Inference about Comparing Two Populations

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Inference about Comparing Two Populations

• Chapter 13

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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 dont.
• 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 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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13.2 Inference about the Difference between Two
Means Independent Samples

• Well look at the relationship between the two
  population means by analyzing the value of m1
  m2.

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The 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 m1 -
  m2
• The variance of is

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Making an inference about m1 m2

• 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 m1 - m2

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Making an inference about m1 m2

• Practically, the Z statistic is hardly used,
  because the population variances are not known.

t
S22
S12
?
?

• Instead, we construct a t statistic using the
  
• sample variances (S12 and S22).

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Making an inference about m1 m2

• Two cases are considered when producing the
  t-statistic.
• The two unknown population variances are equal.
• The two unknown population variances are not
  equal.

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Inference about m1 m2 Equal variances

• If the two variances s12 and s22 are equal to
  one another, then their estimate S12 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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Inference about m1 m2 Equal variances

• Calculate the pooled variance estimate by

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Inference about m1 m2 Equal variances

• Calculate the pooled variance estimate by

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Inference about m1 m2 Equal variances

• Construct the t-statistic as follows

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Inference about m1 m2 Unequal variances
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Which case to useEqual 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
³
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Example Making an inference about m1 m2

• 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.

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Example Making an inference about m1 m2

• 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).

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Example Making an inference about m1 m2

• The hypotheses are
• H0 m1 - m2 0
• H1 m1 - m2 lt 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 S12 4103, and S22 10,670.
• It appears that the variances are unequal.

m1 mean caloric intake for fiber consumers m2
mean caloric intake for fiber non-consumers
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Example Making an inference about m1 m2

• Solving by hand
• From the data we have

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Example Making an inference about m1 m2

• Solving by hand
• H1 m1 - m2 lt 0 The rejection region is t lt
  -ta,df -t.05,123 _at_ -1.658

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Example Making an inference about m1 m2
Xm13-1.xls
At 5 significance level there is sufficient
evidence to reject the null hypothesis.
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Example Making an inference about m1 m2

• Solving by hand
• The confidence interval estimator for the
  differencebetween two means when the variances
  are unequal is

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Example Making an inference about m1 m2
Note that the confidence interval for the
differencebetween the two means falls entirely
in the negativeregion -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 m2 ! This conclusion agrees with
the results of the test performed before.
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Example Making an inference about m1 m2

• 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.

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Example Making an inference about m1 m2

• 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?

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Example Making an inference about m1 m2
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 m1 m2

• Solving by hand
• The hypotheses test is
• H0 m1 - m2 0 H1 m1 - m2 ¹
  0

• To check the relationship between the two
  variances we calculate the value of S12 and S22
  (Xm13-02.xls).
• From the data we have S12 0.8478, and S22
  1.3031.so s12 and s22 appear to be equal.

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Example Making an inference about m1 m2

• Solving by hand

• To calculate the t-statistic we have

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Example Making an inference about m1 m2

• The 2-tail rejection region is t lt -ta/2,n
  -t.025,48 -2.009 or t gt ta/2,n
  t.025,48 2.009
• The test Since t -2.009 lt 0.93 lt 2.009, there
  is insufficient evidence to reject the null
  hypothesis.

For a 0.05
2.009
.093
-2.009
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Example Making an inference about m1 m2
Xm13-02.xls
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Example Making an inference about m1 m2

• Conclusion From this experiment, it is unclear
  at 5 significance level if the two assembly
  methods are different in terms of assembly time

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Example Making an inference about m1
m2Constructing a Confidence Interval
A 95 confidence interval for m1 - m2 when the
two variances areequal is calculated as follows
Thus, at 95 confidence level -0.3176 lt m1 - m2 lt
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)
The data appear to be approximately normal
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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
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.
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13.4 Matched Pairs Experiment
Additional example

• 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?.

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13.4 Matched Pairs Experiment

• Solution
• Compare two populations of quantitative data.
• The parameter tested is m1 - m2

m1
The mean of the highest salaryoffered to Finance
MBAs

• H0 m1 - m2 0 H1 m1 - m2 gt 0

m2
The mean of the highest salaryoffered to
Marketing MBAs
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13.4 Matched Pairs Experiment

• Solution continued

• Let us assume equal variances

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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?

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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 it becomes more difficult to reject
  H0.

Recall that rejection of thenull hypothesis
occurs whent is sufficiently large (tgtta). A
large Sp2 reduces t and therefore it does not
fall inthe rejection region.
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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).

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The matched pairs experiment Are we still
testing m1 m2?

• Note that the difference between the two means is
  equal to the mean difference of pairs of
  observations
• A short example

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The matched pairs experiment Reducing the
variability
The range of observations sample A
Observations might markedly differ...
The range of observations sample B
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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
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The matched pairs experiment

• Example 12.4 (12.3 part II)
• It was suspected that salary offers were affected
  by students GPA, (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?

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The matched pairs hypothesis test

• Solution (by hand)
• The parameter tested is mD (m1 m2)
• The hypothesesH0 mD 0H1 mD gt 0
• The t statistic

The rejection region is t gt t.05,25-1 1.711
Degrees of freedom nD 1
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The matched pairs hypothesis test

• Solution (by hand) continue
• From the data (Xm13-4.xls) calculate

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The matched pairs hypothesis test

• Solution (by hand) continue
• Calculate t

See conclusion later
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The matched pairs hypothesis test
Using Data Analysis in Excel
Xm13-4.xls
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The matched pairs hypothesis test
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.
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The matched pairs mean difference estimation
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The matched pairs mean difference estimation
Using Data Analysis Plus
Xm13-4.xls
First calculate the differences for each pair,
then run the confidence interval procedure in
Data Analysis Plus.
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Checking the required conditionsfor the paired
observations case

• The validity of the results depends on the
  normality of the differences.

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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

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Parameter tested and statistic

• The parameter tested is s12/s22
• The statistic used is

• The Sampling distribution of s12/s22
• The statistic s12/s12 / s22/s22 follows the
  F distribution withNumerator d.f. n1 1, and
  Denominator d.f. n2 1.

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Parameter tested and statistic

• Our null hypothesis is always
• H0 s12 / s22 1

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Testing the ratio of two population variances
Example 13.6 (revisiting 13.1)

• (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.

Calories intake at lunch
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Testing the ratio of two population variances

• Solving by hand
• The rejection region is

• The F statistic value is FS12/S22 .3845
• Conclusion Because .3845lt.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.

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Testing the ratio of two population variances
Example 13.6 (revisiting 13.1)

• (see Xm13.1)

From Data Analysis
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Estimating the Ratio of Two Population Variances

• From the statistic F s12/s12 / s22/s22 we
  can isolate s12/s22 and build the following
  confidence interval

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Estimating the Ratio of Two Population Variances

• Example 13.7
• 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

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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

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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
  (the difference between the sample proportions).
  

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Sampling 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 1 Sample size n1 Number of successes
x1 Sample proportion
Sample 2 Sample size n2 Number of successes
x2 Sample proportion
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Sampling distribution of

• 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)

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The z-statistic
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Testing p1 p2

• There are two cases to consider

Case 1 H0 p1-p2 0 Calculate the pooled
proportion
Case 2 H0 p1-p2 D (D is not equal to 0) Do
not pool the data
Then
Then
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Testing p1 p2 (Case I)

• 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.

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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.

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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)
• The hypotheses test are H0 p1 - p2 0 H1 p1
  - p2 gt 0
• We identify here case 1.

Population 1 purchases of Design A
Population 2 purchases of Design B
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Testing p1 p2 (Case I)

• Solving by hand
• For a 5 significance level the rejection region
  isz gt za z.05 1.645

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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 gt
  1.645).

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Testing p1 p2 (Case I)
Additional example

• Excel (Data Analysis Plus)

• Conclusion
• Since 2.89 gt 1.645, there is sufficient evidence
  in the data to conclude at 5 significance level,
  that design A will outsell design B.

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Testing p1 p2 (Case II)

• 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.

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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.

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Testing p1 p2 (Case II)

• Solution
• The hypotheses to test are H0 p1 - p2
  .03 H1 p1 - p2 gt .03
• We identify case 2 of the test for difference in
  proportions (the difference is not equal to zero).

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Testing p1 p2 (Case II)

• Solving by hand

The rejection region is z gt za z.05
1.645. Conclusion Since 1.58 lt 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.
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Testing p1 p2 (Case II)

• Using Excel (Data Analysis Plus)

Xm13-08.xls
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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.

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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.

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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

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Estimating p1 p2

• Solving by hand
• Sample proportions
• The 95 confidence interval is

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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