copyright © 2011 pearson education, inc. alternative approaches to inference chapter 17
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Copyright © 2011 Pearson Education, Inc.
Alternative Approaches to
Inference
Chapter 17
17.1 A Confidence Interval for the Median
An auto insurance company is thinking about compensating agents by comparing the number of claims they produce to a standard. Annual claims average near $3,200 with a median claim of $2,000.
Claims are highly skewed Use nonparametric methods that don’t rely on a
normal sampling distribution
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17.1 A Confidence Interval for the Median
Distribution of Sample of Claims (n = 42)
For this sample, the average claim is $3,632 with s = $4,254. The median claim is $2,456.
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17.1 A Confidence Interval for the Median
Is Sample Mean Compatible with µ=$3,200?
To answer this question, construct a 95% confidence interval for µ
This interval is $3,632 ± 2.02 x $4,254 / [$2,306 to $4,958]
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42
17.1 A Confidence Interval for the Median
Is Sample Mean Compatible with µ=$3,200?
The national average of $3,200 lies within the 95% confidence t-interval for the mean.
BUT…the sample does not satisfy the sample size condition necessary to use the t-interval.
The t-interval is unreliable with unknown coverage when the conditions are not met.
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17.1 A Confidence Interval for the Median
Nonparametric Statistics
Avoid making assumptions about the shape of the population.
Often rely on sorting the data.
Suited to parameters such as the population median θ (theta).
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17.1 A Confidence Interval for the Median
Nonparametric Statistics
For the claims data that are highly skewed to the right, θ < µ.
If the population distribution is symmetric, then θ = µ.
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17.1 A Confidence Interval for the Median
Nonparametric Confidence Interval
First step in finding a confidence interval for θ is to sort the observed data in ascending order (known as order statistics).
Order statistics are denoted asX(1) < X(2) < … < X(n)
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17.1 A Confidence Interval for the Median
Nonparametric Confidence Interval
If data are an SRS from a population with median θ, then we know
1. The probability that a random draw from the population is less than or equal to θ is ½,
2. The observations in the random sample are independent.
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17.1 A Confidence Interval for the Median
Nonparametric Confidence Interval
Determine the probabilities that the population median lies between ordered observations using the binomial distribution.
To form the confidence interval for θ combine several segments to achieve desired coverage.
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17.1 A Confidence Interval for the Median
Nonparametric Confidence Interval
In general, can’t construct a confidence interval for θ whose coverage is exactly 0.95.
The 94.6% confidence interval for the median claim is [$1,217 to $3,168].
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17.1 A Confidence Interval for the Median
Parametric versus Nonparametric
Limitations of nonparametric methods
1. Coverage is limited to certain values determined by sums of binomial probabilities (difficult to obtain exactly 95% coverage).
2. Median is not equal to the mean if the population distribution is skewed. This prohibits obtaining estimates for the total (total = nµ).
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17.2 Transformations
Transform Data into Symmetric Distributions
Taking base 10 logs of the claims data results in a more symmetric distribution.
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17.2 Transformations
Transform Data into Symmetric Distributions
Taking base 10 logs of the claims data results in data that could be from a normal distribution.
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17.2 Transformations
Transform Data into Symmetric Distributions
If y = log10 x, then = 3.312 with sy = 0.493.
The 95% confidence t-interval for µy is
[3.16 to 3.47].
If we convert back to the original scale of dollars, this interval resembles that for the median rather than that for the mean.
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y
17.3 Prediction Intervals
Prediction Interval: an interval that holds a future
draw from the population with chosen probability.
For the auto insurance example, a prediction interval anticipates the size of the next claim, allowing for the random variation associated with an individual.
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17.3 Prediction Intervals
For a Normal Population
The 100 (1 – α)% prediction interval for an independent draw from a normal population is
where and s estimate µ and σ.
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nstx
n
11
1,2/
x
17.3 Prediction Intervals
Nonparametric Prediction Interval
Relies on the properties of order statistics:
P(X(i) ≤ X ≤ X(i+1)) = 1/(n + 1)
P(X ≤ X(1)) = 1/(n + 1)
P(X(n) ≤ X) = 1/(n + 1)
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17.3 Prediction Intervals
Nonparametric Prediction Interval
Combine segments to get desired coverage.
P (X(2) ≤ X ≤ X(41)) = P ($255 ≤ X ≤ $17,305)
= (41 – 2)/43 0.91
There is a 91% chance that the next claim is between $255 and $17,305.
,
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4M Example 17.1: EXECUTIVE SALARIES
Motivation
Fees earned by an executive placement service are 5% of the starting annual total compensation package. How much can the firm expect to earn by placing a current client as a CEO in the telecom industry?
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4M Example 17.1: EXECUTIVE SALARIES
Method
Obtain data (n = 23 CEOs from telecom industry).
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4M Example 17.1: EXECUTIVE SALARIES
Method
The distribution of total compensation for CEOs in the telecom industry is not normal. Construct a nonparametric prediction interval for the client’s anticipated total compensation package.
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4M Example 17.1: EXECUTIVE SALARIES
Mechanics
Sort the data:
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4M Example 17.1: EXECUTIVE SALARIES
Mechanics
The interval x(3) to x(21) is
$743,801 to $29,863,393
and is a 75% prediction interval.
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4M Example 17.1: EXECUTIVE SALARIES
Message
The compensation package of three out of four placements in this industry is predicted to be in the range from about $750,000 to $30,000,000. The implied fee ranges from $37,500 to $1,500,000.
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17.4 Proportions Based on Small Samples
Wilson’s Interval for a Proportion
An adjustment that moves the sampling distribution of closer to ½ and away from the troublesome boundaries at 0 and 1.
Add four artificial cases (2 successes and 2 failures) to create an adjusted proportion .
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p̂
p~
17.4 Proportions Based on Small Samples
Wilson’s Interval for a Proportion
Add 2 successes and 2 failures to the data and define = (# of successes+2)/n+4 ( = n+4).
The z-interval is
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p~ n~
n
ppzp ~
)~1(~~2/
4M Example 17.2: DRUG TESTING
Motivation
A company is developing a drug to prolong time before a relapse of cancer. The drug must cut the rate of relapse in half. To test this drug, the company first needs to know the current time to relapse.
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4M Example 17.2: DRUG TESTING
Method
Data are collected for 19 patients who were observed for 24 months. Doctors found a relapse in 9 of the 19 patients. While the SRS condition is satisfied, the sample size condition is not. Use Wilson’s interval for a proportion.
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4M Example 17.2: DRUG TESTING
Mechanics
By adding two successes and two failures, we have
The interval is
0.478 ± 1.96 = [0.27 to 0.68]
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478.0)419/()29(~ p
)419/()478.01(478.0
4M Example 17.2: DRUG TESTING
Message
We are 95% confident that the proportion of patients with this cancer that relapse within 24 months is between 27% and 68%. In order to cut this proportion in half, the drug will have to reduce this rate to somewhere between 13% and 34%.
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Best Practices
Check the assumptions carefully when dealing with small samples.
Consider a nonparametric alternative if you suspect non-normal data.
Use the adjustment procedure for proportions from small samples.
Verify that your data are an SRS.
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Pitfalls
Avoid assuming that populations are normally distributed in order to use a t – interval for the mean.
Do not use confidence intervals based on normality just because they are narrower than a nonparametric interval.
Do not think that you can prove normality using a normal quantile plot.
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Pitfalls (Continued)
Do not rely on software to know which procedure to use.
Do not use a confidence interval when you need a prediction interval.
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