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Chapter 13

OMIS 2010 Chapter Notes - Chapter 13: Statistical Inference, Interval Estimation, Test Statistic


Department
Operations Management and Information System
Course Code
OMIS 2010
Professor
Alan Marshall
Chapter
13

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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 means,
µ
1
µ
2. When the descriptive measure is variability, we draw
inferences about the ratio of two variances
σ
1
2/
σ
2
2. In problems where the data are qualitative, the
parameter of interest is the difference between two proportions, p1
p2.
Because three new parameters were introduced in this chapter, the importance of being capable of
identifying the correct technique to use is growing. As you work your way through each example and
exercise below, make certain that you understand how we determine 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
. µ1−µ
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 samples are independently drawn. Hence, the parameter we
estimate and test is µ.
1−µ
2
There are two sets of formulas used to make inferences about
µ
1
µ
2. The formulas and the
conditions that tell us when they are used are listed below.
1. When the population variances are unknown and equal (
σ
1
2
=
σ
2
2), the test statistic is
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t=(
x
1
x
2)(µ1−µ
2)
sp
2(1 / n1+1/n2) d.f. = n1
+
n2
2
The interval estimator is
(x
1x
2)±tα/2 sp
2(1 / n1+1/n2
2. In problems where the population variances are unknown and unequal, the test statistic is
t=(
x
1
x
2)
(µ1−µ2)
s1
2/n1+s2
2/n2 d.f. = (s1
2/n1+s2
2/n2)2
(s1
2/n1)2
n1–1 +(s2
2/n2)2
n2–1
The interval estimate is
(x
1x
2)±t
α
/2 s1
2/n1+s2
2/n2
Example 13.1
A study of the scholastic aptitude test (SAT) revealed that in a random sample of 100 males the
mean SAT score was 431.5 with a standard deviation of 93.7. A random sample of 100 female SAT
scores produced a mean of 423.9 with a standard deviation of 88.6. Can we conclude at the 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 (
µ
1
µ
2). The sample standard deviations are similar, making it
reasonable to assume that σ = σ. Thus, we employ the equal-variances t-test. The complete test
follows. 1
22
2
H
0: (
µ
1
µ
2) = 0
H
1: (
µ
1
µ
2) 0
Rejection region: t >
t
α
/2,n1+n2–2
=
t
.005,198
2.601or t < -2.601
Test statistic: t=(
x
1
x
2)(µ1
µ
2)
sp
21
n1+1
n2
=(431.5
423.9)
0
8315 1
100 +1
100
=.59
Conclusion: Do not reject H0.
There is not enough evidence to conclude that male and female SAT scores differ.
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Example 13.2
Estimate the difference between male and female mean SAT scores (in Example 13.1) with 90%
confidence.
Solution
The interval estimate is
(x
1x
2)±t
α
/2 sp
21
n1+1
n2
=(431.5 423.9)±1.653 8315 1
100+1
100
=7.6±21.3
Example 13.3
A statistician wants to compare the relative success of two 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 below. Do these data provide
sufficient evidence to indicate that the two chains differ? (Test with α = .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 chain with the
population of sales per square foot of another chain. As a result, the parameter of interest is µ1
µ
2.
The sample variances are quite different; we can assume that
σ
1
2
σ
2
2. The appropriate technique is the
unequal variances t-test of µ.
1−µ
2
H
0: ( = 0 µ1−µ
2)
H
1: ( 0 µ1−µ
2)
Test statistic:
t=(
x
1
x
2)(
µ
1
µ
2)
s1
2/n1+s2
2/n2 d.f. = (s1
2/n1+s2
2/n2)2
(s1
2/n1)2
n1–1 +(s2
2/n2)2
n2–1
Rejection region: t >
α
/2,d.f.
.05,6
=
1.943 or t < -1.943
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