inference about two populations
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Inference about Two Populations. Chapter 13. 12.1 Introduction. Variety of techniques are presented whose objective is to compare two populations. We are interested in: The difference between two means. The ratio of two variances. The difference between two proportions. - PowerPoint PPT PresentationTRANSCRIPT
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Inference about Two Populations
Chapter 13
2
12.1 Introduction12.1 Introduction
• Variety of techniques are presented whose objective is to compare two populations.
• We are interested in:– The difference between two means.– The ratio of two variances.– The difference between two proportions.
3
• Two random samples are drawn from the two populations of interest.
• Because we compare two population means, we use the statistic .
13.2 Inference about the Difference between Two Means: Independent Samples
13.2 Inference about the Difference between Two Means: Independent Samples
21 xx
4
1. is normally distributed if the (original) population distributions are normal .
2. is approximately normally distributed if the (original) population is not normal, but the samples’ size is sufficiently large (greater than 30).
3. The expected value of is 1 - 2
4. The variance of is 12/n1 + 2
2/n2
21 xx
21 xx
The Sampling Distribution ofThe Sampling Distribution of21
xx
21xx
21xx
5
• If the sampling distribution of is normal or approximately normal we can write:
• Z can be used to build a test statistic or a confidence interval for 1 - 2
21
21
nn
)()xx(Z
21
21
nn
)()xx(Z
21xx
Making an inference about –Making an inference about –
6
21
21
nn
)()xx(Z
21
21
nn
)()xx(Z
• Practically, the “Z” statistic is hardly used, because the population variances are not known.
? ?
• Instead, we construct a t statistic using the sample “variances” (S1
2 and S22).
S22S1
2t
Making an inference about –Making an inference about –
7
• Two cases are considered when producing the t-statistic.– The two unknown population variances are equal.– The two unknown population variances are not equal.
Making an inference about –Making an inference about –
8
Inference about Inference about ––: Equal variances: Equal variances
2nns)1n(s)1n(
S21
2
22
2
112
p
2nn
s)1n(s)1n(S
21
2
22
2
112
p
Example: s12 = 25; s2
2 = 30; n1 = 10; n2 = 15. Then,
04347.2821510
)30)(115()25)(110(S2
p
• Calculate the pooled variance estimate by:
n2 = 15n1 = 10
21S
22S
The pooledvariance estimator
9
Inference about Inference about ––: Equal variances: Equal variances
2nns)1n(s)1n(
S21
2
22
2
112
p
2nn
s)1n(s)1n(S
21
2
22
2
112
p
Example: s12 = 25; s2
2 = 30; n1 = 10; n2 = 15. Then,
04347.2821510
)30)(115()25)(110(S2
p
• Calculate the pooled variance estimate by:
2pS
n2 = 15n1 = 10
21S
22S
The pooledVariance estimator
10
Inference about Inference about ––: Equal variances: Equal variances
• Construct the t-statistic as follows:
2nn.f.d
)n1
n1
(s
)()xx(t
21
21
2p
21
2nn.f.d
)n1
n1
(s
)()xx(t
21
21
2p
21
• Perform a hypothesis test H0: = 0 H1: > 0
or < 0 or 0
Build a confidence interval
level. confidence the is where
)n1
n1
(st)xx(21
2
p21
11
1
)(
1
)(
)/(d.f.
)(
)()(
2
22
22
1
21
21
22
221
21
2
22
1
21
21
n
ns
n
ns
nsns
n
s
n
s
xxt
1
)(
1
)(
)/(d.f.
)(
)()(
2
22
22
1
21
21
22
221
21
2
22
1
21
21
n
ns
n
ns
nsns
n
s
n
s
xxt
Inference about –: Unequal variancesInference about –: Unequal variances
12
Inference about –: Unequal variancesInference about –: Unequal variances
Conduct a hypothesis test as needed, or, build a confidence interval
level confidence the is where
n
s
n
s2txx
intervalConfidence
)2
22
1
21()21(
13
Which case to use:Equal variance or unequal variance?
Which case to use:Equal variance or unequal variance?
• Whenever there is insufficient evidence that the variances are unequal, it is preferable to perform the equal variances t-test.
• This is so, because for any two given samples
The number of degrees of freedom for the equal variances case
The number of degrees of freedom for the unequal variances case
14
15
• Example 13.1– Do people who eat high-fiber cereal for
breakfast consume, on average, fewer calories for lunch than people who do not eat high-fiber cereal for breakfast?
– A sample of 150 people was randomly drawn. Each person was identified as a consumer or a non-consumer of high-fiber cereal.
– For each person the number of calories consumed at lunch was recorded.
Example: Making an inference about –
Example: Making an inference about –
16
Consmers Non-cmrs568 705498 819589 706681 509540 613646 582636 601739 608539 787596 573607 428529 754637 741617 628633 537555 748
. .
. .
. .
. .
Consmers Non-cmrs568 705498 819589 706681 509540 613646 582636 601739 608539 787596 573607 428529 754637 741617 628633 537555 748
. .
. .
. .
. .
Solution: • The data are interval. • The parameter to be tested is the difference between two means. • The claim to be tested is: The mean caloric intake of consumers (1) is less than that of non-consumers (2).
Example: Making an inference about –
Example: Making an inference about –
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• The hypotheses are:
H0: (1 - 2) = 0H1: (1 - 2) < 0
– To check the whether the population variances are equal, we use (Xm13-01) computer output to find the sample variances
We have s12= 4103, and s2
2 = 10,670.
– It appears that the variances are unequal.
Example: Making an inference about –
Example: Making an inference about –
18
• Compute: Manually– From the data we have:
1236.122
1107
10710670
143
434103
)10710670434103(
670,10,103,423.633,02.604
22
2
22
2121
ssxx
Example: Making an inference about –
Example: Making an inference about –
19
• Compute: Manually– The rejection region is t < -t = -t.05,123 1.658
-2.09
107
10670
43
4103
)0()23.63302.604(
)()(
2
22
1
21
21
n
s
n
s
xxt
Example: Making an inference about –
Example: Making an inference about –
20
Example: Making an inference about –
Example: Making an inference about –
At the 5% significance level there is sufficient evidence to reject the null hypothesis.
-2.09 < -1.6573
Xm13-01
.0193 < .05
t-Test: Two-Sample Assuming Unequal Variances
Consumers NonconsumersMean 604.02 633.23Variance 4102.98 10669.77Observations 43 107Hypothesized Mean Difference 0df 123t Stat -2.09P(T<=t) one-tail 0.0193t Critical one-tail 1.6573P(T<=t) two-tail 0.0386t Critical two-tail 1.9794
21
56.1,86.5665.2721.29107
1067043
41039796.1)239.63302.604(
2n
22
s
1n
21
s
2t)
2x
1x(
• Compute: ManuallyThe confidence interval estimator for the differencebetween two means is
Example: Making an inference about –
Example: Making an inference about –
22
23
• Example 13.2– An ergonomic chair can be assembled using two
different sets of operations (Method A and Method B)
– The operations manager would like to know whether the assembly time under the two methods differ.
Example: Making an inference about –
Example: Making an inference about –
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• Example 13.2– Two samples are randomly and independently selected
• A sample of 25 workers assembled the chair using method A.
• A sample of 25 workers assembled the chair using method B.
• The assembly times were recorded
– Do the assembly times of the two methods differs?
Example: Making an inference about –
Example: Making an inference about –
25
Example: Making an inference about –
Example: Making an inference about –
Method A Method B6.8 5.25.0 6.77.9 5.75.2 6.67.6 8.55.0 6.55.9 5.95.2 6.76.5 6.6. .. .. .. .
Method A Method B6.8 5.25.0 6.77.9 5.75.2 6.67.6 8.55.0 6.55.9 5.95.2 6.76.5 6.6. .. .. .. .
Assembly times in Minutes
Solution
• The data are interval.
• The parameter of interest is the difference between two population means.
• The claim to be tested is whether a difference between the two methods exists.
26
Example: Making an inference about –
Example: Making an inference about –• Compute: Manually
–The hypotheses test is:
H0: (1 - 2) 0 H1: (1 - 2) 0
– To check whether the two unknown population variances areequal we calculate S1
2 and S22 (Xm13-02).
– We have s12= 0.8478, and s2
2 =1.3031.
– The two population variances appear to be equal.
27
Example: Making an inference about –
Example: Making an inference about –• Compute: Manually
4822525.f.d
93.0
251
251
076.1
0)016.6288.6(t
4822525.f.d
93.0
251
251
076.1
0)016.6288.6(t
3031.1s 8478.0s 016.6x 288.6x 22
2121
076.122525
)303.1)(125()848.0)(125(S2
p
– To calculate the t-statistic we have:
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• The rejection region is t < -t =-t.025,48 = -2.009 or t > t = t.025,48 = 2.009
• The test: Since t= -2.009 < 0.93 < 2.009, there is insufficient evidence to reject the null hypothesis.
For = 0.05
2.009.093-2.009
Rejection regionRejection region
Example: Making an inference about –
Example: Making an inference about –
29
Example: Making an inference about –
Example: Making an inference about –
.3584 > .05
-2.0106 < .93 < +2.0106
Xm13-02t-Test: Two-Sample Assuming Equal Variances
Method A Method BMean 6.29 6.02Variance 0.8478 1.3031Observations 25 25Pooled Variance 1.08Hypothesized Mean Difference 0df 48t Stat 0.93P(T<=t) one-tail 0.1792t Critical one-tail 1.6772P(T<=t) two-tail 0.3584t Critical two-tail 2.0106
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• Conclusion: There is no evidence to infer at the 5% significance level that the two assembly methods are different in terms of assembly time
Example: Making an inference about –
Example: Making an inference about –
31
Example: Making an inference about –
Example: Making an inference about –
A 95% confidence interval for 1 - 2 is calculated as follows:
]8616.0,3176.0[5896.0272.0
)25
1
25
11.075(0106.2016.6288.6
)11
()(21
221
nnstxx p
Thus, at 95% confidence level -0.3176 < 1 - 2 < 0.8616
Notice: “Zero” is included in the confidence interval
32
Checking the required Conditions for the equal variances case (Example 13.2)Checking the required Conditions for the equal variances case (Example 13.2)
The data appear to be approximately normal
0
2
4
6
8
10
12
5 5.8 6.6 7.4 8.2 More
Design A
01234567
4.2 5 5.8 6.6 7.4 More
Design B
33
13.4 Matched Pairs Experiment13.4 Matched Pairs Experiment
• What is a matched pair experiment?
• Why matched pairs experiments are needed? • How do we deal with data produced in this way?
The following example demonstrates a situationwhere a matched pair experiment is the correct approach to testing the difference between two population means.
34
35
Example 13.3 – To investigate the job offers obtained by MBA graduates, a
study focusing on salaries was conducted.– Particularly, the salaries offered to finance majors were
compared to those offered to marketing majors.– Two random samples of 25 graduates in each discipline were
selected, and the highest salary offer was recorded for each one. The data are stored in file Xm13-03.
– Can we infer that finance majors obtain higher salary offers
than do marketing majors among MBAs?.
13.4 Matched Pairs Experiment13.4 Matched Pairs Experiment
36
• Solution– Compare two populations of
interval data.
– The parameter tested is 1 - 2
Finance Marketing61,228 73,36151,836 36,95620,620 63,62773,356 71,06984,186 40,203
. .
. .
. .
1
2
The mean of the highest salaryoffered to Finance MBAs
The mean of the highest salaryoffered to Marketing MBAs
– H0: (1 - 2) = 0 H1: (1 - 2) > 0
13.4 Matched Pairs Experiment13.4 Matched Pairs Experiment
37
• Solution – continued
From the data we have:
559,228,262s
,294,433,360s
423,60x624,65x
22
21
2
1
• Let us assume equal variances
13.4 Matched Pairs Experiment13.4 Matched Pairs Experiment
Equal VariancesFinance Marketing
Mean 65624 60423Variance 360433294 262228559Observations 25 25Pooled Variance 311330926Hypothesized Mean Difference 0df 48t Stat 1.04P(T<=t) one-tail 0.1513t Critical one-tail 1.6772P(T<=t) two-tail 0.3026t Critical two-tail 2.0106
There is insufficient evidence to concludethat Finance MBAs are offered higher salaries than marketing MBAs.
38
• Question– The difference between the sample means is
65624 – 60423 = 5,201.– So, why could we not reject H0 and favor H1 where
(1 – 2 > 0)?
The effect of a large sample variabilityThe effect of a large sample variability
39
• Answer: – Sp
2 is large (because the sample variances are large) Sp
2 = 311,330,926. – A large variance reduces the value of the t statistic
and it becomes more difficult to reject H0.
The effect of a large sample variabilityThe effect of a large sample variability
)n1
n1
(s
)()xx(t
21
2p
21
40
Reducing the variabilityReducing the variability
The values each sample consists of might markedly vary...
The range of observationssample B
The range of observationssample A
41
...but the differences between pairs of observations might be quite close to one another, resulting in a small variability of the differences.
0
Differences
The range of thedifferences
Reducing the variabilityReducing the variability
42
The matched pairs experimentThe matched pairs experiment
• Since the difference of the means is equal to the mean of the differences we can rewrite the hypotheses in terms of D (the mean of the differences) rather than in terms of 1 – 2.
• This formulation has the benefit of a smaller variability. Group 1 Group 2 Difference
10 12 - 215 11 +4
Mean1 =12.5 Mean2 =11.5Mean1 – Mean2 = 1 Mean Differences = 1
43
• Example 13.4 – It was suspected that salary offers were affected by
students’ GPA, (which caused S12 and S2
2 to increase).– To reduce this variability, the following procedure was
used:• 25 ranges of GPAs were predetermined.• Students from each major were randomly selected, one from
each GPA range.• The highest salary offer for each student was recorded.
– From the data presented can we conclude that Finance majors are offered higher salaries?
The matched pairs experimentThe matched pairs experiment
44
• Solution (by hand)– The parameter tested is D (=1 – 2)– The hypotheses:
H0: D = 0H1: D > 0
– The t statistic:
Finance Marketing
ns
xt
D
DD
ns
xt
D
DD
The matched pairs hypothesis testThe matched pairs hypothesis test
Degrees of freedom = nD – 1
The rejection region is t > t.05,25-1 = 1.711
45
• Solution – From the data (Xm13-04) calculate:
GPA Group Finance Marketing1 95171 893292 88009 927053 98089 992054 106322 990035 74566 748256 87089 770387 88664 782728 71200 594629 69367 5155510 82618 81591
. .
. .
. .
Difference5842
-4696-11167319-259
100511039211738178121027
.
.
.
Difference
Mean 5065Standard Error 1329Median 3285Mode #N/AStandard Deviation 6647Sample Variance 44181217Kurtosis -0.6594Skewness 0.3597Range 23533Minimum -5721Maximum 17812Sum 126613Count 25
The matched pairs hypothesis testThe matched pairs hypothesis test
46
• Solution
– Calculate t
647,6
065,5
D
D
s
x
81.325664705065
nsx
tD
DD
The matched pairs hypothesis testThe matched pairs hypothesis test
47
3.81 > 1.7109
.0004 < .05
The matched pairs hypothesis testThe matched pairs hypothesis test
Xm13-04t-Test: Paired Two Sample for Means
Finance MarketingMean 65438 60374Variance 444981810 469441785Observations 25 25Pearson Correlation 0.9520Hypothesized Mean Difference 0df 24t Stat 3.81P(T<=t) one-tail 0.0004t Critical one-tail 1.7109P(T<=t) two-tail 0.0009t Critical two-tail 2.0639
48
Conclusion: There is sufficient evidence to infer at 5% significance level that the Finance MBAs’ highest salary offer is, on the average, higher than that ofthe Marketing MBAs.
The matched pairs hypothesis testThe matched pairs hypothesis test
49
The matched pairs mean difference estimation
The matched pairs mean difference estimation
744,2065,525
6647064.250654.13
%95
5.13
1,2/
isExamplein
differencemeantheofintervalconfidenceThe
Example
n
stx
ofEstimatorIntervalConfidence
nD
D
50
The matched pairs mean difference estimation
The matched pairs mean difference estimation
Using Data Analysis Plus Xm13-04GPA Group Finance Marketing
1 95171 893292 88009 927053 98089 992054 106322 990035 74566 748256 87089 770387 88664 782728 71200 594629 69367 5155510 82618 81591
. .
. .
. .
Difference5842
-4696-11167319-259
100511039211738178121027
.
.
.
First calculate the differences,then run the confidence interval procedure in Data Analysis Plus.
t-Estimate: Mean
DifferenceMean 5065Standard Deviation 6647LCL 2321UCL 7808
51
Checking the required conditionsfor the paired observations case
Checking the required conditionsfor the paired observations case
• The validity of the results depends on the normality of the differences.
Histogram
05
10
0 5000 10000 15000 20000
Difference
Fre
qu
en
cy
52
13.5 Inference about the ratio 13.5 Inference about the ratio of two variancesof two variances
13.5 Inference about the ratio 13.5 Inference about the ratio of two variancesof two variances
• In this section we draw inference about the ratio of two population variances.
• This question is interesting because:– Variances can be used to evaluate the consistency
of processes. – The relationship between population variances
determines which of the equal-variances or unequal-variances t-test and estimator of the difference between means should be applied
53
• Parameter to be tested is 12/2
2
• Statistic used is 22
22
21
21
ss
F
Parameter and Statistic Parameter and Statistic
• Sampling distribution of 12/2
2
– The statistic [s12/1
2] / [s22/2
2] follows the F distribution with 1 = n1 – 1, and 2 = n2 – 1.
54
– Our null hypothesis is always
H0: 12 / 2
2 = 1
– Under this null hypothesis the F statistic becomes
F =S1
2/12
S22/2
2
22
21
ss
F 22
21
ss
F
Parameter and Statistic Parameter and Statistic
55
56
(see Xm13-01)In order to perform a test regarding average consumption of calories at people’s lunch in relation to the inclusion of high-fiber cereal in their breakfast, the variance ratio of two samples has to be tested first.
Example 13.6 (revisiting Example 13.1)
Calories intake at lunch
The hypotheses are:
H0:
H1: 1
1
Consmers Non-cmrs568 705498 819589 706681 509540 613646 582636 601739 608539 787596 573607 428529 754637 741617 628633 537555 748
. .
. .
. .
. .
Consmers Non-cmrs568 705498 819589 706681 509540 613646 582636 601739 608539 787596 573607 428529 754637 741617 628633 537555 748
. .
. .
. .
. .
Testing the ratio of two population variances Testing the ratio of two population variances
57
– The F statistic value is F=S12/S2
2 = .3845
– Conclusion: Because .3845<.58 we reject the null hypothesis in favor of the alternative hypothesis, and conclude that there is sufficient evidence at the 5% significance level that the population variances differ.
Testing the ratio of two population variances Testing the ratio of two population variances• Solving by hand
– The rejection region is
F>F2,1,2 or F<1/F
58.72.1
1111
61.1
40,120,025.42,106,025.1,2,2/
120,40,025.106,42,025.2,1,2/
FFFF
FFFF
58
(see Xm13-01)In order to perform a test regarding average consumption of calories at people’s lunch in relation to the inclusion of high-fiber cereal in their breakfast, the variance ratio of two samples has to be tested first.
The hypotheses are:
H0:
H1: 1
1
Example 13.6 (revisiting Example 13.1)
Testing the ratio of two population variances Testing the ratio of two population variances
F-Test Two-Sample for Variances
Consumers NonconsumersMean 604 633Variance 4103 10670Observations 43 107df 42 106F 0.3845P(F<=f) one-tail 0.0004F Critical one-tail 0.6371
59
Estimating the Ratio of Two Population Variances
Estimating the Ratio of Two Population Variances
• From the statistic F = [s12/1
2] / [s22/2
2] we can isolate 1
2/22 and build the following confidence
interval:
1nand1nwhere
Fs
sF
1
s
s
221
1,2,2/22
21
22
21
2,1,2/22
21
1nand1nwhere
Fs
sF
1
s
s
221
1,2,2/22
21
22
21
2,1,2/22
21
60
• Example 13.7– Determine the 95% confidence interval estimate of the ratio of
the two population variances in Example 13.1– Solution
• We find F/2,v1,v2 = F.025,40,120 = 1.61 (approximately)
F/2,v2,v1 = F.025,120,40 = 1.72 (approximately)
• LCL = (s12/s2
2)[1/ F/2,v1,v2 ]
= (4102.98/10,669.77)[1/1.61]= .2388
• UCL = (s12/s2
2)[ F/2,v2,v1 ]
= (4102.98/10,669.77)[1.72]= .6614
Estimating the Ratio of Two Population VariancesEstimating the Ratio of Two Population Variances
61
13.6 Inference about the difference between two population proportions13.6 Inference about the difference between two population proportions• In this section we deal with two populations whose data
are nominal.• For nominal data we compare the population
proportions of the occurrence of a certain event.• Examples
– Comparing the effectiveness of new drug versus older one– Comparing market share before and after advertising
campaign– Comparing defective rates between two machines
62
Parameter and StatisticParameter and Statistic
• Parameter– When the data are nominal, we can only count the
occurrences of a certain event in the two populations, and calculate proportions.
– The parameter is therefore p1 – p2.
• Statistic– An unbiased estimator of p1 – p2 is (the
difference between the sample proportions). 21 p̂p̂
63
Sample 1 Sample size n1
Number of successes x1
Sample proportion
Sample 1 Sample size n1
Number of successes x1
Sample proportion
• Two random samples are drawn from two populations.• The number of successes in each sample is recorded.• The sample proportions are computed.
Sample 2 Sample size n2
Number of successes x2
Sample proportion
Sample 2 Sample size n2
Number of successes x2
Sample proportionx
n1
1
ˆ p1
2
22 n
xp̂
Sampling Distribution ofSampling Distribution of 21 p̂p̂
64
• The statistic is approximately normally distributed if n1p1, n1(1 - p1), n2p2, n2(1 - p2) are all greater than or equal to 5.
• The mean of is p1 - p2.
• The variance of is (p1(1-p1) /n1)+ (p2(1-p2)/n2)
21 p̂p̂
21 p̂p̂
21 p̂p̂
Sampling distribution ofSampling distribution of 21 p̂p̂
65
2
22
1
11
2121
)1()1(
)()ˆˆ(
n
pp
n
pp
ppppZ
2
22
1
11
2121
)1()1(
)()ˆˆ(
n
pp
n
pp
ppppZ
The z-statisticThe z-statistic
Because and are unknown the standard errorBecause and are unknown the standard errormust be estimated using the sample proportions. must be estimated using the sample proportions. The method depends on the null hypothesis The method depends on the null hypothesis
1p 2p
66
Testing the p1 – p2 Testing the p1 – p2
• There are two cases to consider:Case 1: H0: p1-p2 =0
Calculate the pooled proportion
21
21
nn
xxp̂
Then Then
Case 2: H0: p1-p2 =D (D is not equal to 0)Do not pool the data
2
22 n
xp̂
1
11 n
xp̂
)n1
n1
)(p̂1(p̂
)pp()p̂p̂(Z
21
2121
)n1
n1
)(p̂1(p̂
)pp()p̂p̂(Z
21
2121
2
22
1
11
21
n)p̂1(p̂
n)p̂1(p̂
D)p̂p̂(Z
2
22
1
11
21
n)p̂1(p̂
n)p̂1(p̂
D)p̂p̂(Z
67
• Example 13.8– The marketing manager needs to decide which of
two new packaging designs to adopt, to help improve sales of his company’s soap.
– A study is performed in two supermarkets:• Brightly-colored packaging is distributed in supermarket 1.• Simple packaging is distributed in supermarket 2.
– First design is more expensive, therefore,to be financially viable it has to outsell the second design.
Testing p1 – p2 (Case 1) Testing p1 – p2 (Case 1)
68
• Summary of the experiment results– Supermarket 1 - 180 purchasers of Johnson Brothers
soap out of a total of 904
– Supermarket 2 - 155 purchasers of Johnson Brothers soap out of a total of 1,038
– Use 5% significance level and perform a test to find which type of packaging to use.
Testing p1 – p2 (Case 1) Testing p1 – p2 (Case 1)
69
• Solution– The problem objective is to compare the population
of sales of the two packaging designs.– The data are nominal (Johnson Brothers or other
soap) – The hypotheses are
H0: p1 - p2 = 0H1: p1 - p2 > 0
– We identify this application as case 1
Population 1: purchases at supermarket 1Population 2: purchases at supermarket 2
Testing p1 – p2 (Case 1) Testing p1 – p2 (Case 1)
70
Testing p1 – p2 (Case 1) Testing p1 – p2 (Case 1)• Compute: Manually
– For a 5% significance level the rejection region isz > z = z.05 = 1.645
1725.)038,1904()155180()()(ˆ 2121 nnxxp
isproportionpooledThe
90.2
038,1
1
904
1)1725.1(1725.
1493.1991.
11)ˆ1(ˆ
)()ˆˆ(
21
2121
nnpp
ppppZ
becomesstatisticzThe
1493.038,1155ˆ,1991.904180ˆ 21 pandp
aresproportionsampleThe
71
Testing p1 – p2 (Case 1) Testing p1 – p2 (Case 1)• Excel (Data Analysis Plus)
Conclusion: There is sufficient evidence to conclude at the 5% significance level, that brightly-colored design will outsell the simple design.
Xm13-08
z-Test: Two Proportions
Supermarket 1 Supermarket 2Sample Proportions 0.1991 0.1493Observations 904 1038Hypothesized Difference 0z Stat 2.90P(Z<=z) one tail 0.0019z Critical one-tail 1.6449P(Z<=z) two-tail 0.0038z Critical two-tail 1.96
72
• Example 13.9 (Revisit Example 13.8)– Management needs to decide which of two new
packaging designs to adopt, to help improve sales of a certain soap.
– A study is performed in two supermarkets:– For the brightly-colored design to be financially viable it
has to outsell the simple design by at least 3%.
Testing p1 – p2 (Case 2) Testing p1 – p2 (Case 2)
73
• Summary of the experiment results– Supermarket 1 - 180 purchasers of Johnson Brothers’
soap out of a total of 904
– Supermarket 2 - 155 purchasers of Johnson Brothers’ soap out of a total of 1,038
– Use 5% significance level and perform a test to find which type of packaging to use.
Testing p1 – p2 (Case 2) Testing p1 – p2 (Case 2)
74
• Solution– The hypotheses to test are
H0: p1 - p2 = .03H1: p1 - p2 > .03
– We identify this application as case 2 (the hypothesized difference is not equal to zero).
Testing p1 – p2 (Case 2) Testing p1 – p2 (Case 2)
75
• Compute: Manually
The rejection region is z > z = z.05 = 1.645.Conclusion: Since 1.15 < 1.645 do not reject the null hypothesis. There is insufficient evidence to infer that the brightly-colored design will outsell the simple design by 3% or more.
Testing p1 – p2 (Case 2) Testing p1 – p2 (Case 2)
15.1
038,1
)1493.1(1493.
904
)1991.1(1991.
03.038,1
155
904
180
)ˆ1(ˆ)ˆ1(ˆ
)ˆˆ(
2
22
1
11
21
n
pp
n
pp
DppZ
76
Testing p1 – p2 (Case 2) Testing p1 – p2 (Case 2)• Using Excel (Data
Analysis Plus)Xm13-08
z-Test: Two Proportions
Supermarket 1 Supermarket 2Sample Proportions 0.1991 0.1493Observations 904 1038Hypothesized Difference 0.03z Stat 1.14P(Z<=z) one tail 0.1261z Critical one-tail 1.6449P(Z<=z) two-tail 0.2522z Critical two-tail 1.96
77
Estimating p1 – p2 Estimating p1 – p2
• Estimating the cost of life saved– Two drugs are used to treat heart attack victims:
• Streptokinase (available since 1959, costs $460)• t-PA (genetically engineered, costs $2900).
– The maker of t-PA claims that its drug outperforms Streptokinase.
– An experiment was conducted in 15 countries. • 20,500 patients were given t-PA• 20,500 patients were given Streptokinase• The number of deaths by heart attacks was recorded.
78
• Experiment results– A total of 1497 patients treated with Streptokinase
died.– A total of 1292 patients treated with t-PA died.
• Estimate the cost per life saved by using t-PA instead of Streptokinase.
Estimating p1 – p2 Estimating p1 – p2
79
• Solution– The problem objective: Compare the outcomes of
two treatments.– The data are nominal (a patient lived or died)– The parameter to be estimated is p1 – p2.
• p1 = death rate with t-PA
• p2 = death rate with Streptokinase
Estimating p1 – p2 Estimating p1 – p2
80
• Compute: Manually– Sample proportions:
– The 95% confidence interval estimate is
0630.205001292
p̂,0730.205001497
p̂ 21
2
22
1
1121 n
)p̂1(p̂n
)p̂1(p̂)p̂p̂(
2
22
1
1121 n
)p̂1(p̂n
)p̂1(p̂)p̂p̂(
0149.0051.
0049.0100.20500
)0630.1(0630.
20500
)0730.1(0730.96.10630.0730.
UCLLCL
Estimating p1 – p2 Estimating p1 – p2
81
• Interpretation– We estimate that between .51% and 1.49% more
heart attack victims will survive because of the use of t-PA.
– The difference in cost per life saved is 2900-460= $2440.
– The total cost saved by switching to t-PA is estimated to be between 2440/.0149 = $163,758 and 2440/.0051 = $478,431
Estimating p1 – p2 Estimating p1 – p2