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Chapter 13

# Chapter 13 Textbook Study Guide

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York University

Operations Management and Information System

OMIS 2010

Alan Marshall

Fall

Description

Chapter 13: Inference About Two Populations
13.1 Introduction
The methods of drawing inferences when comparing two populations were discussed in this chapter.
When the data are quantitative and we want to compare measures of location, the parameter of interest is
the difference between two m eans, 1 2 . When the descriptive measure is variability, we draw
inferences about the ratio of t wo variances / 2 . In problems where the data are q ualitative, the
1 2
parameter of interest is the difference between two proportiop1 p 2.
Because three new param eters were introduced in th is chapter, the im portance of being capable of
identifying the correct technique to use is growing. As you work your way through each exam ple and
exercise below, make certain that you understand how we det ermine what the parameter is and what
statistical technique needs to be employed.
You are expected to know the following by the time you finish this chapter:
1. How to apply the concepts and techniques of estimation and hypothesis testing introduced in
Chapters 10 and 11.
2. How to recognize when the parameter of interest is the difference between two population
means.
3. How to recognize when the samples were independently drawn and when they were taken from a
matched pairs experiment.
4. How to determine when to use the equal-variances and unequal variances t-test and estimator of
.
1 2
5. How to recognize when the parameter of interest is the ratio of two population variances.
6. How to recognize when the parameter of interest is the difference between two proportions.
7. How to determine when to use the Case 1 test statistic and when to use the Case 2 test statistic
when testing hypotheses about the difference between two proportions.
13.2 Inference About The Difference Between Two Means: Independent
Samples
In this section, the problem objective was to compare two populations. The data are quantitative, the
descriptive measure is location, and the sam ples are independently drawn. Hence, t he parameter we
estimate and test is1 2.
There are t wo sets of form ulas used t o make inferences about 1 2 . The formulas and the
conditions that tell us when they are used are listed below.
2 2
1. When the population variances are unknown and equal ( 1 = 2), the test statistic is
152 (x1 x 2 ( 1 ) 2
t = d.f. = 1 + n22
sp (1/ n1+ 1/ n 2
The interval estimator is
(x1 x )2 t /2 sp(1/ n 1 1/ n 2
2. In problems where the population variances are unknown and unequal, the test statistic is
(x x )( ) (s / n + s / n ) 2
t = 1 2 1 2 d.f. = 1 1 2 2
s / n + s / n (1 / n1)2 (s2 / 2 )
1 1 2 2 +
n11 n21
The interval estimate is
2 2
(x1 x 2t /2 1 / n1+s 2 n 2
Example 13.1
A study of t he scholastic aptitude test (SAT) reveal ed that in a random sample of 100 males the
mean SAT score was 431.5 wi th a st andard deviation of 93.7. A random sample of 100 female SAT
scores produced a m ean of 423.9 wi th a st andard deviation of 88.6. C an we concl ude at t he 1%
significance level that male and female scores differ?
Solution
The problem objective is to compare male and female SAT scores, a quantitative variable.
Therefore, the parameter of interest is (1 2. The sample standard deviations are similar, making it
reasonable to assum e that = . Thus, we em ploy the equal-variances t-test. The complete test
1 2
follows.
H 0: (1 )2= 0
H :1( 1 ) 20
R ejection region: t >/2,n+n 22 = .005,198 2.601or t < -2.601
(x x )( ) (431.5 423.9)0
Test statistic: t = 1 2 1 2 = =.59
1 1
sp 1 + 1 8315 +
n1 n2 100 100
Conclusion: Do not reject H 0.
There is not enough evidence to conclude that male and female SAT scores differ.
153 Example 13.2
Estimate the difference between m ale and fem ale mean SAT scores (i n Example 13.1) wi th 90%
confidence.
Solution
The interval estimate is
1 1 1 1
(x1 x2)t /2 p + = (431.5 423.9)1.653 8315 +
n1 n2 100 100
= 7.6 21.3
Example 13.3
A statistician wants to compare the relative success of t wo large department store chains. She
decides to measure the sales per square foot. She takes a random sample of five stores from chain 1 and
five stores from chain 2. The gross sales per square foot are shown bel ow. Do t hese data provide
sufficient evidence to indicate that the two chains differ? (Test with = .10.)
Chain 1 Chain 2
65.50 82.50
72.00 63.50
103.00 68.00
93.50 70.00
82.60 66.50
Solution
The problem objective is to compare the population of sales per square foot of one chai n with the
population of sales per square foot of another chain. As a resul t, the parameter of interest1 2.
The sample variances are quite different; we can assume that 2 2. The appropriate technique is the
1 2
unequal variances t-test o 1 2 .
H 0: (1 2) = 0
H 1 (1 2) 0
2 2 2
(x 1 x 2( 1 2) (s1/ n 1 s 2 n 2
Test statistict = 2 2 d.f. = (s / n )2 (s2 / n )
s1 / 1 + 2 / 2 1 1 + 2 2
n11 n21
R ejection region: t t t =1.943 or t < -1.943
/2,d.f . .05,6
154 Value of the test statistic:
x1=83.32 x2 = 70.10
s = 234.29 s2 = 53.67
1 2
2
d.f. = ( )4.29/ 5+ 53.67/ 5
234.29 / 52 53.67/ 5 2
( ) + ( )
4 4
= 5.7 6
(x x )( ) (83.32 70.10)0
t= 1 2 1 2 = =1.74
sp(1/ n1+1/ n2 ) 234.29/ 5+ 53.67/ 5
Conclusion: Do not reject H0.
There is not enough evidence to indicate that the mean sales per square foot differ between the two
chains.
EXERCISES
13.1 Test to determine if there is enough evidence to indicate tha1 exceeds 2 for the following
situation. (Use = .05)
x = 28 x = 20
1 2
s1=10 s2=12
n1 = 50 n2 = 40
H0:
H 1
Test statistic:
Rejection region:
155

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