2sided pvalues to 1sided

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Lecture 4

• 2-sided p-values to 1-sided
• Inference about the ratio of two variances
• Inference about two proportions
• Review
• Summary
• The Flow approach to problem solving
• Example

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Understanding P-values

• For a two sample t-test the p-value
  is

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From 2-sided to right(left)- sided

• Given a 2-sided p-value, how do we get a 1-sided
  p-value (JMP gives only the former)?
• Right-sided
• if x-difference gt mu-difference (null)
  right-sided p-value 2-sided p-value /2 (!!)
• If x-difference lt mu-difference
  (null) right-sided p-value gt 0.5, so cant
  reject
• Left-sided practice problem

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P-values 2-sided gt right-sided

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13.5 Inference 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.
• It may help us decide which of the equal- or
  unequal-variances t-test of the difference
  between means to use.
• Because variance can measure risk, it allows us
  to compare risk, for example, between two
  portfolios.

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Parameter and Statistic

• Parameter to be tested is s12/s22
• Statistic used is

• 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 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 Example 13.1)

• (see Xm13-01)
• 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 13.1

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Inference about the difference between two
population proportions

• In this section we deal with two populations
  whose data are nominal.
• For nominal data we compare the population
  proportions of the occurrence of a certain event.
• Examples
• Comparing the effectiveness of new drug versus
  older one
• Comparing market share before and after
  advertising campaign
• Comparing defective rates between two machines

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Parameter and Statistic

• Parameter
• When the data are nominal, we can only count the
  occurrences of a certain event in the two
  populations, and calculate proportions.
• The parameter is therefore p1 p2.
• Statistic
• An unbiased estimator of p1 p2 is
  (the difference between the sample proportions).
  

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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 greater than or equal to 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
Because and are unknown the standard
error must be estimated using the sample
proportions. The method depends on the null
hypothesis
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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

• Example 13.8
• The marketing manager needs to decide which of
  two new packaging designs to adopt, to help
  improve sales of his companys soap.
• A study is performed in two supermarkets
• Brightly-colored packaging is distributed in
  supermarket 1.
• Simple packaging is distributed in supermarket 2.
• First design is more expensive, therefore,to be
  financially viable it has to outsell the second
  design.

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Testing p1 p2

• Summary of the experiment results
• Supermarket 1 - 180 purchasers of Johnson
  Brothers soap out of a total of 904
• Supermarket 2 - 155 purchasers of Johnson
  Brothers soap out of a total of 1,038
• Use 5 significance level and perform a test to
  find which type of packaging to use.

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Testing p1 p2

• Example 13.9 (Revisit Example 13.8)
• Management needs to decide which of two new
  packaging designs to adopt, to help improve sales
  of a certain soap.
• A study is performed in two supermarkets
• For the brightly-colored design to be financially
  viable it has to outsell the simple design by at
  least 3.

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Estimating p1 p2

• Estimating the cost of life saved
• Two drugs are used to treat heart attack victims
• Streptokinase (available since 1959, costs 460)
• t-PA (genetically engineered, costs 2900).
• The maker of t-PA claims that its drug
  outperforms Streptokinase.
• An experiment was conducted in 15 countries.
• 20,500 patients were given t-PA
• 20,500 patients were given Streptokinase
• The number of deaths by heart attacks was
  recorded.

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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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A Review of Chapters 12 and 13

• Summary of techniques seen
• Here (Chapter 14) we build a framework that helps
  decide which technique (or techniques) should be
  used in solving a problem.
• Logical flow chart of techniques for Chapters 12
  and 13 is presented next.

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Summary of statistical inferenceChapters 12 and
13

• Problem objective Describe a population.
• Data type Interval
• Descriptive measurement Central location
• Parameter m
• Test statistic
• Interval estimator
• Required condition Normal population

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Summary - continued

• Problem objective Describe a population.
• Data type Interval
• Descriptive measurement Variability.
• Parameter s2
• Test statistic
• Interval estimator
• Required condition normal population.

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Summary - continued

• Problem objective Describe a population.
• Data type Nominal
• Parameter p
• Test statistic
• Interval estimator
• Required condition

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Summary - continued

• Problem objective Compare two populations.
• Data type Interval
• Descriptive measurement Central location
• Experimental design Independent samples
• population variances
• Parameter m1 - m2
• Test statistic Interval estimator
• Required condition Normal populations

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Summary - continued

• Problem objective Compare two populations.
• Data type Interval.
• Descriptive measurement Central location
• Experimental design Independent samples
• population variances
• Parameter m1 - m2
• Test statistic Interval estimator
• Required condition Normal populations

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Summary - continued

• Problem objective Compare two populations.
• Data type Interval.
• Descriptive measurement Central location
• Experimental design Matched pairs
• Parameter mD
• Test statistic Interval estimator
• Required condition Normal differences

d.f. nD - 1
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Problem objective?
Data type?
Data type?
Z test estimator of p1-p2
Type of descriptive measurement?
Type of descriptive measurement?
F- test estimator of s12/s22
Experimental design?
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Identifying the appropriate technique

• Example 14.1
• Is the antilock braking system (ABS) really
  effective?
• Two aspects of the effectiveness were examined
• The number of accidents.
• Cost of repair when accidents do occur.
• An experiment was conducted as follows
• 500 cars with ABS and 500 cars without ABS were
  randomly selected.
• For each car it was recorded whether the car was
  involved in an accident.
• If a car was involved with an accident, the cost
  of repair was recorded.

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Identifying the appropriate technique

• Example continued
• Data
• 42 cars without ABS had an accident,
• 38 cars equipped with ABS had an accident
• The costs of repairs were recorded (see Xm14-01).
• Can we conclude that ABS is effective?

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Identifying the appropriate technique

• Question 2 Is there sufficient evidence to infer
  that the cost of repairing accident damage in ABS
  equipped cars is less than that of cars without
  ABS?
• Question 3 How much cheaper is it to repair ABS
  equipped cars than cars without ABS?

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Question 2 Compare the mean repair costs per
accident

• Solution

Problem objective?
Describing a single population
Compare two populations
Data type?
Cost of repair per accident
Nominal
Interval
Type of descriptive measurements?
Central location
Variability
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Question 2 Compare the mean repair costs per
accident
Central location

• Solution - continued

Experimental design?
Independent samples
Matched pairs
Run the F test for the ratio of two variances.
Equal
Unequal
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Question 2 Compare the mean repair costs per
accident

• Solution continued
• m1 mean cost of repairing cars without ABS m2
  mean cost of repairing cars with ABS
• The hypotheses tested H0 m1 m2 0 H1 m1
  m2 gt 0
• For the equal variance case we use

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Question 2 Compare the mean repair costs per
accident

• Solution continued
• To determine whether the population variances
  differ we apply the F test
• From JMP we have (Xm14-01)

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Question 2 Compare the mean repair costs per
accident

• Solution continued
• Assuming the variances are really equal we run
  the equal-variances t-test of the difference
  between two means

At 5 significance levelthere is sufficient
evidenceto infer that the cost of repairsafter
accidents for cars with ABS is smaller than the
cost of repairs for cars without ABS.
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Checking required conditions

• The two populations should be normal (or at least
  not extremely nonnormal)

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Question 3 Estimate the difference in repair
costs

• Solution
• Use Estimators Workbook t-Test_2 Means (Eq-Var)
  worksheet

We estimate that the cost of repairing a car not
equipped with ABS is between 71 and 651 more
expensive than to repair an ABS equipped car.