Statistics for business decision making and analysis robert stine and foster chapter 18

18.1 Data for ComparisonsComparison of Two Diets  Frame as a test of the difference between the means of two populations mean number of pounds lost on Atkins versus conventional diets

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Comparison

Chapter 18

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18.1 Data for Comparisons

A fitness chain is considering licensing a

proprietary diet at a cost of $200,000 Is it more effective than the conventional free

government recommended food pyramid?

 Use inferential statistics to test for differences

between two populations

 Test for the difference between two means

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Comparison of Two Diets

 Frame as a test of the difference between the

means of two populations (mean number of pounds lost on Atkins versus conventional diets)

 Let µA denote the mean weight loss in the

population if members go on the Atkins diet and µCdenote the mean weight loss in the population if

members go on the conventional diet.

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 In order to be profitable for the fitness chain, the Atkins diet has to win by more than 2 pounds, on average.

 State the hypotheses as:

H0: µA - µC ≤ 2

H : µ - µ > 2

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 Data used to compare two groups typically arise

in one of three ways:

1 Run an experiment that isolates a specific cause.

2 Obtain random samples from two populations.

3 Compare two sets of observations.

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Experiments

 Experiment: procedure that uses randomization

to produce data that reveal causation.

 Factor: a variable manipulated to discover its

effect on a second variable, the response.

 Treatment: a level of a factor.

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Experiments

 In the ideal experiment, the experimenter

1 Selects a random sample from a population.

2 Assigns subjects at random to treatments defined

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 The factor in the comparison of diets is the diet

offered.

 There are two treatments: Atkins and conventional.

 The response is the amount of weight lost

(measured in pounds)

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Confounding

 Confounding: mixing the effects of two or more

factors when comparing treatments.

 Randomization eliminates confounding.

 If it is not possible to randomize, then sample

independently from two populations.

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) (

2 1

0 2

1

x x

se

D x

x t





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18.2 Two-Sample t - Test

Two-Sample t – Test Summary

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18.2 Two-Sample t - Test

Two-Sample t – Test Checklist

 No obvious lurking variables.

 SRS condition.

 Similar variances While the test allows the

variances to be different, should notice if they are similar.

 Sample size condition Each sample must satisfy this condition.

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4M Example 18.1:

COMPARING TWO DIETS

Motivation

Scientists at U Penn selected 63 subjects

from the local population of obese adults They randomly assigned 33 to the Atkins

diet and 30 to the conventional diet Do the results show that the Atkins diet is worth

licensing?

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COMPARING TWO DIETS

Method – Check Conditions

Since the interquartile ranges of the boxplots appear similar, we can assume similar

variances.

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COMPARING TWO DIETS

 No obvious lurking variables because of randomization.

 SRS condition satisfied.

 Both samples meet the sample size condition.

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COMPARING TWO DIETS

Mechanics

with 60.8255 df and p-value = 0.0308; reject H 0

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91

1369

.3

2)

00

742

.15

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COMPARING TWO DIETS

Message

The experiment shows that the average

weight loss of obese adults on the Atkins

diet does exceed the average weight loss

of obese adults on the conventional diet

Unless the fitness chain’s membership

resembles this population (obese adults),

these results may not apply

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18.3 Confidence Interval for the

Difference

95% Confidence Intervals for µA and µC

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Difference

95% Confidence Intervals for µA and µC

The confidence intervals overlap If they were

nonoverlapping, we could conclude a significant

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Difference

95% Confidence Interval for µ1 - µ2

The 100(1 – α)% two-sample confidence

interval for the difference in means is

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) (

) ( x1  x 2  t / 2se x1  x 2

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Difference

95% Confidence Interval for µA - µc

Since the 95% confidence interval for µA - µB does not include zero, the means are statistically significantly different (those on the Atkins diet lose on average

between 1.7 and 15.2 pounds more than those on a

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EVALUATING A PROMOTION

Motivation

To evaluate the effectiveness of a

promotional offer, an overnight service

pulled records for a random sample of 50

offices that received the promotion and a

random sample of 75 that did not

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Method

Use the two-sample t –interval Let µyes

denote the mean number of packages

shipped by offices that received the

promotion and µno denote the mean

number of packages shipped by offices

that did not

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All conditions are satisfied with the exception

of no obvious lurking variables Since we

don’t know how the overnight delivery

service distributed the promotional offer,

confounding is possible For example, it

could be the case that only larger offices

received the promotion

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Mechanics

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Message

The difference is statistically significant

Offices that received the promotion used

the overnight service to ship from 4 to 21

more packages on average than those

offices that did not receive the promotion There is the possibility of a confounding

effect

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18.4 Other Comparisons

Comparisons Using Confidence Intervals

 Other possible comparisons include comparing

two proportions or comparing two means from

paired data.

 95% confidence intervals generally have the form

Estimated Difference ± 2 Estimated Standard 

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Sample size condition (for proportion).

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) ˆ ˆ

( )

ˆ ˆ

( p1  p2  z / 2se p1  p2

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designs for the coming fall season (one

featuring red and the other violet) If

customers in the two regions differ in their preferences, the buyer will have to do a

special order for each district

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COLOR PREFERENCES

Mechanics

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COLOR PREFERENCES

Mechanics

Based on the data,

and the 95% confidence interval is

0.1389 ±1.96 (0.08645) [-0.031 to 0.308]

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1389

0 4444

0 5833

0 ˆ

ˆE  pW   

p

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interval for the difference between

proportions contains zero

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Paired Comparisons

 Paired comparison: a comparison of two

treatments using dependent samples designed to

be similar (e.g., the same individuals taste test

Coke and Pepsi).

 Pairing isolates the treatment effect by reducing

random variation that can hide a difference.

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Paired Comparisons

 Given paired data, we begin the analysis by

forming the difference within each pair

(i.e., d i = x i – y i ).

 A two-sample analysis becomes a one-sample

analysis Let denote the mean of the

differences and s their standard deviation.

d

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Sample size condition.

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n

s t

d   / 2 ; n  1 d

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SALES FORCE COMPARISON

Motivation

The merger of two pharmaceutical

companies (A and B) allows senior

management to eliminate one of the sales forces Which one should the merged

company eliminate?

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Method

Both sales forces market similar products

and were organized into 20 comparable

geographical districts Use the differences obtained from subtracting sales for Division

B from sales for Division A in each district

to obtain a 95% confidence t-interval for

µA - µB

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Inspect histogram of differences:

All conditions are satisfied

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Mechanics

The 95% t-interval for the mean differences does not include

zero There is a statistically significant difference.

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The high correlation (r = 0.97) of sales between

Sales Force A and Sales Force B in these

districts confirms the benefit of a paired

comparison.

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Best Practices

 Use experiments to discover causal relationships.

 Plot your data

 Use a break-even analysis to formulate the null

hypothesis.

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Best Practices (Continued)

 Use one confidence interval for comparisons

 Compare the variances in the two samples.

 Take advantage of paired comparisons.

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 Don’t forget confounding.

 Do not assume that a confidence interval that

includes zero means that the difference is zero.

 Don’t confuse a two-sample comparison with a

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