mcgraw-hill, bluman, 7th ed., chapter 12

Chapter 12

Analysis of Variance

McGraw-Hill, Bluman, 7th ed., Chapter 12 1

Chapter 12 Overview Introduction 12-1 One-Way Analysis of Variance 12-2 The Scheffé Test and the Tukey Test 12-3 Two-Way Analysis of Variance

Bluman, Chapter 12 2

Chapter 12 Objectives1. Use the one-way ANOVA technique to

determine if there is a significant difference among three or more means.

2. Determine which means differ, using the Scheffé or Tukey test if the null hypothesis is rejected in the ANOVA.

3. Use the two-way ANOVA technique to determine if there is a significant difference in the main effects or interaction.


Introduction The F test, used to compare two variances,

can also be used to compare three of more means.

This technique is called analysis of varianceanalysis of variance or ANOVAANOVA.

For three groups, the F test can only show whether or not a difference exists among the three means, not where the difference lies.

Other statistical tests, Scheffé test Scheffé test and the Tukey testTukey test, are used to find where the difference exists.


12-1 One-Way Analysis of Variance When an F test is used to test a

hypothesis concerning the means of three or more populations, the technique is called analysis of variance analysis of variance (commonly abbreviated as ANOVAANOVA).

Although the t test is commonly used to compare two means, it should not be used to compare three or more.


Assumptions for the F TestThe following assumptions apply when using the F test to compare three or more means.

1. The populations from which the samples were obtained must be normally or approximately normally distributed.

2. The samples must be independent of each other.

3. The variances of the populations must be equal.


The F Test In the F test, two different estimates of

the population variance are made. The first estimate is called the between-between-

group variancegroup variance, and it involves finding the variance of the means.

The second estimate, the within-group within-group variancevariance, is made by computing the variance using all the data and is not affected by differences in the means.


The F Test If there is no difference in the means, the

between-group variance will be approximately equal to the within-group variance, and the F test value will be close to 1—do not reject null hypothesis.

However, when the means differ significantly, the between-group variance will be much larger than the within-group variance; the F test will be significantly greater than 1—reject null hypothesis.


Chapter 12Analysis of Variance

Section 12-1Example 12-1Page #630


Example 12-1: Lowering Blood PressureA researcher wishes to try three different techniques to lower the blood pressure of individuals diagnosed with high blood pressure. The subjects are randomly assigned to three groups; the first group takes medication, the second group exercises, and the third group follows a special diet. After four weeks, the reduction in each person’s blood pressure is recorded. At α = 0.05, test the claim that there is no difference among the means.


Example 12-1: Lowering Blood Pressure


Step 1: State the hypotheses and identify the claim.H0: μ1 = μ2 = μ3 (claim)H1: At least one mean is different from the others.



Step 2: Find the critical value.Since k = 3, N = 15, and α = 0.05,d.f.N. = k – 1 = 3 – 1 = 2d.f.D. = N – k = 15 – 3 = 12The critical value is 3.89, obtained from Table H.



Step 3: Compute the test value.a.Find the mean and variance of each sample (these were provided with the data).

b.Find the grand meangrand mean, the mean of all values in the samples.

c. Find the between-group variancebetween-group variance, .

10 12 9 4 116 7.7315 15

GM

XX

N

2Bs

2

2

1

i i GMB

n X Xs

k



Step 3: Compute the test value. (continued)c. Find the between-group variancebetween-group variance, .

d.Find the within-group variancewithin-group variance, .2Ws

22 1

1

i iB

i

n ss

n

4 5.7 4 10.2 4 10.3 104.80 8.734 4 4 12

2 2 22 5 11.8 7.73 5 3.8 7.73 5 7.6 7.73

3 1160.13 80.07

2

Bs

2Bs

Step 3: Compute the test value. (continued)e. Compute the F value.

Step 4: Make the decision.Reject the null hypothesis, since 9.17 > 3.89.

Step 5: Summarize the results.There is enough evidence to reject the claim and conclude

that at least one mean is different from the others.



2

2 B

W

sFs

80.07 9.178.73

ANOVA The between-group variance is sometimes

called the mean square, MSmean square, MSBB. The numerator of the formula to compute MSB

is called the sum of squares between sum of squares between groups, SSgroups, SSBB.

The within-group variance is sometimes called the mean square, MS mean square, MSWW.

The numerator of the formula to compute MSW is called the sum of squares within groups, sum of squares within groups, SSSSWW.


ANOVA Summary Table


Source Sum of Squares

d.f. MeanSquares

F

Between

Within (error)

SSB

SSW

k – 1

N – k

MSB

MSW

Total

MSMS

B

W

ANOVA Summary Table for Example 12-1



d.f. MeanSquares

F

Between

Within (error)

160.13

104.80

2

12

80.07

8.73

9.17

Total 264.93 14

Example 12-2: Toll Road EmployeesA state employee wishes to see if there is a significant difference in the number of employees at the interchanges of three state toll roads. The data are shown. At α = 0.05, can it be concluded that there is a significant difference in the average number of employees at each interchange?


Example 12-2: Toll Road Employees


Step 1: State the hypotheses and identify the claim.H0: μ1 = μ2 = μ3 H1: At least one mean is different from the others (claim).



Step 2: Find the critical value.Since k = 3, N = 18, and α = 0.05,d.f.N. = 2, d.f.D. = 15The critical value is 3.68, obtained from Table H.



Step 3: Compute the test value.a.Find the mean and variance of each sample (these were provided with the data).

b.Find the grand meangrand mean, the mean of all values in the samples.

c. Find the between-group variancebetween-group variance, .

7 14 32 11 152 8.415 18

GM

XX

N

2Bs

2

2

1

i i GMB

n X Xs

k



Step 3: Compute the test value. (continued)c. Find the between-group variancebetween-group variance, .

d.Find the within-group variancewithin-group variance, . 2Ws

22 1

1

i iB

i

n ss

n

5 81.9 5 25.6 5 29.0 682.5 45.54 4 4 15

2 2 22 6 15.5 8.4 6 4 8.4 6 5.8 8.4

3 1459.18 229.59

2

Bs

2Ws

Step 3: Compute the test value. (continued)e. Compute the F value.

Step 4: Make the decision.Reject the null hypothesis, since 5.05 > 3.68.

Step 5: Summarize the results.There is enough evidence to support the claim that there

is a difference among the means.



2

2 B

W

sFs

229.59 5.0545.5

ANOVA Summary Table for Example 12-2



d.f. MeanSquares

F

Between

Within (error)

459.18

682.5

2

15

229.59

45.5

5.05

Total 1141.68 17

12-2 The Scheffé Test and the Tukey Test

When the null hypothesis is rejected using the F test, the researcher may want to know where the difference among the means is.

The Scheffé test Scheffé test and the Tukey test Tukey test are procedures to determine where the significant differences in the means lie after the ANOVA procedure has been performed.


The Scheffé Test In order to conduct the Scheffé testScheffé test,

one must compare the means two at a time, using all possible combinations of means.

For example, if there are three means, the following comparisons must be done:


1 2 1 3 2 3 versus versus versus X X X X X X

Formula for the Scheffé Test

where and are the means of the samples being compared, and are the respective sample sizes, and the within-group variance is .


2

2 1 1

i jS

W i j

X XF

s n n

iX jXin jn

2Ws

F Value for the Scheffé Test To find the critical value F for the Scheffé test, multiply the critical value for the F test

by k 1:

There is a significant difference between the two means being compared when Fs is greater than F.


1 C.V. F k

Example 12-3: Lowering Blood PressureUsing the Scheffé test, test each pair of means in Example 12–1 to see whether a specific difference exists, at α = 0.05.


1 2

2

1 22

1 2

a. For versus ,

1 1

SW

X X

X XF

s n n

211.8 3.818.33

8.73 1 5 1 5

2 3

2

2 32

2 3

b. For versus ,

1 1

SW

X X

X XF

s n n

23.8 7.64.14

8.73 1 5 1 5

Example 12-3: Lowering Blood PressureUsing the Scheffé test, test each pair of means in Example 12–1 to see whether a specific difference exists, at α = 0.05.


1 3

2

1 32

1 3

c. For versus ,

1 1

SW

X X

X XF

s n n

211.8 7.65.05

8.73 1 5 1 5



The critical value for the ANOVA for Example 12–1 was F = 3.89, found by using Table H with α = 0.05, d.f.N. = 2, and d.f.D. = 12.

In this case, it is multiplied by k – 1 as shown.

Since only the F test value for part a ( versus ) is greater than the critical value, 7.78, the only significant difference is between and , that is, between medication and exercise.

3.89F

1 C.V. 2 3.89 7.78 SF k

1X 2X

1X 2X

An Additional Note


On occasion, when the F test value is greater than the critical value, the Scheffé test may not show any significant differences in the pairs of means. This result occurs because the difference may actually lie in the average of two or more means when compared with the other mean. The Scheffé test can be used to make these types of comparisons, but the technique is beyond the scope of this book.

The Tukey Test The Tukey test Tukey test can also be used after

the analysis of variance has been completed to make pairwise comparisons between means when the groups have the same sample size.

The symbol for the test value in the Tukey test is q.


Formula for the Tukey Test

where and are the means of the samples being compared, is the size of the sample, and the within-group variance is .


2

i j

W

X Xq

s n

iX jXn

2Ws

Example 12-4: Lowering Blood PressureUsing the Tukey test, test each pair of means in Example 12–1 to see whether a specific difference exists, at α = 0.05.


1 2

1 22

a. For versus ,

W

X X

X Xqs n

11.8 3.8 6.068.73 5

1 3

1 32

b. For versus ,

W

X X

X Xqs n

11.8 7.6 3.188.73 5

Example 12-3: Lowering Blood PressureUsing the Tukey test, test each pair of means in Example 12–1 to see whether a specific difference exists, at α = 0.05.


3.8 7.6 2.888.73 5

2 3

2 32

c. For versus ,

W

X X

X Xqs n



To find the critical value for the Tukey test, use Table N.The number of means k is found in the row at the top, and

the degrees of freedom for are found in the left column (denoted by v). Since k = 3, d.f. = 12, and α = 0.05, the critical value is 3.77.



Hence, the only q value that is greater in absolute value than the critical value is the one for the difference between and . The conclusion, then, is that there is a significant difference in means for medication and exercise.

These results agree with the Scheffé analysis.

1X 2X

12-3 Two-Way Analysis of Variance In doing a study that involves a two-two-

way analysis of varianceway analysis of variance, the researcher is able to test the effects of two independent variables or factors on one dependent variable.

In addition, the interaction effect of the two variables can be tested.


Two-Way Analysis of Variance Variables or factors are changed

between two levelslevels (i.e., two different treatments).

The groups for a two-way ANOVA are sometimes called treatment groupstreatment groups.

A two-way ANOVA has several null hypotheses. There is one for each independent variable and one for the interaction.


Two-Way ANOVA Summary Table



d.f. MeanSquares

F

AB

A X BWithin (error)

SSA

SSB

SSAXB

SSW

a – 1b – 1

(a – 1)(b – 1)ab(n – 1)

MSA

MSB

MSAXB

MSW

FA

FB

FAXB

Total

Assumptions for Two-Way ANOVA1. The populations from which the samples

were obtained must be normally or approximately normally distributed.

2. The samples must be independent.3. The variances of the populations from

which the samples were selected must be equal.

4. The groups must be equal in sample size.


Example 12-5: Gasoline ConsumptionA researcher wishes to see whether the type of gasoline used and the type of automobile driven have any effect on gasoline consumption. Two types of gasoline, regular and high-octane, will be used, and two types of automobiles, two-wheel- and four-wheel-drive, will be used in each group. There will be two automobiles in each group, for a total of eight automobiles used. Use a two-way analysis of variance at α = 0.05.


Example 12-5: Gasoline ConsumptionStep 1: State the hypotheses.The hypotheses for the interaction are these:

H0: There is no interaction effect between type of gasoline used and type of automobile a person drives on gasoline consumption.

H1: There is an interaction effect between type of gasoline used and type of automobile a person drives on gasoline consumption.


Example 12-5: Gasoline ConsumptionStep 1: State the hypotheses.The hypotheses for the gasoline types are

H0: There is no difference between the means of gasoline consumption for two types of gasoline.

H1: There is a difference between the means of gasoline consumption for two types of gasoline.


Example 12-5: Gasoline ConsumptionStep 1: State the hypotheses.The hypotheses for the types of automobile driven are

H0: There is no difference between the means of gasoline consumption for two-wheel-drive and four-wheel-drive automobiles.

H1: There is a difference between the means of gasoline consumption for two-wheel-drive and four-wheel-drive automobiles.


Example 12-5: Gasoline Consumption


Step 2: Find the critical value for each.Since α = 0.05, d.f.N. = 1, and d.f.D. = 4 for each of the factors, the critical values are the same, obtained from Table H as

Step 3: Find the test values.Since the computation is quite lengthy, we will use the summary table information obtained using statistics software such as Minitab.

C.V. 7.71

Example 12-5: Gasoline Consumption


Two-Way ANOVA Summary Table


d.f. MeanSquares

F

Gasoline AAutomobile B

Interaction A X BWithin (error)

3.920 9.680 54.080 3.300

1114

3.920 9.680 54.080 0.825

4.75211.73365.552

Total 70.890 7

Step 4: Make the decision.Since FB = 11.733 and FAXB = 65.552 are greater than the

critical value 7.71, the null hypotheses concerning the type of automobile driven and the interaction effect should be rejected.

Step 5: Summarize the results.Since the null hypothesis for the interaction effect was

rejected, it can be concluded that the combination of type of gasoline and type of automobile does affect gasoline consumption.



mcgraw-hill, bluman, 7th ed., chapter 12

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