 1 2 0: H Dµ µ− ≤0 1 2 0: H Dµ µ− ≤0 1 2 0: H D
µ µ− >1 2 0: aH Dµ µ− >1 2 0: aH Dµ µ− ≥0 1 2 0: H Dµ µ− ≥0 1 2 0: H Dµ µ− <1 2 0: aH Dµ µ− <1 2 0: aH D
Left-tailed Right-tailed Two-tailedn Test Statistic
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
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13Slide
n Example: Par, Inc.
Hypothesis Tests About µ 1 − µ 2:σ 1 and σ 2 Known
Can we conclude, usingα = .01, that the mean drivingdistance of Par, Inc. golf ballsis greater than the mean drivingdistance of Rap, Ltd. golf balls?
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
14Slide
H0: µ1 - µ2 < 0 Ha: µ1 - µ2 > 0
where: µ1 = mean distance for the population
of Par, Inc. golf ballsµ2 = mean distance for the population
of Rap, Ltd. golf balls
1. Develop the hypotheses.
n p –Value and Critical Value Approaches
Hypothesis Tests About µ 1 − µ 2:σ 1 and σ 2 Known
2. Specify the level of significance. α = .01
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
8
15Slide
3. Compute the value of the test statistic.
Hypothesis Tests About µ 1 − µ 2:σ 1 and σ 2 Known
n p –Value and Critical Value Approaches
σ σ
− −=
+
1 2 02 21 2
1 2
( )x x Dz
n nσ σ
− −=
+
1 2 02 21 2
1 2
( )x x Dz
n n
2 2
(235 218) 0 17 6.492.62(15) (20)
120 80
z − −= = =
+2 2
(235 218) 0 17 6.492.62(15) (20)
120 80
z − −= = =
+
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
16Slide
n p –Value Approach
4. Compute the p–value.
For z = 6.49, the p –value < .0001.
Hypothesis Tests About µ 1 − µ 2:σ 1 and σ 2 Known
5. Determine whether to reject H0.
Because p–value < α = .01, we reject H0.
At the .01 level of significance, the sample evidenceindicates the mean driving distance of Par, Inc. golfballs is greater than the mean driving distance of Rap,Ltd. golf balls.
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
9
17Slide
Hypothesis Tests About µ 1 − µ 2:σ 1 and σ 2 Known
5. Determine whether to reject H0.
Because z = 6.49 > 2.33, we reject H0.
n Critical Value Approach
For α = .01, z.01 = 2.33
4. Determine the critical value and rejection rule.
Reject H0 if z > 2.33
The sample evidence indicates the mean drivingdistance of Par, Inc. golf balls is greater than the meandriving distance of Rap, Ltd. golf balls.
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
18Slide
In-class exercise
n #2 (p.400) (confidence interval)n #3 (p.400) (hypothesis testing)
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
10
19Slide
Interval Estimation of µ1 - µ2:σ 1 and σ 2 Unknown
When σ 1 and σ 2 are unknown, we will:
• replace zα/2 with tα/2.
• use the sample standard deviations s1 and s2
as estimates of σ 1 and σ 2 , and
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
20Slide
2 21 2
1 2 / 21 2
s sx x tn nα− ± +
2 21 2
1 2 / 21 2
s sx x tn nα− ± +
Where the degrees of freedom for tα/2 are:
Interval Estimation of µ1 - µ2:σ 1 and σ 2 Unknown
n Interval Estimate
22 21 2
1 22 22 2
1 2
1 1 2 2
1 11 1
s sn n
dfs s
n n n n
+
=
+ − −
22 21 2
1 22 22 2
1 2
1 1 2 2
1 11 1
s sn n
dfs s
n n n n
+
=
+ − −
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
11
21Slide
n Example: Specific Motors
Difference Between Two Population Means:σ 1 and σ 2 Unknown
Specific Motors of Detroithas developed a new automobileknown as the M car. 24 M carsand 28 J cars (from Japan) were roadtested to compare miles-per-gallon (mpg) performance. The sample statistics are shown on the next slide.
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
22Slide
Difference Between Two Population Means:σ 1 and σ 2 Unknown
n Example: Specific Motors
Sample SizeSample MeanSample Std. Dev.
Sample #1M Cars
Sample #2J Cars
24 cars 28 cars29.8 mpg 27.3 mpg2.56 mpg 1.81 mpg
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
12
23Slide
Difference Between Two Population Means:σ 1 and σ 2 Unknown
Let us develop a 90% confidenceinterval estimate of the differencebetween the mpg performances ofthe two models of automobile.
n Example: Specific Motors
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
24Slide
Point estimate of µ1 − µ2 = x x1 2−x x1 2−
Point Estimate of µ 1 − µ 2
where:µ1 = mean miles-per-gallon for the
population of M carsµ2 = mean miles-per-gallon for the
population of J cars
= 29.8 - 27.3
= 2.5 mpg
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
13
25Slide
Interval Estimation of µ 1 − µ 2:σ 1 and σ 2 Unknown
The degrees of freedom for tα/2 are:22 2
2 22 2
(2.56) (1.81)24 28
24.07 241 (2.56) 1 (1.81)
24 1 24 28 1 28
df
+
= = =
+ − −
22 2
2 22 2
(2.56) (1.81)24 28
24.07 241 (2.56) 1 (1.81)
24 1 24 28 1 28
df
+
= = =
+ − −
With α/2 = .05 and df = 24, tα/2 = 1.711
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
26Slide
Interval Estimation of µ 1 − µ 2:σ 1 and σ 2 Unknown
2 2 2 21 2
1 2 /21 2
(2.56) (1.81) 29.8 27.3 1.71124 28
s sx x tn nα− ± + = − ± +
2 2 2 21 2
1 2 /21 2
(2.56) (1.81) 29.8 27.3 1.71124 28
s sx x tn nα− ± + = − ± +
We are 90% confident that the difference betweenthe miles-per-gallon performances of M cars and J carsis 1.431 to 3.569 mpg.
2.5 + 1.069 or 1.431 to 3.569 mpg
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
14
27Slide
Hypothesis Tests About µ 1 − µ 2:σ 1 and σ 2 Unknown
n Hypotheses
1 2 02 21 2
1 2
( )x x Dts sn n
− −=
+
1 2 02 21 2
1 2
( )x x Dts sn n
− −=
+
µ µ− ≠1 2 0: aH Dµ µ− ≠1 2 0: aH Dµ µ− =0 1 2 0: H Dµ µ− =0 1 2 0: H Dµ µ− ≤0 1 2 0: H Dµ µ− ≤0 1 2 0: H D
µ µ− >1 2 0: aH Dµ µ− >1 2 0: aH Dµ µ− ≥0 1 2 0: H Dµ µ− ≥0 1 2 0: H Dµ µ− <1 2 0: aH Dµ µ− <1 2 0: aH D
Left-tailed Right-tailed Two-tailedn Test Statistic
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
28Slide
n Example: Specific Motors
Hypothesis Tests About µ 1 − µ 2:σ 1 and σ 2 Unknown
Can we conclude, using a.05 level of significance, that themiles-per-gallon (mpg) performanceof M cars is greater than the miles-per-gallon performance of J cars?
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
15
29Slide
H0: µ1 - µ2 < 0 Ha: µ1 - µ2 > 0where:
µ1 = mean mpg for the population of M carsµ2 = mean mpg for the population of J cars
1. Develop the hypotheses.
n p –Value and Critical Value Approaches
Hypothesis Tests About µ 1 − µ 2:σ 1 and σ 2 Unknown
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
30Slide
2. Specify the level of significance.
3. Compute the value of the test statistic.
α = .05
n p –Value and Critical Value Approaches
Hypothesis Tests About µ 1 − µ 2:σ 1 and σ 2 Unknown
1 2 02 2 2 21 2
1 2
( ) (29.8 27.3) 0 4.003(2.56) (1.81)
24 28
x x Dts sn n
− − − −= = =
+ +
1 2 02 2 2 21 2
1 2
( ) (29.8 27.3) 0 4.003(2.56) (1.81)
24 28
x x Dts sn n
− − − −= = =
+ +
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
16
31Slide
Hypothesis Tests About µ 1 − µ 2:σ 1 and σ 2 Unknown
n p –Value Approach
4. Compute the p –value.
The degrees of freedom for tα are:22 2
2 22 2
(2.56) (1.81)24 28
24.07 241 (2.56) 1 (1.81)
24 1 24 28 1 28
df
+
= = =
+ − −
22 2
2 22 2
(2.56) (1.81)24 28
24.07 241 (2.56) 1 (1.81)
24 1 24 28 1 28
df
+
= = =
+ − −
Because t = 4.003 > t.005 = 2.797, the p–value < .005.
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
32Slide
5. Determine whether to reject H0.
We are at least 95% confident that the miles-per-gallon (mpg) performance of M cars is greater than the miles-per-gallon performance of J cars?.
n p –Value Approach
Because p–value < α = .05, we reject H0.
Hypothesis Tests About µ 1 − µ 2:σ 1 and σ 2 Unknown
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
17
33Slide
4. Determine the critical value and rejection rule.
n Critical Value Approach
Hypothesis Tests About µ 1 − µ 2:σ 1 and σ 2 Unknown
For α = .05 and df = 24, t.05 = 1.711
Reject H0 if t > 1.711
5. Determine whether to reject H0.
Because 4.003 > 1.711, we reject H0.We are at least 95% confident that the miles-per-
gallon (mpg) performance of M cars is greater than the miles-per-gallon performance of J cars?.
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
34Slide
In-class exercise
n #9 (p.406) (confidence interval)n #10 (p.407) (hypothesis testing)
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
18
35Slide
n With a matched-sample design each sampled itemprovides a pair of data values.
n This design often leads to a smaller sampling errorthan the independent-sample design becausevariation between sampled items is eliminated as asource of sampling error.
Inferences About the Difference BetweenTwo Population Means: Matched Samples
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
36Slide
n Example: Express Deliveries
Inferences About the Difference BetweenTwo Population Means: Matched Samples
A Chicago-based firm hasdocuments that must be quicklydistributed to district officesthroughout the U.S. The firmmust decide between two deliveryservices, UPX (United Parcel Express) and INTEX(International Express), to transport its documents.
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
19
37Slide
n Example: Express Deliveries
Inferences About the Difference BetweenTwo Population Means: Matched Samples
In testing the delivery timesof the two services, the firm senttwo reports to a random sampleof its district offices with onereport carried by UPX and theother report carried by INTEX. Do the data on thenext slide indicate a difference in mean deliverytimes for the two services? Use a .05 level ofsignificance.
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
38Slide
3230191615181410
716
25241515131515
89
11
UPX INTEX DifferenceDistrict OfficeSeattleLos AngelesBostonClevelandNew YorkHoustonAtlantaSt. LouisMilwaukeeDenver
Delivery Time (Hours)
7 6 4 1 2 3
-1 2
-2 5
Inferences About the Difference BetweenTwo Population Means: Matched Samples
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
20
39Slide
H0: µd = 0 Ha: µd ≠ 0
Let µd = the mean of the difference values for thetwo delivery services for the populationof district offices
1. Develop the hypotheses.
Inferences About the Difference BetweenTwo Population Means: Matched Samples
n p –Value and Critical Value Approaches
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
40Slide
2. Specify the level of significance. α = .05
Inferences About the Difference BetweenTwo Population Means: Matched Samples
n p –Value and Critical Value Approaches
3. Compute the value of the test statistic.
d dn
i= ∑ =+ + +
=( ... ) .7 6 5
102 7d d
ni= ∑ =
+ + +=
( ... ) .7 6 510
2 7
s d dnd
i= −∑−
= =( ) . .2
176 1
92 9s d d
ndi= −∑−
= =( ) . .2
176 1
92 9
2.7 0 2.942.9 10
d
d
dts n
µ− −= = =2.7 0 2.94
2.9 10d
d
dts n
µ− −= = =
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
21
41Slide
5. Determine whether to reject H0.
We are at least 95% confident that there is a difference in mean delivery times for the two services?
4. Compute the p –value.
For t = 2.94 and df = 9, the p–value is between.02 and .01. (This is a two-tailed test, so we double the upper-tail areas of .01 and .005.)
Because p–value < α = .05, we reject H0.
Inferences About the Difference BetweenTwo Population Means: Matched Samples
n p –Value Approach
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
42Slide
4. Determine the critical value and rejection rule.
Inferences About the Difference BetweenTwo Population Means: Matched Samples
n Critical Value Approach
For α = .05 and df = 9, t.025 = 2.262.Reject H0 if t > 2.262
5. Determine whether to reject H0.
Because t = 2.94 > 2.262, we reject H0.
We are at least 95% confident that there is a difference in mean delivery times for the two services?
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
22
43Slide
In-class exercise
n #21 (p.414) (hypothesis testing)n #22 (p.414) (confidence interval)
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
44Slide
n Expected Value
Sampling Distribution of p p1 2−
E p p p p( )1 2 1 2− = −E p p p p( )1 2 1 2− = −
σ p pp p
np p
n1 21 1
1
2 2
2
1 1− =
−+
−( ) ( )σ p pp p
np p
n1 21 1
1
2 2
2
1 1− =
−+
−( ) ( )
where: n1 = size of sample taken from population 1n2 = size of sample taken from population 2
n Standard Deviation (Standard Error)
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
23
45Slide
If the sample sizes are large, the sampling distributionof can be approximated by a normal probabilitydistribution.
p p1 2−p p1 2−
The sample sizes are sufficiently large if all of theseconditions are met:
n1p1 > 5 n1(1 - p1) > 5
n2p2 > 5 n2(1 - p2) > 5
Sampling Distribution of p p1 2−•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
46Slide
Sampling Distribution of p p1 2−
p1 – p2
σ p pp p
np p
n1 21 1
1
2 2
2
1 1− = − + −( ) ( )σ p p
p pn
p pn1 2
1 1
1
2 2
2
1 1− = − + −( ) ( )
p p1 2−p p1 2−
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
24
47Slide
Interval Estimation of p1 - p2
n Interval Estimate
1 1 2 21 2 /2
1 2
(1 ) (1 )p p p pp p zn nα
− −− ± +1 1 2 2
1 2 /21 2
(1 ) (1 )p p p pp p zn nα
− −− ± +
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
48Slide
Market Research Associates isconducting research to evaluate theeffectiveness of a client’s new adver-tising campaign. Before the newcampaign began, a telephone surveyof 150 households in the test marketarea showed 60 households “aware” ofthe client’s product.
Interval Estimation of p1 - p2
n Example: Market Research Associates
The new campaign has been initiated with TV andnewspaper advertisements running for three weeks.
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
25
49Slide
A survey conducted immediatelyafter the new campaign showed 120of 250 households “aware” of theclient’s product.
Interval Estimation of p1 - p2
n Example: Market Research Associates
Does the data support the positionthat the advertising campaign has provided an increased awareness ofthe client’s product?
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
50Slide
Point Estimator of the Difference BetweenTwo Population Proportions
= sample proportion of households “aware” of theproduct after the new campaign
= sample proportion of households “aware” of theproduct before the new campaign
1p1p
2p2p
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
1 2120 60 .48 .40 .08250 150
p p− = − = − =1 2120 60 .48 .40 .08250 150
p p− = − = − =
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
26
51Slide
.08 + 1.96(.0510).08 + .10
.48(.52) .40(.60).48 .40 1.96250 150
− ± +.48(.52) .40(.60).48 .40 1.96
250 150− ± +
Interval Estimation of p1 - p2
Hence, the 95% confidence interval for the differencein before and after awareness of the product is-.02 to +.18.
For α = .05, z.025 = 1.96:
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
52Slide
Hypothesis Tests about p1 - p2
n Hypotheses
H0: p1 - p2 < 0Ha: p1 - p2 > 0 − ≠1 2: 0aH p p− ≠1 2: 0aH p p
− =0 1 2: 0H p p− =0 1 2: 0H p p− ≤0 1 2: 0H p p− ≤0 1 2: 0H p p− >1 2: 0aH p p− >1 2: 0aH p p
− ≥0 1 2: 0H p p− ≥0 1 2: 0H p p− <1 2: 0aH p p− <1 2: 0aH p p
Left-tailed Right-tailed Two-tailed
We focus on tests involving no difference betweenthe two population proportions (i.e. p1 = p2)
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
27
53Slide
Hypothesis Tests about p1 - p2
1 2p p−1 2p p−n Pooled Estimate of Standard Error of
1 21 2
1 1(1 )p p p pn n
σ −
= − +
1 2
1 2
1 1(1 )p p p pn n
σ −
= − +
1 1 2 2
1 2
n p n ppn n
+=
+1 1 2 2
1 2
n p n ppn n
+=
+
where:
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
54Slide
Hypothesis Tests about p1 - p2
−=
− +
1 2
1 2
( )1 1(1 )
p pzp p
n n
−=
− +
1 2
1 2
( )1 1(1 )
p pzp p
n n
n Test Statistic
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
28
55Slide
Can we conclude, using a .05 levelof significance, that the proportion ofhouseholds aware of the client’s productincreased after the new advertisingcampaign?
Hypothesis Tests about p1 - p2
n Example: Market Research Associates
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
56Slide
Hypothesis Tests about p1 - p2
1. Develop the hypotheses.
n p -Value and Critical Value Approaches
H0: p1 - p2 < 0Ha: p1 - p2 > 0
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
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
29
57Slide
Hypothesis Tests about p1 - p2
2. Specify the level of significance. α = .05
3. Compute the value of the test statistic.
n p -Value and Critical Value Approaches
p =++
= =250 48 150 40
250 150180400
45(. ) (. ) .p =++
= =250 48 150 40
250 150180400
45(. ) (. ) .
sp p1 245 55 1
2501150 0514− = + =. (. )( ) .sp p1 2
45 55 1250
1150 0514− = + =. (. )( ) .
(.48 .40) 0 .08 1.56.0514 .0514
z − −= = =
(.48 .40) 0 .08 1.56.0514 .0514
z − −= = =
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
58Slide
Hypothesis Tests about p1 - p2
5. Determine whether to reject H0.
We cannot conclude that the proportion of householdsaware of the client’s product increased after the newcampaign.
4. Compute the p –value.
For z = 1.56, the p–value = .0594
Because p–value > α = .05, we cannot reject H0.
n p –Value Approach
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
30
59Slide
Hypothesis Tests about p1 - p2
n Critical Value Approach
5. Determine whether to reject H0.
Because 1.56 < 1.645, we cannot reject H0.
For α = .05, z.05 = 1.645
4. Determine the critical value and rejection rule.
Reject H0 if z > 1.645
We cannot conclude that the proportion of householdsaware of the client’s product increased after the newcampaign.
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
60Slide
In-class exercise
n #30 (p.420) (confidence interval) n #31 (p.421) (hypothesis testing)
•L.O.•Test for means
−σ known−σ unknown−Matched sample
•Test for proportions
31
61Slide
End of Chapter 10
62Slide
Proportion
Same as one population caseMeans:matched
Means:σ unknown
Means:σ known
Otherformula
Test Statistics
ConfidenceInterval
2 21 2
1 2 /21 2
x x zn nασ σ
− ± +1 2 0
2 21 2
1 2
( )x x Dz
n nσ σ
− −=
+
2 21 2
1 2 /21 2
s sx x tn nα− ± +
1 2 02 2
1 2
1 2
( )x x Dts sn n
− −=
+
22 21 2
1 22 22 2
1 2
1 1 2 2
1 11 1
s sn n
dfs s
n n n n
+
=
+ − −
1 1 2 21 2 /2
1 2
(1 ) (1 )p p p pp p z
n nα
− −− ± +
−=
− +
1 2
1 2
( )
1 1(1 )
p pzp p
n n
1 1 2 2
1 2
n p n ppn n
+=
+
Formulas for two population case