Statistical techniques in business ecohomics chap012

TWO Conduct a test of hypothesis to determine whether the variances of two populations are equal... Test for Equal Variances of Two Populations2 2 2 1 s s 2 2 s For the two tail test, t

Analysis of Variance

GOALS

When you have completed this chapter, you

will be able to:

ONE

List the characteristics of the F distribution.

TWO

Conduct a test of hypothesis to determine whether the variances of two

populations are equal.

Chapter Twelve continued

Characteristics of

F-Distribution

Its values range from 0

to  As F   the

curve approaches the

X-axis but never touches it.

There is a “family” of F Distributions.

Each member of the family is determined by two parameters: the numerator degrees of freedom and the denominator degrees of

freedom

F cannot be

negative, and

it is a continuous

distribution.

The F

distribution is positively skewed

4.5

1

Test for Equal Variances of Two Populations

2 2

2 1

s

s

2 2

s

For the two tail test, the test statistic is given by

Test for Equal Variances of Two Populations

and are the

sample variances for

the two samples The

larger s is placed in

the denominator.

s 2

1

The degrees of freedom are

-1 for the denominator.

The null hypothesis is rejected

if the computed value of the test statistic is greater than the critical value.

Example 1

The mean rate of return on a sample

of 8 utility stocks was 10.9 percent

with a standard deviation of 3.5

percent At the 05 significance level,

can Colin conclude that there is more

variation in the software stocks?

Securities, reported that the mean rate of return on a sample of 10 internet stocks was 12.6 percent with a standard deviation of 3.9 percent

Example 1 continued

2

2 1

2

2 0

:

:

U I

U I

Step 3: The test statistic is the F distribution.

Step 1: The hypotheses

are

Step 2: The significance level is 05.

Example 1 continued

2416

1 )

5 3 (

) 9 3

insufficient evidence to show more variation in the internet stocks.

The ANOVA Test of Means

The null and alternate hypotheses for four sample

means is given as:

Ho: 1 = 2 = 3 = 4

H1: 1 = 2 = 3 = 4

The ANOVA Test of Means

The F distribution is also

used for testing whether two

or more sample means came

from the same or equal

populations

This technique is called analysis of variance or

ANOVA

The populations have equal standard deviations.

ANOVA requires the following conditions

Underlying assumptions for

If there are k populations

being sampled, the numerator

degrees of freedom is k – 1

If there are a total of n

observations the denominator

degrees of freedom is n – k.

In the following table,

i stands for the i th observation

k is the number of treatment groups

ANOVA Test of Means

ANOVA divides the Total Variation into the

ANOVA Table

Source of

Variation

Sum of Squares

Degrees

of Freedom

Mean Square

Rosenbaum Restaurants specialize in meals for

families Katy Polsby, President, recently

developed a new meat loaf dinner Before

making it a part of the regular menu she decides

to test it in several of her restaurants

Example 2

She would like to know if

there is a difference in the

mean number of dinners sold

per day at the Anyor, Loris,

and Lander restaurants Use

the 05 significance level

Number of Dinners Sold by Restaurant

10 12 13 11

18 16 17 17 17

Example 2 continued

Ho: Aynor = Loris = Landis

H1: Aynor = Loris = Landis

Step Two: Select the level of significance This is

given in the problem statement as 05

Step Three: Determine the test statistic The test

statistic follows the F distribution

Example 2 continued

Step Five: Select the sample, perform the calculations,

and make a decision

Step Four: Formulate the decision rule.

The numerator degrees of freedom, k-1, equal 3-1 or 2 The denominator degrees of freedom, n-k, equal 13-3 or

10 The value of F at 2 and 10 degrees of freedom is 4.10

Example 2 continued

Using the data provided, the ANOVA calculations follow

= 76.25 Example 2 continued

Computation of SST

ANOVA Table

Source of

Variation

Sum of Squares

Degrees

of Freedom

Mean Square

=12

Example 2 continued

same

Since an F of 39.103 > the

critical F of 4.10, the p

of 000018 < a of 05, the

decision is to reject the

null hypothesis and

conclude that

At least two of the treatment means are not the same

Individual 95% CIs For Mean

Based on Pooled StDev

Level N Mean StDev

-

+ -Aynor 4 12.750 0.957 ( -* -)

Loris 4 11.500 1.291 ( -* -)

Lander 5 17.000 0.707 ( -* -) -+ -+ -

+ -Pooled StDev = 0.987 12.5 15.0 17.5

Source of Variation SS df MS F P-value F crit

Between Groups 76.25 2 38.13 39.10 2E-05 4.10

Example 2 continued

Inferences About Treatment Means

One of the simplest procedures

is through the use of confidence

intervals around the difference

in treatment means.

When I reject the null

hypothesis that the

means are equal, I want

to know which

treatment means differ.

Confidence Interval for the Difference Between Two Means

If the confidence interval around the difference

in treatment means includes zero, there is not a

difference between the treatment means.

95% confidence interval for the difference

in the mean number of meat loaf dinners

sold in Lander and Aynor

Can Katy conclude that there is a difference

between the two restaurants?

Example 3continued

The mean number of

meals sold in Aynor

is different from

Lander

in the interval, we conclude that this pair of means

differs.

Two-Factor ANOVA

SSB = r (Xb – XG)2

where r is the number of blocks

Xb is the sample mean of block b

XG is the overall or grand mean

In the following ANOVA table, all sums of squares are computed as before, with the addition of the SSB.

Sometimes there are other causes of variation For the

two-factor ANOVA we test whether there is a significant difference between the treatment effect and whether there is a difference

in the blocking effect (a second treatment variable).

Source of

Variation

Sum of Squares Degrees

of Freedom

Mean Square

Example 4

At the 05 significance level, can we conclude there is a difference in the mean

production by shift and in the mean production by employee?

The Bieber Manufacturing

Co operates 24 hours a

day, five days a week The

workers rotate shifts each

week Todd Bieber, the

owner, is interested in

whether there is a

difference in the number of

units produced when the

employees work on

various shifts A sample

of five workers is selected

and their output recorded

on each shift

Example 4 continued

Output

Evening Output

Night Output

Step 5: Perform the calculations

and make a decision

decision rule

Ho is rejected if F > 4.46, the degrees of freedom are

2 and 8, or if p < 05

Example 4 continued

the alternate hypothesis

H1: Not all means are equal

test statistic The test

statistic follows the F

distribution.

Step 1: State the null hypothesis and

the alternate hypothesis

test statistic The test

statistic follows the F

12-35

Block Sums of Squares

Effects of time of day and worker on productivity

Day Evening Night Employee x SSB

Example 4 continued

Compute the remaining sums of squares as before:

TSS = 139.73 SST = 62.53 SSE = 43.47 (139.73-62.53-33.73)

df(block) = 4 (b-1) df(treatment) = 2 (k-1) df(error)=8 (k-1)(b-1)

Degrees of Freedom

Mean Square

Example 4 continued

Block Effect

Since the computed F of 1.55

< the critical F of 3.84, the p

of 28>  of 05, H0 is not rejected since there is no significant difference in the average number of units

produced for the different employees

Since the computed

units produced for

the different time

periods

Example 4 continued

Two-way ANOVA: Units versus Worker, Shift

Analysis of Variance for Units

Source DF SS MS F P Worker 4 33.73 8.43 1.55 0.276 Shift 2 62.53 31.27 5.75 0.028 Error 8 43.47 5.43

Total 14 139.73

Minitab output

SUMMARY Count Sum Average Variance