chapter 11 hypothesis tests and estimation for population ... · population variance ... chapter 11...

QMIS 220, by Dr. M. Zainal

Chapter 11 Student Lecture Notes 11-1

Business Statistics

Department of Quantitative Methods & Information Systems

Dr. Mohammad Zainal QMIS 220

Chapter 11

Hypothesis Tests and Estimation

for Population Variances

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

QMIS 220, by Dr. M. Zainal 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


Single Population


Tests for a Single

Population Variance

Chi-Square test statistic

H0: σ2 = σ0

2

HA: σ2 ≠ σ02

H0: σ2 σ0

2

HA: σ2 < σ02

H0: σ2 ≤ σ0

2

HA: σ2 > σ02

* Two tailed test

Lower tail test

Upper tail test




Chi-Square Test Statistic


Tests for a Single

Population Variance

Chi-Square test statistic *

The chi-squared test statistic for

a Single Population Variance is:

2

22

σ

1)s(n

where

2 = standardized chi-square variable

n = sample size

s2 = sample variance

σ2 = hypothesized variance


The Chi-square Distribution

The chi-square distribution is a family of

distributions, depending on degrees of freedom

(Like the t distribution).

The chi-square distribution curve starts at the

origin and lies entirely to the right of the vertical

axis.

The chi-square distribution assumes

nonnegative values only, and these are denoted

by the symbol 2 (read as “chi-square”).




The Chi-square Distribution

The shape of a specific chi-square distribution

depends on the number of degrees of freedom.

peak of a 2 distribution curve with 1 or 2

degrees of freedom occurs at zero and for a

curve with 3 or more degrees of freedom at

(df−2).

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

d.f. = 1 d.f. = 5 d.f. = 15

2 2 2


Finding the Critical Value

The critical value, , is found from the

chi-square table

Do not reject H0 Reject H0

2

2

2

H0: σ2 ≤ σ0

2

HA: σ2 > σ02

Upper tail test:








Example: Find the value of 2 for 7 degrees of freedom and an

area of .10 in the right tail of the chi-square distribution curve.





Example: Find the value of 2 for 9 degrees of freedom and an

area of .05 in the left tail of the chi-square distribution curve.

Example

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





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

Do not reject H0 Reject H0

= .05

2

2

2

= 24.9958

= 24.9958 ( = .05 and 16 – 1 = 15 d.f.)

22.516

1)24(16

σ

1)s(n2

22

The test statistic is:

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


Lower Tail or Two Tailed Chi-square Tests

H0: σ2 = σ0

2

HA: σ2 ≠ σ02

H0: σ2 σ0

2

HA: σ2 < σ02

2/2

Do not reject H0 Reject

21-

2

Do not reject H0

Reject

/2

21-/2

2

/2

Reject

Lower tail test: Two tail test:

(2U) (2

L)




Confidence Interval Estimate for σ2

2/2

/2

21-/2

2

/2

(2U) (2

L)

The confidence interval estimate for σ2 is

2

L

22

2

U

2 1)s(nσ

1)s(n

χχ

Where 2L and 2

U are from the

2 distribution with n -1 degrees

of freedom


Example

A sample of 16 freezers yields a sample

variance of s2 = 24.

Form a 95% confidence interval for the

population variance.




Use the chi-square table to find 2L and 2

U :

6.2621

( = .05 and 16 – 1 = 15 d.f.)

Example (continued)

2.025

/2=.025

2.975

/2=.025

(2U) (2

L)

27.4884 57.489σ13.096

6.2621

1)24(16σ

27.4884

1)24(16

1)s(nσ

1)s(n

2

2

2

L

22

2

U

2

χχ

We are 95% confident that the population variance is between 13.096

and 57.489 degrees2. (Taking the square root, we are 95% confident that

the population standard deviation is between 3.619 and 7.582 degrees.)



Tests for Two

Population Variances

F test statistic

*

F Test for Difference in Two Population Variances

H0: σ12 = σ2

2

HA: σ12 ≠ σ2

2 Two tailed test

Lower tail test

Upper tail test

H0: σ12 σ2

2

HA: σ12 < σ2

2

H0: σ12 ≤ σ2

2

HA: σ12 > σ2

2





F test statistic *

F Test for Difference in Two Population Variances

Tests for Two

Population Variances 2

2

2

1

s

sF

The F test statistic is:

= Variance of Sample 1

D1 = n1 - 1 = numerator degrees of freedom

D2 = n2 - 1 = denominator degrees of freedom

= Variance of Sample 2

2

1s

2

2s

Where F has D1

numerator and D2

denominator

degrees of freedom


The F critical value is found from the F table

The are two appropriate degrees of freedom:

D1 (numerator) and D2 (denominator)

In the F table,

numerator degrees of freedom determine the column

denominator degrees of freedom determine the row

The F Distribution

where D1 = n1 – 1 ; D2 = n2 – 1 2

2

2

1

s

sF




The F distribution is continuous and skewed to the right.

The units of an F distribution, denoted by FD1,D2, are

nonnegative.

The F Distribution


Finding the F Value


Example: Find the F value for 4 degrees of freedom for the

numerator, 10 degrees of freedom for the denominator, and

0.1 area in the right tail of F distribution curve. (F .1, 4,10)



For a two-tailed test, always place the larger

sample variance in the numerator

For a one-tailed test, consider the alternative

hypothesis: place in the numerator the sample

variance for the population that is predicted

(based on HA) to have the larger variance

Formulating the F Ratio

where D1 = n1 – 1 ; D2 = n2 – 1 2

2

2

1

s

sF


F 0

rejection region for a one-tail test is


F 0

2/2

2

2

1 Fs

sF F

s

sF

2

2

2

1

(where the larger sample variance in the numerator)

rejection region for a two-tailed test is

/2

F F/2 Reject H0 Do not

reject H0

Reject H0 Do not reject H0

H0: σ12 = σ2

2

HA: σ12 ≠ σ2

2

H0: σ12 σ2

2

HA: σ12 < σ2

2

H0: σ12 ≤ σ2

2

HA: σ12 > σ2

2




F Test: An Example

You are a financial analyst for a brokerage firm. You

want to compare dividend yields between stocks listed

on the NYSE & NASDAQ. You collect the following data:

NYSE NASDAQ

Number 21 25

Mean 3.27 2.53

Std dev 1.30 1.16

Is there a difference in the

variances between the NYSE

& NASDAQ at the = 0.05 level?


F Test: Example Solution

Form the hypothesis test:

H0: σ21 = σ2

2 (there is no difference between variances)

HA: σ21 ≠ σ2

2 (there is a difference between variances)

Find the F critical value for = .05:

Numerator:

D1 = n1 – 1 = 21 – 1 = 20

Denominator:

D2 = n2 – 1 = 25 – 1 = 24

F.05/2, 20, 24 = 2.327




The test statistic is:

0

256.116.1

30.1

s

sF

2

2

2

2

2

1

/2 = .025

F/2

=2.327

Reject H0 Do not reject H0

H0: σ12 = σ2

2

HA: σ12 ≠ σ2

2

F Test: Example Solution

F = 1.256 is not greater than

the critical F value of 2.327, so

we do not reject H0

(continued)

Conclusion: There is no evidence of a

difference in variances at = .05


Using EXCEL

EXCEL

F test for two variances:

Data | Data Analysis | F-test: Two Sample for Variances




Chapter Summary

Performed chi-square tests for the variance

Used the chi-square table to find chi-square

critical values

Performed F tests for the difference between two

population variances

Used the F table to find F critical values


Problems


Suppose a sample of 30 ECC students are given an IQ test. If

the sample has a standard deviation of 12.23 points, find a

90% confidence interval for the population standard deviation.

Chap 11-30



Problems


Problems


A random sample of 10 hot drinks from Dispenser A had a

mean volume of 203 ml and a standard deviation of 3 ml.

Another random sample of 15 hot drinks from Dispenser B

gave corresponding values of 206 ml and 5 ml. The amount

dispensed by each machine may be assumed to be normally

distributed. Test, at the 5% significance level, the hypothesis

that there is no difference in the variability of the volume

dispensed by the two machines.

Chap 11-32



Copyright

The materials of this presentation were mostly

taken from the PowerPoint files accompanied

Business Statistics: A Decision-Making Approach,

7e © 2008 Prentice-Hall, Inc.


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