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Bluman, Chapter 12 1

Chapter 12

Analysis of Variance

McGraw-Hill, Bluman, 7th ed., Chapter 12 2

3

21Per 5-8; Wed Schedule Sec 12.1

23ACT testing; Classes don’t meet

25 Sec12.2

28 Sec 12.3

30 Sec 12.3 continued.

May 2 Various topics Chapter 13

5 Various topics Chapter 13




14Test 12-14

16 Review for Final Exam

19 Review for Final Exam

2 Review for Final Exam 1

23 Final Exam

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


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.


What is the ANOVA Test?

• Remember the 2-Mean T-Test?• For example: A salesman in car sales wants

to find the difference between two types of cars in terms of mileage:

• Mid-Size Vehicles

• Sports Utility Vehicles

Car Salesman’s Sample

The salesman took an independent SRS from each population of vehicles:

Level n Mean StDevMid-size 28 27.101 mpg 2.629 mpg SUV 26 20.423 mpg 2.914 mpgIf a 2-Mean tTest were done on this data:t = 8.15 P-value = ~0

What if the salesman wanted to compare another type of car, Pickup Trucks in addition to the SUV’s and Mid-size vehicles?

Level n Mean StDevMidsize 28 27.101 mpg 2.629

mpg SUV 26 20.423 mpg 2.914 mpg Pickup 8 23.125 mpg 2.588

mpg

This is an example of when we would use the ANOVA Test.In a 2-Mean TTest, we see if the

difference between the 2 sample means is significant.The ANOVA is used to compare multiple means, and see if the

difference between multiple sample means is significant.

Let’s Compare the Means…

Do these sample means look significantly different from each other?

Yes, we see that no two of these confidence intervals overlap, therefore the means are significantly different.

This is the question that the ANOVA test answers mathematically.

More Confidence Intervals

What if the confidence intervals were different? Would these confidence intervals be significantly different?

SignificantNot Significant

ANOVA Test HypothesesH0: µ1 = µ2 = µ3 (All of the means are equal)

H1: Not all of the means are equal

For Our Example:H0: µMid-size = µSUV = µPickup

The mean mileages of Mid-size vehicles, Sports Utility Vehicles, and Pickup trucks are all equal.

H1: At least one of the mean mileages of Mid-size

vehicles, Sports Utility Vehicles, and Pickup trucks is different.

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 variance or ANOVA.

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 and the Tukey 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 (commonly abbreviated as ANOVA).

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-

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

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.


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 mean, the mean of all values in the samples.

c. Find the between-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 variance, .

d. Find the within-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 Please see page 634 The between-group variance is sometimes

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

is called the sum of squares between groups, SSB.

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

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



Choice HIn ( ) enter the lists… follow each list by a comma


Scroll down to see the rest of the values.

ANOVA Summary Table: see page 633


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, page 634

Compare the calculator results with the values on the chart.

29


d.f. MeanSquares

F

Between

Within (error)

160.13

104.80

2

12

80.07

8.73

9.17

Total 264.93 14

ANOVA Summary Table for Example 12-1, page 634



d.f. MeanSquares

F

Between

Within (error)

160.13

104.80

2

12

80.07

8.73

9.17

Total 264.93 14

ANOVA on your Calculator


Enter the data on the lists on your calculator.

Chapter 12Analysis of Variance

Section 12-1Example 12-2Page #632



Full solution for ANOVA

For ANOVA testing in addition to performing hypothesis testing, an ANOVA summary table with proper values a and symbols should be included.

For examples of ANOVA summary table see page 634 and 635 of your text.


ANOVA Summary Table



d.f. MeanSquares

F

Between

Within (error)

SSB

SSW

k – 1

N – k

MSB

MSW

Total

MSMS

B

W

Use the values displayed by your calculator to determine the value of each of the following.

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 mean, the mean of all values in the samples.

c. Find the between-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 variance, .

d. Find the within-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

Homework

Read section 12.1 and take notes

Sec 12.1 page 637 #1-7 and 9, 11, 13


Thanks For Watching

A special thanks to Mr. Coons for all the help and advice.

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