Post on 21-Dec-2015
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Inference about Comparing Two
Populations
Chapter 13
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13.1 Introduction13.1 Introduction
• In previous discussions we presented methods designed to make an inference about characteristics of a single population. We estimated, for example the population mean, or hypothesized on the value of the standard deviation.
• However, in the real world we encounter many times the need to study the relationship between two populations. – For example, we want to compare the effects of a new drug on blood
pressure, in which case we can test the relationship between the mean blood pressure of two groups of individuals: those who take the drug, and those who don’t.
– Or, we are interested in the effects a certain ad has on voters’ preferences as part of an election campaign. In this case we can estimate the difference in the proportion of voters who prefer one candidate before and after the ad is televised.
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13.1 Introduction13.1 Introduction
• Variety of techniques are presented whose objective is to compare two populations.
• These techniques are designed to study the…– difference between two means.– ratio of two variances.– difference between two proportions.
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• The reason we are looking at the difference between the two means is that is strongly related to a normal distribution, whose mean is 1 – 2. See next for details.
21 xx
• Two random samples are therefore drawn from the two populations of interest and their means and are calculated.
1x 2x
• We’ll look at the relationship between the two population means by analyzing the value of 1 – 2.
13.2 Inference about the Difference between Two Means: Independent Samples13.2 Inference about the Difference between Two Means: Independent Samples
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The Sampling Distribution ofThe Sampling Distribution of is normally distributed if the (original)
population distributions are normal .
is approximately normally distributed if the (original) population is not normal, but the samples’ size is sufficiently large (greater than 30).
The expected value of is 1 - 2
The variance of is
21 xx
21 xx
21xx
21 xx
21 xx
2
22
1
21
nσ
nσ
6
• 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 –
7
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 –
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• 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 –
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Inference about Inference about ––: Equal variances: Equal variances
• If the two variances 12 and 2
2 are equal to one another, then their estimate S1
2 and S22
estimate the same value.• Therefore, we can pool the two sample
variances and provide a better estimate of the common populations’ variance, based on a larger amount of information.
• This is done by forming the pooled variance estimate. See next.
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To get some intuition about this pooled estimate,
note that we can re-write it as
which has the form of a weighted average of the two sample variances. The weights are the relative sample sizes. A larger sample provides larger weight and thus influences the pooled estimate more (it might be easier to eliminate the values ‘-1’ and ‘-2’ from the formula in order to see the structuremore easily
22
21
221
21
12p S
2nn1n
S2nn
1nS
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
• Calculate the pooled variance estimate by:
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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:
12
Note how Sp2 replaces both
S12 and S2
2.
2
2p
1
2p
2121
ns
ns
)μ(μ)xx(t
2
2p
1
2p
2121
ns
ns
)μ(μ)xx(t
Inference about Inference about ––: Equal variances: Equal variances
• Construct the t-statistic as follows:
2nnd.f.
n1
n1
s
)μ(μ)xx(t
21
21
2p
2121
2nnd.f.
n1
n1
s
)μ(μ)xx(t
21
21
2p
2121
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1n)ns(
1n)ns(
)nsn(sd.f.
ns
ns
)μ(μ)xx(t
2
2
222
1
21
21
22
221
21
2
22
1
21
2121
1n)ns(
1n)ns(
)nsn(sd.f.
ns
ns
)μ(μ)xx(t
2
2
222
1
21
21
22
221
21
2
22
1
21
2121
Inference about –: Unequal variancesInference about –: Unequal variances
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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 run 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
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• 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 –
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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 quantitative. • 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 relationships between the variances, we use a computer output to find the sample
variances (Xm13-1.xls). From the data we have S1
2= 4103, and S22 =
10,670.
– It appears that the variances are unequal.
Example: Making an inference about –
Example: Making an inference about –
11= mean caloric intake for fiber consumers= mean caloric intake for fiber consumers
22= mean caloric intake for fiber non-consumers= mean caloric intake for fiber non-consumers
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• Solving by hand– From the data we have:
123122.6
110710710670
143434103
10710670434103df
10,670s4,103,s633.23x604.2,x
22
22
2121
Example: Making an inference about –
Example: Making an inference about –
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• Solving by hand– H1: 1 - 2 < 0
The rejection region is t < -tdf = -t.05,123 1.658
-2.09
10710670
434103
)0()23.6332.604(
ns
ns
)()xx(t
2
22
1
21
21
Example: Making an inference about –
Example: Making an inference about –
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Example: Making an inference about –
Example: Making an inference about –
t-Test: Two-Sample Assuming Unequal Variances
ConsumersNonconsumersMean 604.023 633.234Variance 4102.98 10669.8Observations 43 107Hypothesized Mean Difference0df 123t Stat -2.09107P(T<=t) one-tail 0.01929t Critical one-tail 1.65734P(T<=t) two-tail 0.03858t Critical two-tail 1.97944
At 5% significance level there is sufficient evidence to reject the null hypothesis.
-2.09107 < -1.65734
Xm13-1.xls
.01929 < .05
22
56.1,86.5665.2721.29107
1067043
41039796.1)239.63302.604(
2n
22
s
1n
21
s
2t)
2x
1x(
• Solving by handThe confidence interval estimator for the differencebetween two means when the variances are unequal is
Example: Making an inference about –
Example: Making an inference about –
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Note that the confidence interval for the differenceNote that the confidence interval for the differencebetween the two means falls entirely in the negativebetween the two means falls entirely in the negativeregion: [-56.86, -1.56]; even at best the difference region: [-56.86, -1.56]; even at best the difference between the two means is mbetween the two means is m11 – m – m22 = -1.56, so we = -1.56, so we
can be 95% confident mcan be 95% confident m1 1 is smaller than mis smaller than m2 2 !!
This conclusion agrees with the results of the test This conclusion agrees with the results of the test performed before.performed before.
Example: Making an inference about –
Example: Making an inference about –
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• 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 design A.
• A sample of 25 workers assembled the chair using design 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 –
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Example: Making an inference about –
Example: Making an inference about –
Design-A Design-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. .. .. .. .
Design-A Design-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 quantitative.
• The parameter of interest is the difference between two population means.
• The claim to be tested is whether a difference between the two designs exists.
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Example: Making an inference about –
Example: Making an inference about –• Solving by hand
–The hypotheses test is:
H0: 1 - 2 0 H1: 1 - 2 0
– To check the relationship between the two variances we calculate the value of S1
2 and S22 (Xm13-02.xls).
– From the data we have S12= 0.8478, and S2
2 =1.3031.
so 12 and 2
2 appear to be equal.
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Example: Making an inference about –
Example: Making an inference about –• Solving by hand
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 2-tail 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 –
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Example: Making an inference about –
Example: Making an inference about –
t-Test: Two-Sample Assuming Equal Variances
Design-A Design-BMean 6.288 6.016Variance 0.847766667 1.3030667Observations 25 25Pooled Variance 1.075416667Hypothesized Mean Difference0df 48t Stat 0.927332603P(T<=t) one-tail 0.179196744t Critical one-tail 1.677224191P(T<=t) two-tail 0.358393488t Critical two-tail 2.01063358
t-Test: Two-Sample Assuming Equal Variances
Design-A Design-BMean 6.288 6.016Variance 0.847766667 1.3030667Observations 25 25Pooled Variance 1.075416667Hypothesized Mean Difference0df 48t Stat 0.927332603P(T<=t) one-tail 0.179196744t Critical one-tail 1.677224191P(T<=t) two-tail 0.358393488t Critical two-tail 2.01063358
.35839 > .05-2.0106 < .9273 < +2.0106
Xm13-02.xls
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• Conclusion: From this experiment, it is unclear at 5% significance level if the two assembly methods are different in terms of assembly time
Example: Making an inference about –
Example: Making an inference about –
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Example: Making an inference about –Constructing a Confidence Interval
Example: Making an inference about –Constructing a Confidence IntervalA 95% confidence interval for 1 - 2 when the two variances are
equal is calculated as follows:
]8616.0,3176.0[5896.0272.0
)251
251
1.075(0106.2016.6288.6
)n1
n1
(st)xx(21
2
p21
Thus, at 95% confidence level -0.3176 < 1 - 2 < 0.8616
Notice: “Zero” is included in the confidence interval and therefore the two mean values could be equal.
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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
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13.4 Matched Pairs Experiment -Dependent samples
13.4 Matched Pairs Experiment -Dependent samples
• What is a matched pair experiment?• A matched pairs experiment is a sampling design in which every two
observations share some characteristic. For example, suppose we are interested in increasing workers productivity. We establish a compensation program and want to study its efficiency. We could select two groups of workers, measure productivity before and after the program is established and run a test as we did before.
• But, if we believe workers’ age is a factor that may affect changes in productivity, we can divide the workers into different age groups, select a worker from each age group, and measure his or her productivity twice. One time before and one time after the program is established. Each two observations constitute a matched pair, and because they belong to the same age group they are not independent.
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13.4 Matched Pairs Experiment -Dependent samples
13.4 Matched Pairs Experiment -Dependent samples
Why matched pairs experiments are needed?
The following example demonstrates a situationwhere a matched pair experiment is the correct approach to testing the difference between two population means.
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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.
– From the data, can we infer that finance majors obtain higher
salary offers than marketing majors among MBAs?.
13.4 Matched Pairs Experiment13.4 Matched Pairs Experiment
Additional example
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• Solution– Compare two populations of
quantitative 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
38
• Solution – continued
From Xm13-3.xls 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
t-Test: Two-Sample Assuming Equal Variances
Finance MarketingMean 65624 60423Variance 360433294 262228559Observations 25 25Pooled Variance 311330926Hypothesized Mean Difference 0df 48t Stat 1.04215119P(T<=t) one-tail 0.15128114t Critical one-tail 1.67722419P(T<=t) two-tail 0.30256227t Critical two-tail 2.01063358
There is insufficient evidence to concludethat Finance MBAs are offered higher salaries than marketing MBAs.
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• Question– The difference between the sample means is
65624 – 60423 = 5,201.– So, why could not we reject H0 and favor H1?
The effect of a large sample variabilityThe effect of a large sample variability
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• 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
Recall that rejection of thenull hypothesis occurs when‘t’ is sufficiently large (t>t).A large Sp
2 reduces ‘t’ and therefore it does not fall inthe rejection region.
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The matched pairs experimentThe matched pairs experiment
• We are looking for hypotheses formulation where the variability of the two samples has been reduced.
• By taking matched pair observations and testing the differences per pair we achieve two goals:– We still test 1 – 2 (see explanation next)– The variability used to calculate the t-statistic is
usually smaller (see explanation next).
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The matched pairs experiment – Are we still testing 1 – 2?
The matched pairs experiment – Are we still testing 1 – 2?
• Note that the difference between the two means is equal to the mean difference of pairs of observations
• A short exampleGroup 1 Group 2 Difference
10 12 - 215 11 +4
Mean1 =12.5 Mean2 =11.5Mean1 – Mean2 = 1 Mean Differences = 1
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The matched pairs experiment – Reducing the variability
The matched pairs experiment – Reducing the variability
Observations might markedly differ...
The range of observationssample B
The range of observationssample A
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...but the differences between pairs of observations might have much smaller variability.
0
Differences
The range of thedifferences
The matched pairs experiment – Reducing the variability
The matched pairs experiment – Reducing the variability
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• Example 12.4 (12.3 part II)– 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
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• Solution (by hand)– The parameter tested is D (=1 – 2)– The hypotheses:
H0: D = 0H1: D > 0
– The t statistic:
ns
xt
D
DD
ns
xt
D
DD
The matched pairs hypothesis testThe matched pairs hypothesis test
–
The rejection region is t > t.05,25-1 = 1.711
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• Solution (by hand) – continue – From the data (Xm13-4.xls) 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
.
.
.
The matched pairs hypothesis testThe matched pairs hypothesis test
Difference
Mean 5064.52Standard Error 1329.3791Median 3285Mode #N/AStandard Deviation 6646.8953Sample Variance 44181217Kurtosis -0.659419Skewness 0.359681Range 23533Minimum -5721Maximum 17812Sum 126613Count 25
Using Descriptive Statistics in Excel we get:
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• Solution (by hand) – continue
– Calculate t
647,6s065.5x
D
D
81.325664705065
nsx
tD
DD
The matched pairs hypothesis testThe matched pairs hypothesis test
See conclusion laterSee conclusion later
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t-Test: Paired Two Sample for Means
Finance MarketingMean 65438.2 60373.68Variance 4.45E+08 4.69E+08Observations 25 25Pearson Correlation 0.952025Hypothesized Mean Difference0df 24t Stat 3.809688P(T<=t) one-tail 0.000426t Critical one-tail 1.710882P(T<=t) two-tail 0.000851t Critical two-tail 2.063898
Recall: The rejection regionis t > t. Indeed, 3.809 > 1.7108
.000426 < .05
The matched pairs hypothesis testThe matched pairs hypothesis test
Xm13-4.xls
Using Data Analysis in Excel
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Conclusion: There is sufficient evidence to infer at 5% significance level that the Finance MBAs’ highest salary offer is, on the average, higher than this ofthe Marketing MBAs.
The matched pairs hypothesis testThe matched pairs hypothesis test
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The matched pairs mean difference estimation
The matched pairs mean difference estimation
744,2065,525
6647064.25065is4.13examplein
differencemeantheofervalintconfidence%95The5.13Example
ns
tx
ofEstimatorervalintConfidence
1n,2/D
D
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The matched pairs mean difference estimation
The matched pairs mean difference estimation
Using Data Analysis Plus Xm13-4.xlst-Estimate:Mean
DifferenceMean 5065Standard Deviation 6647LCL 2321UCL 7808
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
.
.
.
First calculate the differences for each pair, then run the confidence interval procedure in Data Analysis Plus.
53
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.
Diffrences
0
2
4
6
-300
0 030
0060
0090
00
1200
0
1500
0
1800
0M
ore
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13.5 Inferences about the ratio 13.5 Inferences about the ratio of two variancesof two variances
13.5 Inferences about the ratio 13.5 Inferences about the ratio of two variancesof two variances
• In this section we draw inference about the relationship between two population variances.
• This question is interesting because:– Variances can be used to evaluate the consistency
of processes. – The relationships between variances determine the
technique used to test relationships between mean values
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• The parameter tested is 12/2
2
• The statistic used is 22
22
21
21
ss
F
Parameter tested and statistic Parameter tested and statistic
• The Sampling distribution of 12/2
2
– The statistic [s12/1
2] / [s22/2
2] follows the F distribution with…Numerator d.f. = n1 – 1, and Denominator d.f. = n2 – 1.
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– 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
F22
21
ss
F
Parameter tested and statistic Parameter tested and statistic
57
58
(see example 13.1)In order to test whether having a rich-in-fiber breakfast reduces the amount of caloric intake at lunch, we need to decide whether the variances are equal or not.
Example 13.6 (revisiting 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
59
1n,1n,2
1n,1n,2
12
21
F1
F
FF
– The F statistic value is F=S12/S2
2 = .3845
– Conclusion: Because .3845<.63 we can reject the null hypothesis in favor of the alternative hypothesis, and conclude that there is sufficient evidence in the data to argue at 5% significance level that the variance of the two groups differ.
Testing the ratio of two population variances Testing the ratio of two population variances• Solving by hand
– The rejection region is
63.F
1F
1
61.1FF
40,120,025.42,106,025.
120,40,025.106,42,025.
60
(see Xm13.1)
The hypotheses are:
H0:
H1: 1
1
F-Test Two-Sample for Variances
Consumers NonconsumersMean 604.0232558 633.2336449Variance 4102.975637 10669.76565Observations 43 107df 42 106F 0.384542245P(F<=f) one-tail0.000368433F Critical one-tail0.637072617
F-Test Two-Sample for Variances
Consumers NonconsumersMean 604.0232558 633.2336449Variance 4102.975637 10669.76565Observations 43 107df 42 106F 0.384542245P(F<=f) one-tail0.000368433F Critical one-tail0.637072617
Example 13.6 (revisiting 13.1)
Testing the ratio of two population variances Testing the ratio of two population variances
From Data AnalysisFrom Data Analysis
61
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
62
• Example 13.7– Determine the 95% confidence interval estimate of the ratio of
the two population variances in example 12.1– 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/ Fa/2,v1,v2 ]
= (4102.98/10,669.770)[1/1.61]= .2388
• UCL = (s12/s2
2)[ Fa/2,v2,v1 ]
= (4102.98/10,669.770)[1.72]= .6614
Estimating the Ratio of Two Population VariancesEstimating the Ratio of Two Population Variances
63
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 vs.old one– Comparing market share before and after advertising
campaign– Comparing defective rates between two machines
64
Parameter tested and statisticParameter tested and statistic
• Parameter– When the data is nominal, we can only count the
occurrences of a certain event in the two populations, and calculate proportions.
– The parameter tested is therefore p1 – p2.
• Statistic– An unbiased estimator of p1 – p2 is (the
difference between the sample proportions). 21 p̂p̂
65
Sample 1 Sample size n1
Number of successes x1
Sample proportion
Sample 1 Sample size n1
Number of successes x1
Sample proportion
Sampling distribution ofSampling distribution of
• 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 proportion
2
22 n
xp̂
21 p̂p̂
1
11 n
xp ˆ
66
• The statistic is approximately normally distributed if n1p1, n1(1 - p1), n2p2, n2(1 - p2) are all equal to or greater than 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
67
Because p1 and p2 are unknown, we use their estimates instead. Thus,
should all be equal to or greater than 5.
)p̂1(n,p̂n),p̂1(n,p̂n 22221111
2
22
1
11
2121
n)p1(p
n)p1(p
)pp()p̂p̂(Z
2
22
1
11
2121
n)p1(p
n)p1(p
)pp()p̂p̂(Z
The z-statisticThe z-statistic
68
Testing p1 – p2 Testing 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
)p̂p̂(Z
21
21
)n1
n1
)(p̂1(p̂
)p̂p̂(Z
21
21
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
69
• 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 communities:• Design A is distributed in Community 1.• Design B is distributed in Community 2.• The old design packages is still offered in both
communities.– Design A is more expensive, therefore,to be
financially viable it has to outsell design B.
Testing p1 – p2 (Case I) Testing p1 – p2 (Case I)
70
• Summary of the experiment results– Community 1 - 580 packages with new design A sold
324 packages with old design sold– Community 2 - 604 packages with new design B sold
442 packages with old design sold
– Use 5% significance level and perform a test to find which type of packaging to use.
Testing p1 – p2 (Case I) Testing p1 – p2 (Case I)
71
• Solution– The problem objective is to compare the population
of sales of the two packaging designs.– The data is qualitative (yes/no for the purchase of
the new design per customer)– The hypotheses test are
H0: p1 - p2 = 0H1: p1 - p2 > 0
– We identify here case 1.
Population 1 – purchases of Design APopulation 2 – purchases of Design B
Testing p1 – p2 (Case I) Testing p1 – p2 (Case I)
72
• Solving by hand– For a 5% significance level the rejection region is
z > z = z.05 = 1.645
6072.)1046904()604580()nn()xx(p̂
isproportionpooledThe
2121
89.2
10461
9041
)6072.1(6072.
5774.6416.
n1
n1
)p̂1(p̂
)pp()p̂p̂(Z
becomesstatisticzThe
21
2121
5774.1046604p̂and,6416.904580p̂
aresproportionsampleThe
21 From Xm13-08.xls we have:
Testing p1 – p2 (Case I) Testing p1 – p2 (Case I)
73
• Conclusion: At 5% significance level there sufficient evidence to infer that the proportion of sales with design A is greater that the proportion of sales with design B (since 2.89 > 1.645).
Testing p1 – p2 (Case I) Testing p1 – p2 (Case I)
74
• Excel (Data Analysis Plus)
Testing p1 – p2 (Case I) Testing p1 – p2 (Case I)
z-Test: Two ProportionsCommunity 1 Community 2
sample proportions 0.6416 0.5774Observations 904 1046Hypothesized Difference 0z Stat 2.89P(Z<=z) one tail 0.0019z Critical one-tail 1.6449P(Z<=z) two-tail 0.0038z Critical two-tail 1.96
Xm13-08.xls
• ConclusionSince 2.89 > 1.645, there is sufficient evidence in the data to conclude at 5% significance level, that design A will outsell design B.
Additional example
75
• Example 13.9 (Revisit example 13.08)– 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 communities:• Design A is distributed in Community 1.• Design B is distributed in Community 2.• The old design packages is still offered in both communities.
– For design A to be financially viable it has to outsell design B by at least 3%.
Testing p1 – p2 (Case II) Testing p1 – p2 (Case II)
76
• Summary of the experiment results– Community 1 - 580 packages with new design A sold
324 packages with old design sold– Community 2 - 604 packages with new design B sold
442 packages with old design sold• Use 5% significance level and perform a test to
find which type of packaging to use.
Testing p1 – p2 (Case II) Testing p1 – p2 (Case II)
77
• Solution– The hypotheses to test are
H0: p1 - p2 = .03H1: p1 - p2 > .03
– We identify case 2 of the test for difference in proportions (the difference is not equal to zero).
Testing p1 – p2 (Case II) Testing p1 – p2 (Case II)
78
58.1
1046)577.1(577.
904)642.1(642.
03.442604
604324580
580
n)p̂1(p̂
n)p̂1(p̂
D)p̂p̂(Z
2
22
1
11
21
• Solving by hand
The rejection region is z > z = z.05 = 1.645.Conclusion: Since 1.58 < 1.645 do not reject the null hypothesis. There is insufficient evidence to infer that packaging with Design A will outsell this of Design B by 3% or more.
Testing p1 – p2 (Case II) Testing p1 – p2 (Case II)
79
• Using Excel (Data Analysis Plus)
Testing p1 – p2 (Case II) Testing p1 – p2 (Case II)
Xm13-08.xls
z-Test: Two Proportions
Community 1 Community 2Sample Proportion 0.6416 0.5774Observations 904 1046Hypothesized Difference 0.03z stat 1.5467P(Z<=z) one-tail 0.061z Critical one-tail 1.6449P(Z<=z) two-tail 0.122z Critical two-tail 1.96
80
Estimating p1 – p2 Estimating p1 – p2
• Example (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.
81
• 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
82
• Solution– The problem objective: Compare the outcomes of
two treatments.– The data is nominal (a patient lived/died)– The parameter estimated is p1 – p2.
• p1 = death rate with t-PA
• p2 = death rate with Streptokinase
Estimating p1 – p2 Estimating p1 – p2
83
• Solving by hand– Sample proportions:
– The 95% confidence interval 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.UCL0051.LCL
0049.100.20500
)0630.1(0630.20500
)0730.1(0730.96.10630.0730.
Estimating p1 – p2 Estimating p1 – p2
84
• 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