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BACS 2213 Chapter 9 : ANOVA .

OBJECTIVES

To use the one-way ANOVA technique to determine if there is a significant difference among three or more means.To determine the difference among means when Ho in ANOVA is rejected.

10.0 Analysis of Variance (ANOVA)When F test is used to test a hypothesis concerning the means of three or more

populations, the technique is called analysis of variance (ANOVA).

10.1 One Way ANOVA

With the F test, two different estimates of the population variance are made.i) The first estimate is called the between-group variance, and it involves

finding the variance of the means.ii) The second estimate, the within-group variance, is made by computing the

variance using all the data and is not affected by differences in the means.

For a test of the difference among three or more means, the following hypotheses should be used:

H0: = = … = H1: At least two of the means are not equal

Steps to conduct test of the difference among three or more means:

Step 1 : State the hypothesis and identify the claim.Step 2 : Find the critical value.Step 3 : Compute the F test value.Step 4 : Make the Decision.Step 5 : Summarize the results.

Procedure for finding the F test value for the Analysis of Variance:

Step 1 : Find the mean and variance of each sample

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CHAPTER 10 :ANALYSIS OF VARIANCE ( ANOVA )


Step 2 : Find the grand mean

=

Step 3 : Find the between-group variance.

MSB = SSB/ k-1 = =

Step 4 : Find the within-group variance.

MSW= SSW/ N-k = =

Step 5 : Find the F test value.

F =

The degrees of freedom are d.f.N = k-1, and d.f.D = N-kWhere k is the number of groups

N is the sum of the sample sizes of the groups,

ANOVA table for one way classification:

Source of variation

Sum of Squares

Degree of Freedom

Mean of Squares F test Statistic

Between SSB k - 1 MSB = F =

Within SSWN-k

MSW =

Total

SSB – Sum of Squares between groupsSSW – Sum of Squares within groups** Critical Region : Reject if F > (refer to upper critical values of F-distribution)

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Example 10.1 :

A 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 next. At

can it be concluded that there is a significant difference in the average number of employees at each interchange ?

PA Turnpike Greensburg Bypass Valley Express7 10 114 1 1232 1 119 0 910 11 111 1 11

=15.5 =4.0 =5.8=81.9 =25.6 =29.0

Step 1 : State the hypothesis and identify the claim. H0: = = … =

H1: At least two of the means are not equal

Step 2 : Find the critical value. Since k=3, N= 18, and , d.f.N = k-1 = 3-1 = 2 d.f.D = N-k = 18-3 = 15 The critical value is 3.68

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Step 3 : Compute the test value.

a) Find the mean and variance of each sample.

b) Find the grand mean

= =

c) Find the between-group variance.

MSB= =

=459.18/2 =229.59 d) Find the within-group variance.

MSW = = =

= = 45.5 e) Find the F test value.

F = =

Step 4 : Make the Decision. Since 5.05 >3.68, the decision is to reject the null Hypothesis.

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

The ANOVA Summary table for this example is shown below :Source of variation

Sum of Squares

Degree of Freedom

Mean of Squares

F test Statistic

Between 459.18 2 F = =

5.05Within 682.5 15

Total 1141.68 17(i)

H1: At least two of the means are not equal

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(ii)(iii) Critical region : Reject if

(iv) Test statistics F = = 5.05

Since F > 3.68 , Reject

Example 10.2:

In an effort to improve the quality of recording tapes, the effects of four kinds of coatings A,B,C,D on the reproducing quality of sound are compared. Suppose that the measurements of sound distortion (given in the table below) are obtained from tapes treated with four coatings.

A B C D45 58 54 4243 56 50 3751 62 55 49

Does any significant difference exist among the mean distortions obtained using the four coatings? Use a 0.01 level of significance.

Solution:

A B C D45 58 54 4243 56 50 3751 62 55 49

=46.33 =58.67 =36.33 =42.67

=17.3 =9.3 =7.0 =36.3

Step 1 : State the hypothesis and identify the claim.H0: = = … = H1: At least two of the means are not equal

Step 2 : Find the critical value. Since k=4, N= 12, and , d.f.N = k-1 = 4-1 = 3 d.f.D = N-k = 12-4 = 8 The critical value is 7.59Step 3 : Compute the test value.

c) Find the mean and variance of each sample.d) Find the grand mean

= =


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MSB= =

= 453.67/3 = 151.22d) Find the within-group variance.

MSW = =

=

=

e) Find the F test value.

F = =


Step 5 : Summarize the results. We can conclude that there is a significant difference in treatment effects.


Sum of Squares

Degree of Freedom

Mean of Squares

F test Statistic

Between 453.67 3 F = =

8.64Within 139.8 8

Total 1148.18 11(v)

H1: At least two of the means are not equal(vi)(vii) Critical region : Reject if

(viii) Test statistics F = = 8.64

(ix) Since F > 7.59 , Reject

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Example 10.3:

A 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 = 0.05, test the claim that there is no difference among the means.

Medication Exercise Diet10 6 512 8 99 3 1215 0 813 2 4

=11.8 =3.8 =7.6

=5.7 =10.2 =10.3

Step 1 : State the hypothesis and identify the claim.

H0: = = 3

H1: At least one mean is different from the others

Step 2 : Find the critical value. Since k=3, N= 15, and ,d.f.N = k-1 = 3-1 = 2d.f.D = N-k = 15-3 = 12The critical value is 3.89

Step 3 : Compute the test value.a) Find the mean and variance of each sample.

=11.8 =3.8 =7.6

=5.7 =10.2 =10.3

b) Find the grand mean

= =


MSB= = =80.07

d) Find the within-group variance.

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MSW = =

= = 8.73

e) Find the F test value.

F = =


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


Sum of Squares

Degree of Freedom

Mean of Squares F test Statistic

Between 160.13 2F = = 9.17

Within 104.8012

Total 264.93 14(x)

H1: At least two of the means are not equal(xi)(xii) Critical region : Reject if

(xiii) Test statistics F = = 9.17

Since F > 3.89 , Reject

Exercises

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1. Six different machines are being considered for use in manufacturing rubber seals the machines are being compared with respect to tensile strength of the product. A random sample of 4 seals from each machine is used to determine whether the mean tensile strength varies from machine to machine. The following are the tensile-strength measurements in kilograms per square centimeter 10-1.

Machine1 2 3 4 5 6

17.5 16.4 20.3 14.6 17.5 18.316.9 19.2 15.7 16.7 19.2 16.215.8 17.7 17.8 20.8 16.5 17.518.6 15.4 18.9 18.9 20.5 20.1

Perform the analysis of variance at the 0.05 level of significance and indicate whether or not the mean tensile strengths differ significantly for the 6 machines.

2. The data in the following table represent the number of hours of relief provided by 5 different brands of headache tablets administrated to 25 subjects of variance fever of 38C or more. Perform the analysis of variance and test the hypothesis at the level of 0.05 level of significance that the mean number of hours if relief provided by the tablets is the same for all 5 brands. Discuss the results.

TabletA B C D E

5.2 9.1 3.2 2.4 7.14.7 7.1 5.8 3.4 6.68.1 8.2 2.2 4.1 9.36.2 6.0 3.1 1.0 4.23.0 9.1 7.2 4.0 7.6

3. Immobilization of free-ranging white-tailed deer by drugs allows researchers the opportunity to closely examine deer and gather valuable physiological information. In the study Influence of Physical Restraint and Restraint-Facilitating Drugs on Blood Measurements of White-Tailed Deer and Other Selected Mammals conducted at the Virginia Polytechnic Institute and State University, wildlife biologists tested the "knockdown" time (time from injection to immobilization) of three different immobilizing drugs. Immobilization, in this case, is defined as the point to remain standing thirty male white-tailed deer were randomly assigned to each of three treatments. Group A received 5 mg of liquid succinylcholine chloride (SCC); group B received 8 mg of powdered SCC; and group C received 200 mg of phencyclidine hydrochloride. Knockdown times, in minutes, were recorded as follows:

Group

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A B C11 10 45 7 414 16 67 7 310 7 57 5 623 10 84 10 311 6 711 12 3

Perform an analysis of variance at the 0.01 level of significance and determine whether or not the average knockdown time for the 3 drugs is the same.

Solution for the Exercises

Solution 1:

at least two of the means are not equal

Critical region: with and dof.Source of Variation

Sum of Squares

Degree of Freedom

Mean of Squares

F-Test Statistics

Between 5.34 5 1.07 F=0.31Within 62.64 18 3.48Total 67.98 23

F<2.77Decision: The treatment means do not differ significantly.

Solution 2:



Sum of Squares

Degree of Freedom

Mean of Squares

F-Test Statistics


F>2.87Decision: Reject . The mean number of hours of relief differ.

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Solution 3:



Sum of Squares

Degree of Freedom

Mean of Squares

F-Test Statistics


F<5.49Decision: The average knockdown time did not differ significantly.

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