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

17.1 A Confidence Interval for the Median Nonparametric Statistics population... 17.1 A Confidence Interval for the Median Nonparametric Confidence Interval  First step in finding a co

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Alternative Approaches to

Inference

Chapter 17

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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

sampling distribution

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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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Median

Is Sample Mean Compatible with µ=$3,200?

interval for µ

 This interval is

$3,632 ± 2.02 x $4,254 /

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Median

Is Sample Mean Compatible with µ=$3,200?

confidence t-interval for the mean.

condition necessary to use the t-interval.

the conditions are not met

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Median

Nonparametric Statistics

population

(theta)

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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)

X(1) < X(2) < … < X(n)

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Median

we know

less than or equal to θ is ½,

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Median

between ordered observations using the binomial

distribution

segments to achieve desired coverage

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Median

whose coverage is exactly 0.95

[$1,217 to $3,168]

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Median

Parametric versus Nonparametric

of binomial probabilities (difficult to obtain exactly 95%

coverage)

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

 If y = log10 x, then = 3.312 with s y = 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

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17.3 Prediction Intervals

from the population with chosen probability

anticipates the size of the next claim, allowing for the

random variation associated with an individual

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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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n

s t

x ± α / 2,n−1 1 + 1

x

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Nonparametric Prediction Interval

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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Nonparametric Prediction Interval

 Combine segments to get desired coverage

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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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EXECUTIVE SALARIES Method

Obtain data (n = 23 CEOs from telecom industry)

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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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EXECUTIVE SALARIES Mechanics

Sort the data:

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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

closer to ½ and away from the troublesome boundaries at

0 and 1

create an adjusted proportion

pˆ

p

~

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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

p ~ ( 1 ~ ~ )

α

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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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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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0 )

4 19

/(

) 2 9

(

~ p = + + ≈

)419

/(

)478

01(478

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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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in order to use a t – interval for the mean.

because they are narrower than a nonparametric interval

quantile plot

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Pitfalls (Continued)

prediction interval

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