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Chapter 13
Inference about
Comparing Two
Populations
1
13.1 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.
– 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.
2
13.1 Introduction
• Variety of techniques are presented whose
objective is to compare two populations.
• These techniques are designed to compare:
– two population means.
– two population variances.
– two proportions.
3
13.2 Inference about the Difference
between Two Means: Independent Samples
• We’ll look at the relationship between the two
population means by analyzing the value of m1 – m2.
• The reason we are looking at the difference between
the two means is that x1  x 2 is strongly related to a
normal distribution, whose mean is m1 – m2. See next
for details.
• Two random samples are therefore drawn from the
two populations of interest and their means x1 and x2
are calculated.
4
The Sampling Distribution of x1  x 2




x1  x2
is normally distributed if the (original)
population distributions are normal .
x1  x2
is approximately normally distributed if the
(original) population is not normal, but the samples’ size
is sufficiently large (greater than 30).
The mean value of
The variance of
x1  x 2 is m1 - m2
x1  x 2
σ12 σ 22
is

n1 n2
5
Making an inference about m – m
• The Z – score of x1  x 2 is
( x 1  x 2 )  (m  m  )
Z


  

n1 n2
• However, if x1  x 2 is normal or approximately
normal, then Z is standard normal. So…
• Z can be used to build a confidence interval or test
a hypothesis about m1-m2. See next.
6
Making an inference about m – m
• Practically, the “Z” statistic is hardly used,
because the population variances are not known.
• Instead, we construct a “t” statistic using the
sample “variances” (S12 andS22 ).
( x 1  x 2 )  (m  m  )
Zt 


2
2
?
?


S
S1
2

n1 n2
7
Making an inference about m – m
• Two cases are considered when producing the
t-statistic.
– The two unknown population variances are equal.
– The two unknown population variances are not equal.
8
Inference about m – m: Equal variances
•
If the two variances σ12
2
2
another, thenS1 and S2
2
and σ 2
are equal to one
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.
9
Inference about m – m: Equal variances
• Calculate the pooled variance estimate by:
(n1  1)s1  (n2  1)s 2
Sp 
n1  n2  2
2
2
2
To get some intuition about this pooled estimate,
note that we can re-write it as
S p2 
n1  1
n2  1
S12 
S 22
n1  n2  2
n1  n2  2
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 structure more easily.
10
Inference about m – m: Equal variances
• Calculate the pooled variance estimate by:
(n1  1)s1  (n2  1)s 2
Sp 
n1  n2  2
2
2
2
Example: S12 = 25; S22 = 30; n1 = 10; n2 = 15. Then,
S p2 
(10  1)( 25)  (15  1)( 30)
 28.04347
10  15  2
11
Inference about m – m: Equal variances
• Construct the t-statistic as follows:
t
(x1  x 2 )  (μ1  μ2 )
1 1
s   
 n1 n2 
d.f.  n1  n2  2
2
p
Note how Sp replaces both
S12 and S22.
2
t
(x1  x 2 )  (μ1  μ2 )
 s p2 s p2 
  
n n 
2 
 1
12
Inference about m – m: Unequal variances
t
(x1  x 2 )  (μ1  μ2 )
2
1
s s

n1 n2
2
• Since σ12 and σ 2 are unequal
we can’t produce a single estimate
for both variances.
• Thus we use the sample variances
in the ‘t’ formula
d.f. 
2
2
(s12 n1  s 22 n2 ) 2
2
1
2
2
2
( s n1 ) ( s n2 )

n1  1
n2  1
2
13
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
14
15
Example: Making an inference about m – m
• Example 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.
16
Example: Making an inference about m – m
Consmers Non-cmrs
568
498
589
681
540
646
636
739
539
596
607
529
637
617
633
555
.
.
.
.
705
819
706
509
613
582
601
608
787
573
428
754
741
628
537
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 (m1)
is less than that of non-consumers (m2).
17
Example: Making an inference about m – m
•
The hypotheses are:
H0: m1 - m2 = 0
H1: m1 - m2 < 0
m1= mean caloric intake for fiber consumers
m2= mean caloric intake for fiber non-consumers
– To check the relationships between the variances,
we use a computer output to find the sample
variances (Xm13-1.xlsx).
From the data we have S12= 4103, and S22 =
10,670.
– It appears that the variances are unequal.18
Example: Making an inference about m – m
• Solving by hand
– From the data we have:
x1  604.2,
x 2  633.23
s12  4,103, s 22  10,670
4103 43  10670 107
df 
 122.6  123
2
2
4103 43  10670 107 
43  1
107  1
19
Example: Making an inference about m – m
• Solving by hand
– H1: m1 - m2 < 0
The rejection region is t < -ta,df = -t.05,123  1.658
t
( x1  x 2 )  (m  m  )
s s

n1 n2
2
1
2
2
(604.2  633.23)  (0)

 -2.09
4103 10670

43
107
20
Example: Making an inference about m – m
t-Test: Two-Sample Assuming
Unequal Variances
Consumers
Nonconsumers
Mean
604.023 633.234
Variance
4102.98 10669.8
Observations
43
107
Hypothesized Mean Difference
0
df
123
t Stat
-2.09107
P(T<=t) one-tail 0.01929
t Critical one-tail 1.65734
P(T<=t) two-tail 0.03858
t Critical two-tail 1.97944
Conclusion:
At 5% significance level there is
sufficient evidence to reject the
null hypothesis, and argue that
m1 < m2. Click.
The p-value approach: .01929 < .05
The rejection region approach
-2.09107 < -1.65734
21
Example: Making an inference about m – m
• Solving by hand
The confidence interval estimator for the difference
between two means when the variances are unequal is
 s2 s2 
 1

2
(x  x )  t
 

1 2
a 2 n
n 
2
 1
4103 10670
 (604 .02  633 .239 )  1.9796

43
107
 29.21  27.65   56.86,  1.56
22
Example: Making an inference about m – m
Note that the confidence interval for the difference
between the two means falls entirely in the negative
region: [-56.86, -1.56]; even at best the difference
between the two means is m1 – m2 = -1.56, so we
can be 95% confident m1 is smaller than m !
This conclusion agrees with the results of the test
performed before.
23
Example: Making an inference about m – m
• Example 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.
24
Example: Making an inference about m – m
• 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?
25
Example: Making an inference about m – m
Assembly times in Minutes
Design-A Design-B
6.8
5.2
Solution
5.0
6.7
7.9
5.7
5.2
6.6
• The data are quantitative.
7.6
8.5
5.0
6.5
• The parameter of interest is the difference
5.9
5.9
5.2
6.7
between two population means.
6.5
6.6
.
.
.
.
• The claim to be tested is whether a difference
.
.
between the two designs exists.
.
.
26
Example: Making an inference about m – m
• Solving by hand
–The hypotheses test is:
H0: m1 - m2  0
H1: m1 - m2  0
Since we ask whether or not the
assembly times are the same on
the average, the alternative
hypothesis is of the form m1  m
– To check the relationship between the two variances
we run the F test. (Xm13-02).
2
S
– From the data we have
0.8478, and 2 =1.3031.
The p-value of the F test = 2(.1496) = .299
S12 =
Conclusion: 12 and 22 appear to be equal.
27
Example: Making an inference about m – m
• Solving by hand
– To calculate the t-statistic we have:
x1  6.288 x2  6.016 s12  0.8478 s22  1.3031
(25  1)( 0.848)  (25  1)(1.303)
S 
 1.076
25  25  2
2
p
t
(6.288  6.016)  0
1
 1
1.076  
 25 25 
d.f .  25  25  2  48
 0.93
28
Example: Making an inference about m – m
• The 2-tail rejection region is t < -ta/,n =-t.025,48 = -2.009 or
t > ta/,n = t.025,48 = 2.009
For a = 0.05
• The test: Since t= -2.009 < 0.93 < 2.009, there is insufficient
evidence to reject the null hypothesis.
Rejection region
Rejection region
-2.009
.093 2.009
29
Example: Making an inference about m – m
t-Test: Two-Sample Assuming Equal Variances
Design-A
Mean
6.288
Variance
0.847766667
Observations
25
Pooled Variance
1.075416667
Hypothesized Mean Difference
0
df
48
t Stat
0.927332603
P(T<=t) one-tail
0.179196744
t Critical one-tail
1.677224191
P(T<=t) two-tail
0.358393488
t Critical two-tail
2.01063358
Design-B
6.016
1.3030667
25
Conclusion:
From this experiment, it is
unclear at 5% significance
level if the two assembly
methods are different in
terms of assembly time
-2.0106 < .9273 < +2.0106
.35839 > .05
30
Example: Making an inference about m – m:
A 95% confidence interval for m1 - m2 when the two variances are
equal is calculated as follows:
( x1  x 2 )  t a  s p2 (
1 1
 )
n1 n2
1
1
 6.288  6.016  2.0106 1.075(  ) 
25 25
 0.272  0.5896  [ 0.3176 , 0.8616 ]
Thus, at 95% confidence level -0.3176 < m1 - m2 < 0.8616
Notice: “Zero” is included in the confidence interval and
therefore the two mean values could be equal.
31
Checking the required Conditions for the equal
variances case (example 13.2)
Design A
12
10
The data appear to be
approximately normal
8
6
4
2
0
5
5.8
6.6
Design B
7.4
8.2
More
4.2
5
5.8
7
6
5
4
3
2
1
0
6.6
7.4
More
32
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. Click.
• 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.
33
13.4 Matched Pairs Experiment Dependent samples
Why matched pairs experiments are needed?
The following example demonstrates a situation
where a matched pair experiment is the correct
approach to testing the difference between two
population means.
34
Additional example
13.4 Matched Pairs Experiment
Example 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?.
35
13.4 Matched Pairs Experiment
• Solution
– Compare two populations of
quantitative data.
– The parameter tested is
m1 - m2
– H0: m1 - m2 = 0
H1: m1 - m2 > 0
Finance
61,228
51,836
20,620
73,356
84,186
.
.
.
Marketing
73,361
36,956
63,627
71,069
40,203
.
.
.
m1 The mean of the highest salary
offered to Finance MBAs
m2 The mean of the highest salary
offered to Marketing MBAs
36
13.4 Matched Pairs Experiment
• Solution – continued
From Xm13-3.xls we have:
x1  65,624
x 2  60,423
s12  360,433,294,
s 22  262,228,559
• Let us assume equal
variances
t-Test: Two-Sample Assuming
Equal Variances
Finance
Mark eting
Mean
65624
60423
Variance
360433294 262228559
Observations
25
25
Pooled Variance
311330926
Hypothesized Mean Difference
0
df
48
t Stat
1.04215119
P(T<=t) one-tail
0.15128114
t Critical one-tail
1.67722419
P(T<=t) two-tail
0.30256227
t Critical two-tail
2.01063358
There is insufficient evidence to conclude
that Finance MBAs are offered higher
37
salaries than marketing MBAs.
The effect of a large sample variability
• Question
– The difference between the sample means is
65624 – 60423 = 5,201.
– So, why could not we reject H0 and favor H1?
38
The effect of a large sample variability
• Answer:
– Sp2 is large (because the sample variances are
large) Sp2 = 311,330,926.
– A large variance reduces the value of the t statistic
and this is why t does not fall in the rejection region.
( x1  x 2 )  (m  m  )
t
1
2 1
sp (  )
n1 n2
Recall that rejection of H0
in this problem occurs when
‘t’ is sufficiently large (t>ta).
A large Sp2 reduces ‘t’ and
therefore it does not fall in
the rejection region.
39
The 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 m1 – m2 (see explanation next)
– The variability used to calculate the t-statistic is
usually smaller (see explanation next).
40
The matched pairs experiment –
Are we still testing m1 – m2?
• Yes. Note that the difference between the two
means is equal to the mean difference of pairs
of observations
• A short example
Group 1
10
15
Group 2
12
11
Difference
-2
+4
Mean1 =12.5 Mean2 =11.5
Mean1 – Mean2 = 1
Mean Differences = 1
41
The matched pairs experiment –
Reducing the variability
The range of observations
sample A
Observations might markedly differ...
The range of observations
sample B
42
The matched pairs experiment –
Reducing the variability
Differences
...but the differences between pairs of observations
might have much smaller variability.
The range of the
differences
0
43
The matched pairs experiment
• Example 4 (Example 3 part II)
– It was suspected that salary offers were affected by
students’ GPA. Since GPAs were different, so were the
salaries (which caused S12 and S22 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?
44
The matched pairs hypothesis test
• Solution (by hand)
– The parameter tested is mD (=m1 – m2)
Finance Marketing
– The hypotheses:
H0: mD = 0
The rejection region is
H1: mD > 0
t > t.05,25-1 = 1.711
– The t statistic:
t
xD  mD
sD
Degrees of freedom  nD – 
n
45
The matched pairs hypothesis test
• Solution (by hand) – continue
– From the data (Xm13-4.xls) calculate:
Using Descriptive Statistics in Excel we get:
GPA Group Finance Marketing Difference
1
95171
89329
5842
2
88009
92705
-4696
3
98089
99205
-1116
4
106322
99003
7319
5
74566
74825
-259
6
87089
77038
10051
7
88664
78272
10392
8
71200
59462
11738
9
69367
51555
17812
10
82618
81591
1027
.
.
.
.
.
.
.
.
.
Difference
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
5064.52
1329.3791
3285
#N/A
6646.8953
44181217
-0.659419
0.359681
23533
-5721
17812
126613
25
46
The matched pairs hypothesis test
• Solution (by hand) – continue
x D  5.065
sD  6,647
See conclusion later
– Calculate t
x D  mD
5065  0
t

 3.81
sD n 6647 25
47
The matched pairs hypothesis test
Using Data Analysis in Excel
t-Test: Paired Two Sample for Means
Finance Mark eting
Mean
65438.2 60373.68
Variance
4.45E+08 4.69E+08
Observations
25
25
Pearson Correlation 0.952025
Hypothesized Mean Difference0
df
24
t Stat
3.809688
P(T<=t) one-tail
0.000426
t Critical one-tail
1.710882
P(T<=t) two-tail
0.000851
t Critical two-tail
2.063898
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 of the Marketing
MBAs.
Recall: The rejection region
is t > ta. Indeed, 3.809 > 1.7108
.000426 < .05
48
The matched pairs mean difference
estimation
Confidence int erval Estimator of mD
x D  t a / 2 ,n1
s
n
Example 13.5
The 95% confidence int erval of the mean difference
6647
 5,065  2,744
in example 13.4 is 5065  2.064
25
49
The matched pairs mean difference
estimation
Using Data Analysis Plus
GPA Group Finance Marketing Difference
1
95171
89329
5842
2
88009
92705
-4696
3
98089
99205
-1116
4
106322
99003
7319
5
74566
74825
-259
6
87089
77038
10051
7
88664
78272
10392
8
71200
59462
11738
9
69367
51555
17812
10
82618
81591
1027
.
.
.
.
.
.
.
.
.
t-Estimate:Mean
Mean
Standard Deviation
LCL
UCL
Difference
5065
6647
2321
7808
First calculate the differences for each
pair, then run the confidence interval
procedure in Data Analysis Plus.
50
Checking the required conditions
for the paired observations case
• The results validity depends on the normality of
the differences.
Diffrences
6
4
2
e
or
M
0
18
00
0
15
00
0
12
00
90
00
60
00
0
30
00
-3
00
0
0
51
13.5 Inferences about the ratio
of 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
52
Parameter tested and statistic
• The parameter tested is 12/22
2
1
2
2
s 
• The statistic used is F 
s 
2
1
2
2
• The Sampling distribution of 12/22
– The statistic [s12/12] / [s22/22] follows the F distribution
with…
Numerator d.f. = n1 – 1, and Denominator d.f. = n2 –531.
Parameter tested and statistic
– Our null hypothesis is always
H0: 12 / 22 = 1
S12/12
– Under this null hypothesis the F statistic F = 2 2
S2 /2
becomes
s
F
s
2
1
2
2
54
55
Testing the ratio of two population variances
Example 6 (revisiting example 1)
(see example 1)
Calories intake at lunch
Consmers Non-cmrs
In order to test whether
568
705
The hypotheses are:
498
819
having a rich-in-fiber
589
706

681
509

breakfast reduces the
540
613
H0:   1
646
582
amount of caloric intake
636
601
 
739
608

at lunch, we need to
539
787


596
573
1
H
:
decide whether the
607
428
1

529
754

variances are equal or
637
741
617
628
not.
633
537
555
.
.
.
.
748
.
.
.
.
56
Testing the ratio of two population
variances
• Solving by hand
– The rejection region is
Note: The numerator degrees
of freedom
and the denominator degrees
of freed are
replacing one another!
F  Fa
2
F
,n1 1,n 2 1
1
Fa
2
,n 2 1,n1 1
 F.025,42,106  F.025,40,120  1.61

1
F.025,106,42

1
F.025,120,40
 .63
– The F statistic value is F=S12/S22 = .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%
57
significance level that the variance of the two groups differ.
Testing the ratio of two population variances
Example 6 (revisiting example 1)
(see Xm13.1)
The hypotheses are:
H0:
H1:


 

 
1
1
From Data Analysis
F-Test Two-Sample for Variances
Consumers Nonconsumers
Mean
604.0232558
633.2336449
Variance 4102.975637
10669.76565
Observations
43
107
df
42
106
F
0.384542245
P(F<=f) one-tail
0.000368433
F Critical one-tail
0.637072617
58
Estimating the Ratio of Two Population
Variances
• From the statistic F = [s12/12] / [s22/22] we can
isolate 12/22 and build the following confidence
interval:
2
2 
 s12 


s
1
1
 
 1 Fa / 2,n 2,n1


2
 s2  F
 s2 

2
 2  a / 2,n1,n 2
 2
where n1  n  1 and n 2  n2  1
59
Estimating the Ratio of Two Population Variances
• Example 7
– Determine the 95% confidence interval estimate of the ratio
of the two population variances in example 12.1
– Solution
• We find Fa/2,v1,v2 = F.025,40,120 = 1.61 (approximately)
Fa/2,v2,v1 = F.025,120,40 = 1.72 (approximately)
• LCL = (s12/s22)[1/ Fa/2,v1,v2 ]
= (4102.98/10,669.770)[1/1.61]= .2388
• UCL = (s12/s22)[ Fa/2,v2,v1 ]
= (4102.98/10,669.770)[1.72]= .6614
60
13.6 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
61
Parameter 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 p̂1  p̂ 2 (the
difference between the sample proportions).
62
Sampling distribution of p̂1  p̂ 2
• Two random samples are drawn from two populations.
• The number of successes in each sample is recorded.
• The sample proportions are computed.
Sample 1
Sample size n1
Number of successes x1
Sample proportion
x
ˆp1  1
n1
Sample 2
Sample size n2
Number of successes x2
Sample proportion
x2
p̂ 2 
n2
63
Sampling distribution of p̂1  p̂ 2
• The statistic p̂1  p̂ 2 is approximately normally distributed
if n1p1, n1(1 - p1), n2p2, n2(1 - p2) are all equal to or greater
than 5.
• The mean of p̂1  p̂ 2 is p1 - p2.
• The standard deviation of p̂1  p̂ 2 is
p1 (1 p1 ) p 2 (1 p 2 )

n1
n2
64
The z-statistic
Z
(p̂1  p̂ 2 )  (p1  p 2 )
p1 (1  p1 ) p 2 (1  p 2 )

n1
n2
Because p1 and p2 are unknown, we use
their estimates instead. Thus,
n1p̂1 , n1 (1 p̂1 ), n2p̂2 , n2 (1 p̂2 )
should all be equal to or greater than 5.
65
Testing p1 – p2
• There are two cases to consider:
Case 1:
H0: p1-p2 =0
Case 2:
H0: p1-p2 =D (D is not equal to 0)
Pool the samples and determine the pooled proportion
Keep the sample proportions separate
x1  x 2
p̂ 
n1  n 2
Then
Z
ˆp1  x1 ; ˆp 2  x 2
n1
n2
(p̂1  p̂ 2 )
p̂(1  p̂)(
1
1
 )
n1 n 2
Then
Z
(p̂1  p̂ 2 )  D
p̂1 (1  p̂1 ) p̂ 2 (1  p̂ 2 )

n1
n2
66
Testing p1 – p2 (Case I)
• Example 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.
67
Testing p1 – p2 (Case I)
• 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.
68
Testing p1 – p2 (Case I)
• 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)
Population 1 – purchases of Design A
– The hypotheses test are
Population 2 – purchases of Design B
H0: p1 - p2 = 0
H1: p1 - p2 > 0
– We identify here case 1.
69
Testing p1 – p2 (Case I)
• Solving by hand
– For a 5% significance level the rejection region is
z > za = z.05 = 1.645
From Xm13-08.xls we have:
The sample proportion s are
p̂1  580 904  .6416, and p̂ 2  604 1046  .5774
The pooled proportion is
p̂  ( x 1  x 2 ) (n1  n2 )  (580  604) (904  1046)  .6072
The z statistic becomes
Z
(p̂1  p̂ 2 )  (p1  p 2 )
1 1
p̂(1  p̂)  
 n1 n2 

.6416  .5774
1 
 1
.6072(1  .6072)


904
1046


 2.89
70
Testing p1 – p2 (Case I)
• 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).
71
Additional example
Testing p1 – p2 (Case I)
• Excel (Data Analysis Plus)
z-Test: Two Proportions
sample proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
Community 1 Community 2
0.6416
0.5774
904
1046
0
2.89
0.0019
1.6449
0.0038
1.96
• Conclusion
Since 2.89 > 1.645, there is sufficient evidence in the
data to conclude at 5% significance level, that design
A will outsell design B.
72
Testing p1 – p2 (Case II)
• Example 9 (modifying example 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.
– For design A to be financially viable it has to outsell
design B by at least 3%.
73
Testing p1 – p2 (Case II)
• 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.
74
Testing p1 – p2 (Case II)
• Solution
– The hypotheses to test are
H0: p1 - p2 = .03
H1: p1 - p2 > .03
– We identify case 2 of the test for difference in
proportions (the difference is not equal to zero).
75
Testing p1 – p2 (Case II)
• Solving by hand
Z
(p̂1  p̂ 2 )  D
p̂1 (1  p̂1 ) p̂ 2 (1  p̂ 2 )

n1
n2
 580   604 


  .03
580  324   604  442 

 1.58
.642(1  .642) .577(1  .577)

904
1046
The rejection region is 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.
76
Testing p1 – p2 (Case II)
• Using Excel (Data Analysis Plus)
z-Test: Two Proportions
Sample Proportion
Observations
Hypothesized Difference
z stat
P(Z<=z) one-tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
Community 1
0.6416
904
0.03
1.5467
0.061
1.6449
0.122
1.96
Community 2
0.5774
1046
77
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.
78
Estimating p1 – p2
• 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.
79
Estimating p1 – p2
• 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
80
Estimating p1 – p2
• Solving by hand
1497
1292
 .0730, p̂ 2 
 .0630
– Sample proportions: p̂1 
20500
20500
(p̂1  p̂ 2 ) 
p̂1 (1  p̂1 ) p̂ 2 (1  p̂ 2 )

n1
n2
– The 95% confidence interval is
.0730(1  .0730 ) .0630(1  .0630 )

 .100  .0049
20500
20500
UCL  .0149
.0730  .0630  1.96
LCL  .0051
81
Estimating p1 – p2
• 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
82