Statistics for business decision making and analysis robert stine and foster chapter 17

Chapter 17

Alternative
Approaches to
Inference
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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]

42

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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 as
X(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,
The observations in the random sample are independent.

2.

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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).
Median is not equal to the mean if the population
distribution is skewed. This prohibits obtaining estimates
for the total (total = nµ).

2.

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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

=y3.312 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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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

1
x ± tα / 2 ,n−1 s 1 +
n

and s estimate µ and σ.

x
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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
Motivation SALARIES
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
Mechanics SALARIES
Sort the data:

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4M Example 17.1:
Mechanics
EXECUTIVE
SALARIES
The interval x(3) to x(21) is
$743,801 to $29,863,393
and is a 75% prediction interval.

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