Inference about Comparing Two Populations

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Inference about Comparing Two Populations

• Chapter 13

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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.

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13.2 Inference about the Difference between Two
Means Independent Samples

• Two random samples are drawn from the two
  populations of interest.
• Because we compare two population means, we use
  the statistic .

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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 s12/n1
  s22/n2

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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

• Calculate the pooled variance estimate by

The pooled Variance estimator
n2 15
n1 10
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Inference about m1 m2 Equal variances

• Calculate the pooled variance estimate by

The pooled Variance estimator
n2 15
n1 10
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Inference about m1 m2 Equal variances

• Construct the t-statistic as follows

• Perform a hypothesis test
• H0 m1 - m2 0
• Ha m1 - m2 gt 0

or lt 0
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Inference about m1 m2 Unequal variances
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Inference about m1 m2 Unequal variances
Run a hypothesis test as needed, or, build a
confidence interval
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Which case to useEqual variance or unequal
variance?

• Whenever there is sufficient 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). We have S12
  4103, and S22 10,670.
• It appears that the variances are unequal.

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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
• The rejection region is t lt -ta,n -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 is

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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).
• We have S12 0.8478, and S22 1.3031.
• The two variances 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 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 m2
A 95 confidence interval for m1 - m2 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
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Checking the required Conditions for the equal
variances case (example 13.2)
Testing Normality
Testing m1-m2, non-normal populations
The data appear to be approximately normal
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13.4 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
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 do 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
  where(m1 m2 gt 0)?

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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.

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Reducing the variability
The range of observations sample A
The values each sample consists of might markedly
vary...
The range of observations sample B
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Reducing the variability
Differences
...but the differences between pairs of
observations might be quite close to one
another, resulting in a small variability of the
differences.
The range of the differences
0
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The 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 mD (the mean of the
  differences) rather in terms of m1 m2.
• This formulation has the benefit of a smaller
  variability.

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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

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The matched pairs hypothesis test
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, then run the
confidence interval procedure in Data Analysis
Plus.
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Checking the required conditionsfor the paired
observations case
Testing Normality
Testing mD non-normal populations

• 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 ratio
  of 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 with n1 n1 1, and n2 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 perform a test regarding average
  consumption of calories at peoples lunch in
  relation to the inclusion of high-fiber cereal in
  their breakfast, the variance ratio of two
  samples has to be tested first.

Calories intake at lunch
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Testing the ratio of two population variances

• Solving by hand
• The rejection region is FgtFa/2,n1,n2
  or Flt1/Fa/2,n2,n1

• 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)
• In order to perform a test regarding average
  consumption of calories at peoples lunch in
  relation to the inclusion of high-fiber cereal in
  their breakfast, the variance ratio of two
  samples has to be tested first.

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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 the 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)
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