# Variance and Decision-making Approach

Topics: Variance, Chi-square distribution, Normal distribution Pages: 20 (1582 words) Published: February 27, 2013
A Decision-Making Approach
7th Edition

Chapter 11
Hypothesis Tests and Estimation
for Population Variances

Chap 11-1

Chapter Goals
After completing this chapter, you should be
able to:

Formulate and complete hypothesis tests for a single
population variance
Find critical chi-square distribution values from the
chi-square table
Formulate and complete hypothesis tests for the
difference between two population variances
Use the F table to find critical F values

Chap 11-2

Hypothesis Tests for Variances
Hypothesis Tests
for Variances

Tests for a Single
Population Variance

Tests for Two
Population Variances

Chi-Square test statistic

F test statistic

Chap 11-3

Single Population
Hypothesis Tests for Variances

Tests for a Single
Population Variance

*

Chi-Square test statistic

H0: σ2 = σ02
HA: σ2 ≠ σ02

Two tailed test

H0: σ2  σ02
HA: σ2 < σ02

Lower tail test

H0: σ2 ≤ σ02
HA: σ2 > σ02

Upper tail test

Chap 11-4

Chi-Square Test Statistic
Hypothesis Tests for Variances
The chi-squared test statistic for
a Single Population Variance is:

Tests for a Single
Population Variance

Chi-Square test statistic

(n  1)s

2
σ
2

*

2

where
2 = standardized chi-square variable
n = sample size
s2 = sample variance
σ2 = hypothesized variance

Chap 11-5

The Chi-square Distribution

The chi-square distribution is a family of
distributions, depending on degrees of freedom:
d.f. = n - 1

0 4 8 12 16 20 24 28

d.f. = 1

2 0 4 8 12 16 20 24 28 2 0 4 8 12 16 20 24 28 2

d.f. = 5

d.f. = 15
Chap 11-6

Finding the Critical Value

2 , is found from the
The critical value,
chi-square table

Upper tail test:

Ho: σ2 16

H0: σ2 ≤ σ02
HA: σ2 > σ02

2
Do not reject H0

2

Reject H0

Chap 11-7

Example

Ho: σ2 16

A commercial freezer must hold the selected
temperature with little variation. Specifications call
for a standard deviation of no more than 4 degrees
(or variance of 16 degrees2). A sample of 16
freezers is tested and
yields a sample variance
of s2 = 24. Test to see
whether the standard
deviation specification
is exceeded. Use
 = .05

Chap 11-8

Finding the Critical Value

Use the chi-square table to find the critical value:

2 = 24.9958 ( = .05 and 16 – 1 = 15 d.f.)
The test statistic is:

(n  1)s2 (16  1)24
2 

 22.5
2
σ
16
Since 22.5 < 24.9958,
do not reject H0
There is not significant
evidence at the  = .05 level
that the standard deviation
specification is exceeded

 = .05

2
Do not reject H0

2 Reject H0
= 24.9958

Chap 11-9

Lower Tail or Two Tailed
Chi-square Tests
Lower tail test:

Two tail test:

H0: σ2  σ02
HA: σ2 < σ02

H0: σ2 = σ02
HA: σ2 ≠ σ02
/2

/2

2
Reject

21-

Do not reject H0

2
Reject

Do not
reject H0

21-/2
(2L)

Reject

2/2
(2U)
Chap 11-10

Confidence Interval Estimate
for σ2

The confidence interval estimate for σ2 is

(n  1)s2
(n  1)s 2
 σ2 
2
2
χU
χL...