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Chapter 12

Analysis of Variance

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Goals1. List the characteristics of the F distribution

2. Conduct a test of hypothesis to determine whether the variances of two populations are equal

3. Discuss the general idea of analysis of variance

4. Organize data into a ANOVA table

5. Conduct a test of hypothesis among three or more treatment means

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F Distribution1. Used to test whether two

samples are from populations having equal variances

2. Applied when we want to compare several population means simultaneously to determine if they came from equal population ANOVA

Analysis of variance In both situations:

Populations must be Normally distributed

Data must be Interval-scale or higher

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Characteristics Of The F Distribution1. Family of F Distributions

1. Each is determined by:1. df in numerator

comes from pop. 1 which has larger sample variation

2. df in denominator comes from pop. 2 which has smaller sample variation

2. F distribution is continuous F Value can assume an infinite number of values from 0 to ∞

3. Value for F Distribution cannot be negative Smallest value = 0

4. Positively skewed1. Long tail is always to right2. As # of df increases in both the numerator and the denominator,

the distribution approaches normal

5. Asymptotic1. As X increases the F curve approaches the X-axis

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Why Do We Want To Compare To See If Two Population Have Equal Variances?

What if two machines are making the same part for an airplane?

Do we want the parts to be identical or nearly identical? Yes!

We would test to see if the means are the same: Chapter 10 & 11

We would test to see if the variation is the same for the two machines: Chapter 12

What if two stocks have similar mean returns? Would we like to test and see if one stock has more

variation than the other?

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Why Do We Want To Compare To See If Two Population Have Equal Variances?

Remember Chapter 11: Assumptions for small sample tests of means:

1. Sample populations must follow the normal distribution

2. Two samples must be from independent (unrelated) populations

3. The variances & standard deviations of the two populations are equal

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Conduct A Test Of Hypothesis To Determine Whether The Variances Of Two Populations Are Equal

To conduct a test: Conduct two random samples List population 1 as the sample with the largest

variance: n1 = # of observations s1^2 = sample variance n1 – 1 = df1 = degree of freedom (numerator for

critical value lookup) List population 2 as the sample with the smaller

variance: n2 = # of observations s2^2 = sample variance n2 – 1 = df2 = degree of freedom (denominator for

critical value lookup)

Always list the sample

with thelarger

sample variance

as population

1 (allows us

touse fewer

tables)

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Step 1: State null and alternate hypotheses

• List the population with the suspected largest variance as population 1• Because we want to limit the number of F tables

we need to use to look up values, we always put the larger variance in the numerator and the smaller variance in the denominator

• This will force the F value to be at least 1• We will only use the right tail of the F distribution

• Examples of Step 1:

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211

22

210

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:

H

H22

211

22

210

:

:

H

H

9

22

211

22

210

:

:

H

H22

211

22

210

:

:

H

H

Step 2: Select a level of significance:

• Appendix G only lists significance levels: .05 and .01

Significance level = .10.10/2 = .05

Use .05 table inAppendix G

Significance level = .05

Use .05 table inAppendix G

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Step 3: Identify the test statistic (F), find critical value and draw picture

• Look up Critical value in Appendix G and draw your pictureLevel of Significance 0.05

Degrees of Freedom for Numerator (From Pop 1)

1 2 3 4 5 6 71 161 200 216 225 230 234 237

2 18.5 19.0 19.2 19.3 19.3 19.3 19.4

3 10.1 9.55 9.28 9.12 9.01 8.94 8.89

4 7.7 6.94 6.59 6.39 6.26 6.16 6.09

5 6.61 5.79 5.41 5.19 5.05 4.95 4.88

6 5.99 5.14 4.76 4.53 4.39 4.28 4.21

7 5.59 4.74 4.35 4.12 3.97 3.87 3.79

8 5.32 4.46 4.07 3.84 3.69 3.58 3.5

9 5.12 4.26 3.86 3.63 3.48 3.37 3.29

10 4.96 4.1 3.71 3.48 3.33 3.22 3.14

11 4.84 3.98 3.59 3.36 3.2 3.09 3.0112 4.75 3.89 3.49 3.26 3.11 3 2.9113 4.67 3.81 3.41 3.18 3.03 2.92 2.8314 4.6 3.74 3.34 3.11 2.96 2.85 2.7615 4.54 3.68 3.29 3.06 2.9 2.79 2.7116 4.49 3.63 3.24 3.01 2.85 2.74 2.6617 4.45 3.59 3.2 2.96 2.81 2.7 2.61

Deg

rees

of

Fre

edo

m f

or

Den

om

inat

or

(Fro

m P

op

2)

If you have a dfthat is not listedin the border,

calculate your Fby estimating

a valuebetween

two values.HW #5:df = 11,

use valueBetween10 & 12

Book says:(3.14+3.07)/2 =3.105 3.10

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Step 4

• Step 4: Formulate a decision rule:• Example:• If our calculated test statistic is greater than

3.87, reject Ho and accept H1, otherwise fail to reject Ho

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Step 5

• Step 5: Take a random sample, compute the test statistic, compare it to critical value, and make decision to reject or not reject null and hypotheses

Test Statistic F:

• Example Conclusion for a two tail test:• Fail to reject null

• “The evidence suggests that there is not a difference in variation”

• Reject null and accept alternate• “The evidence suggests that there is a difference

in variation”

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s

sF

Larger variancein numerator,

always!!

Let’s Look atHandout

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?

Colin, a stockbroker at Critical Securities, reported that the mean rate of return on a sample of 10 software stocks was 12.6 percent with a standard deviation of 3.9 percent.

Example 1 continued

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220

:

:

UI

UI

H

H

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(2

2

F

Step 5: The value of F is computed as follows.

H0 is not rejected. There is insufficient evidence to show more variation in the software stocks.

Step 4: H0 is rejected if F>3.68 or if p < .05. The degrees of freedom are n1-1 or 9 in the numerator and n1-1 or 7 in the denominator.

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ANOVAAnalysis Of Variance

Technique in which we compare three or more population means to determine whether they could be equal Assumptions necessary:

Populations follow the normal distribution Populations have equal standard deviations () Populations are independent

Why ANOVA? Using t-distribution leads to build up of type 1 error

“Treatment” = different populations being examined

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Case Where Treatment Means Are Different

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Case Where Treatment Means Are The Same

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Example Of ANOVA Test To See If Four Treatment Means Are The Equal

22 students earned the following grades in Professor Rad’s class. The grades are listed under the classification the student gave to the instructor

Is there a difference in the mean score of the students in each of the four categories?

Use significance level α = .01# of "Treatments" 1 2 3 4Rating of Instructor Excellent Good Fair Poor

Course Grades94 75 70 6878 90 80 8281 77 76 7280 83 89 73

88 80 7475 6565

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Conduct A Test Of Hypothesis Among Four Treatment Means

Step 1: State H0 and H1

H0 : µ1 = µ2 = µ3 = µ4

H1 : The Mean scores are not all equal (at least one treatment mean is different)

Step 2: Significance Level?

α = .01

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Step 3: Determine Test Statistic And Select Critical Value

# of "Treatments" 1 2 3 4Rating of Instructor Excellent Good Fair Poor

Course Grades94 75 70 6878 90 80 8281 77 76 7280 83 89 73

88 80 7475 6565

k = Number of treatments = 4n = Total number of observations from all the treatments = 22Degrees of Freedom in the numerator = k - 1 3Degrees of Freedom in the denominator = n - k 18α = Level of significance = 0.01F from appendix G ((df = 3, 18), α = 0.01) 5.09

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Step 4: State Decision Rule

If our calculated test statistic is greater than we reject H0 and accept H1, otherwise we fail to reject H0

Now we move on to Step 5: Select the sample, perform calculations, and make a decision…

Are you ready for a lot of procedures?!!

5.09

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ANOVA TableSources of Variations Sum of Squares

Degrees of Freedoms

Mean Square (Estimate of Variation) F

Treatments SST (k - 1) SST/(k - 1) = MSTError SSE (n - k) SSE/(n - k) = MSE

TotalSS Total(Total Variation) (n - 1)

MST/MSE

The idea is: If we estimate variation in two ways and use one estimate in the numerator and the other estimate in the denominator: If we divide and get 1 or close to 1, the sample means

are assumed to be the same If we get a number far from 1, we say that the means

are assumed to be different The F critical value will determined whether we

are close to 1 or not

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ANOVA Table So Far

ANOVA Table

Sources of Variations Sum of Squares

Degrees of Freedoms

Mean Square (Estimate of Variation) F

Treatments SST 3 SST/3 = MSTError SSE 18 SSE/18 = MSE

TotalSS Total(Total Variation) 21

MST/MSE

Let’s go calculate this!

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Calculation 1: Treatment Means and Overall Mean

Mean Grand Mean Overall X G

# of "Treatments" 1 2 3 4Rating of Instructor Excellent Good Fair Poor

Course Grades94 75 70 6878 90 80 8281 77 76 7280 83 89 73

88 80 7475 6565

Treatment Mean 83.25 82.60 76.43 72.33Grand Mean = Overall Mean for all the data = 77.95

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Calculation 2: Total Variation

Mean OverallX

nObservatio ParticularA X

Variation Total Total Squares of Sum

Total SS )X - (X 2

G

G

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Calculation 2: Total VariationExample of 1st Example of 2nd94-77.95 = 16.05 16.05 2̂ = 257.6025

Excellent (X - 77.95) (X - 77.95) 2̂ Good (X - 77.95) (X - 77.95) 2̂94 16.05 257.6025 75 -2.95 8.702578 0.05 0.0025 90 12.05 145.202581 3.05 9.3025 77 -0.95 0.902580 2.05 4.2025 83 5.05 25.5025

88 10.05 101.0025

Totals 271.11 281.3125

Fair (X - 77.95) (X - 77.95) 2̂ Poor (X - 77.95) (X - 77.95) 2̂70 -7.95 63.2025 68 -9.95 99.002580 2.05 4.2025 82 4.05 16.402576 -1.95 3.8025 72 -5.95 35.402589 11.05 122.1025 73 -4.95 24.502580 2.05 4.2025 74 -3.95 15.602575 -2.95 8.7025 65 -12.95 167.702565 -12.95 167.7025

Totals 373.9175 358.615

Total Variation = SS Total = 1284.96

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ANOVA Table So FarANOVA Table

Sources of Variations Sum of Squares

Degrees of Freedoms

Mean Square (Estimate of Variation) F

Treatments SST 3 SST/3 = MSTError SSE 18 SSE/18 = MSE

Total 1284.96 21

MST/MSE

Let’s go calculate this!

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Calculation 3: Random Variation

Treatment Particular C

CTreatment for Mean SampleX

nObservatio ParticularA X

Variation Random Error Squares of Sum

SSE )X - (X 2

C

C

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Calculation 3: Random VariationExcellent (X - 83.25) (X - 83.25) 2̂ Good (X - 82.6) (X - 82.6) 2̂

94 10.75 115.5625 75 -7.6 57.7678 -5.25 27.5625 90 7.4 54.7681 -2.25 5.0625 77 -5.6 31.3680 -3.25 10.5625 83 0.4 0.16

88 5.4 29.16

Xbar 83.25 82.6

Totals 158.75 173.2

Fair (X - 76.43) (X - 76.43) 2̂ Poor (X - 72.33) (X - 72.33) 2̂70 -6.43 41.3449 68 -4.33 18.748980 3.57 12.7449 82 9.67 93.508976 -0.43 0.1849 72 -0.33 0.108989 12.57 158.0049 73 0.67 0.448980 3.57 12.7449 74 1.67 2.788975 -1.43 2.0449 65 -7.33 53.728965 -11.43 130.6449

Xbar 76.43 72.33

Totals 357.7143 169.3334

Sum of Squares Error = SSE = 859

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ANOVA TableSources of Variations Sum of Squares

Degrees of Freedoms

Mean Square (Estimate of Variation) F

Treatments SST 3 SST/3 = MSTError 859 18 859/18 = 47.72

Total 1284.96 21

MST/47.72

ANOVA Table So Far

Let’s go calculate this!

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Calculation 4: Treatment Variation

ANOVA TableSources of Variations Sum of Squares

Degrees of Freedoms

Mean Square (Estimate of Variation) F

Treatments 425.96 3 425.96/3 = 141.99Error 859 18 859/18 = 47.72

Total 1284.96 21

141.99/47.72 = 2.98

mean overall grand the

andmean ment each treatbetween sdifference

Sqaure theof Sum The Variation Treatment

Treatment Squares of Sum SSE - Total SS SST

Simple Subtraction!

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Calculation 5: Mean Square (Estimate of Variation)

ANOVA TableSources of Variations Sum of Squares

Degrees of Freedoms

Mean Square (Estimate of Variation) F

Treatments 425.96 3 425.96/3 = 141.99Error 859 18 859/18 = 47.72

Total 1284.96 21

141.99/47.72 = 2.98

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Calculation 6: F

ANOVA TableSources of Variations Sum of Squares

Degrees of Freedoms

Mean Square (Estimate of Variation) F

Treatments 425.96 3 425.96/3 = 141.99Error 859 18 859/18 = 47.72

Total 1284.96 21

141.99/47.72 = 2.98

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Step 5: Make A Decision Because is less than 5.09, we fail to

reject H0

The evidence suggests that the mean score of the students in each of the four categories are equal (no difference)

2.98

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Summarize Chapter 121. List the characteristics of the F distribution

2. Conduct a test of hypothesis to determine whether the variances of two populations are equal

3. Discuss the general idea of analysis of variance

4. Organize data into a ANOVA table

5. Conduct a test of hypothesis among three or more treatment means