ANOVA Examples STAT 314

ANOVA Examples

1.

STAT 314

If we define s = MSE , then of which parameter is s an estimate?

If we define s = MSE , then s is an estimate of the common population standard deviation,

¦Ò, of the populations under consideration. (This presumes, of course, that the equalstandard-deviations assumption holds.)

2.

Explain the reason for the word variance in the phrase analysis of variance.

The reason for the word variance in the phrase analysis of variance is because the analysis

of variance procedure for comparing means involves analyzing the variation in the sample

data.

3.

The null and alternative hypotheses for a one-way ANOVA test are¡­

H 0 : ?1 = ?2 = L = ?k

H a : Not all means are equal.

Suppose in reality that the null hypothesis is false. Does this mean that no two of the populations

have the same mean? If not, what does it mean?

If, in reality, the null hypothesis is false, this translates into ¡°Not all of the means are the

same¡± or equivalently ¡°At least two of the means are not the same.¡± (Notice that ¡°at least

two¡± applies to 2 or 3 or ¡­ or k means.) This last statement in quotes is not equivalent to

saying ¡°No two of the populations have the same mean¡± since this is equivalent to saying,

¡°All of the population means are different.¡±

4.

In a one-way ANOVA, identify the statistic used¡­

a. as a measure of variation among the sample means.

MSTr (or SSTr) is a statistic that measures the variation among the sample means for a

one-way ANOVA.

b. as a measure of variation within the samples.

MSE (or SSE) is a statistic that measures the variation within the samples for a one-way

ANOVA.

c. to compare the variation among the sample means to the variation within the samples.

The statistic that compares the variation among the sample means to the variation within

MSTr

the samples is F =

.

MSE

5.

The times required by three workers to perform an assembly-line task were recorded on five

randomly selected occasions. Here are the times, to the nearest minute.

Hank

8

10

9

11

10

Joseph

8

9

9

8

10

Susan

10

9

10

11

9

(Note: y1 = 9.6, y 2 = 8.8 , y 3 = 9.8 , s12 = 1.3, s22 = 0.7, s32 = 0.7, and y = 9.4 .) Construct the oneway ANOVA table for the data. Compute SSTr and SSE using the defining formulas.

(

)

2

(

)

2

(

)

2

SSTr = n1 y1 ? y + n2 y 2 ? y + n3 y3 ? y = 5(9.6 ? 9.4) + 5(8.8 ? 9.4) + 5(9.8 ? 9.4 ) = 2.8

2

2

2

2

2

2

SSE = (n1 ? 1)s1 + ( n2 ? 1) s2 + (n 3 ? 1) s3 = 4(1.3) + 4 (0.7) + 4(0.7) = 10.8

SSTo = SSTr + SSE = 2.8 + 10.8 = 13.6

SSTr 2.8

SSE

10.8

MSTr 1.4

MSTr =

=

= 1.4 ; MSE =

=

= 0.9; F =

=

= 1.5556

k ?1 3 ?1

N ? k 15 ? 3

MSE 0.9

Source

Treatments

Error

Total

6.

df

2

12

14

SS

2.8

10.8

13.6

MS = SS/df

1.4

0.9

p-value

p-value > 0.10

F-statistic

1.5556

Fill in the missing entries of the partially completed one-way ANOVA table.

Source

Treatments

Error

Total

df

_____

20

_____

SS

2.124

_____

_____

MS = SS/df

0.708

_____

F-statistic

0.75

SSTr

SSTr 2.124

? df Tr =

=

=3

df Tr

MSTr 0.708

MSTr

MSTr 0.708

F=

? MSE =

=

= 0.944

MSE

F

0.75

SSE

MSE =

? SSE = MSE ? df E = 0.944(20) = 18.880

df E

SSTo = SSTr + SSE = 2.124 + 18.880 = 21.004

dfTo = dfTr + df E = 3 + 20 = 23

MSTr =

Source

Treatments

Error

Total

df

3

20

23

SS

2.124

18.880

21.004

MS = SS/df

0.708

0.944

F-statistic

0.75

7.

Data on Scholastic Aptitude Test (SAT) scores are published by the College Entrance Examination

Board in National College-Bound Senior. SAT scores for randomly selected students from each of

four high-school rank categories are displayed in the following table.

Top Tenth

528

586

680

718

Second Tenth

514

457

521

370

532

Second Fifth

649

506

556

413

470

Third Fifth

372

440

495

321

424

330

( N o t e : y1 = 628.0, y 2 = 478.8, y 3 = 518.8 , y 4 = 397.0 , s12 = 7522.667, s22 = 4540.700 ,

s32 = 8018.700, s42 = 4614.400 , and y ?? = 494.1.) Construct the one-way ANOVA table for the

data. Compute SSTr and SSE using the defining formulas.

(

)

2

(

)

2

(

)

2

(

SSTr = n1 y1 ? y + n2 y 2 ? y + n3 y3 ? y + n4 y 4 ? y

)

2

= 4(628.0 ? 494.1) 2 + 5(478.8 ? 494.1)2 + 5(518.8 ? 494.1)2 + 6(397.0 ? 494.1)2 = 132,508.2

SSE = (n1 ? 1)s12 + ( n2 ? 1) s22 + (n 3 ? 1) s32 + (n4 ? 1) s42

= 3(7522.667) + 4 (4540.700) + 4 (8018.700) + 5(4614.400) = 95,877.6

SSTo = SSTr + SSE = 132,508.2 + 95,877.6 = 228,385.8

SSTr 132,508.2

SSE

95,877.6

MSTr =

=

= 44,169.40; MSE =

=

= 5,992.35;

k ?1

4 ?1

N?k

20 ? 4

MSTr 44,169.40

F=

=

= 7.37

MSE

5,992.35

Source

Treatments

Error

Total

df

3

16

19

SS

132,508.2

95,877.6

228,385.8

MS = SS/df

44,169.40

5,992.35

F-statistic

7.37

p-value

0.001 < p-value < 0.01

8.

Four brands of flashlight batteries are to be compared by testing each brand in five flashlights.

Twenty flashlights are randomly selected and divided randomly into four groups of five flashlights

each. Then each group of flashlights uses a different brand of battery. The lifetimes of the

batteries, to the nearest hour, are as follows.

Brand A

42

30

39

28

29

Brand B

28

36

31

32

27

Brand C

24

36

28

28

33

Brand D

20

32

38

28

25

Preliminary data analyses indicate that the independent samples come from normal populations with

equal standard deviations. At the 5% significance level, does there appear to be a difference in mean

lifetime among the four brands of batteries?

Let the subscripts 1, 2, 3, and 4 refer to Brand A, Brand B, Brand C, and Brand D, respectively. Each

sample size is 5, and the total number of pieces of data is 20.

Step 0 :

Step 1 :

Check Assumptions

Hypotheses

H0 : ?1 = ?2 = ?3 = ? 4 (The mean lifetimes are equal.)

Ha : Not all of the means are equal.

Step 2 :

Significance Level

Step 3 :

Critical Value and Rejection Region

Step 4 :

Reject the null hypothesis if F ¡Ý 3.24 (P-value ¡Ü 0.05).

Construct the One-way ANOVA Table

T1 = 168 ; T2 = 154 ; T3 = 149 ; T4 = 143

¦Á = 0.05

F¦Á ,( df 1 =k ?1,df 2 =N ? k) = F0.05,( df 1 =4 ?1,df 2 =20 ?4 ) = F0.05,( df 1 =3,df 2 =16) = 3.24

k

T = 614 ;

k

ni

ni

ni

ni

j =1

2

j =1

j =1

ni

¡Æ ¡Æ y ij2 = ¡Æ y12j + ¡Æ y 22j + ¡Æ y 23 j + ¡Æ y 24 j = 19,410

i=1 j =1

ni

SSTo = ¡Æ ¡Æ y ij ?

2

i=1 j =1

2

(T )

N

= 19,410 ?

(614)

20

j =1

2

= 560.2

2

2

2

2

? (T )

T2 )

T3 )

T4 ) ? (T )

(

(

(

1

SSTr = ?

+

+

+

??

n2

n3

n4 ?

N

? n1

? (168) 2 (154) 2 (149)2 (143)2 ? (614) 2

??

=?

+

+

+

= 68.2

5

5

5 ?

20

? 5

SSE = SSTo ? SSTr = 560.2 ? 68.2 = 492.0

Source

Treatments

Error

Total

Step 5 :

Step 6 :

df

3

16

19

SS

68.2

492.0

560.2

MS = SS/df

22.7333

30.75

F-statistic

0.7393

p-value

p-value > 0.10

Decision

Since 0.7393 < 3.24 (p-value > 0.05), fail to reject the null hypothesis.

State conclusion in words

At the ¦Á = 0.05 level of significance, there is not enough evidence to conclude that the

mean lifetimes of the brands of batteries differ.

9.

Manufacturers of golf balls always seem to be claiming that their ball goes the farthest. A writer for

a sports magazine decided to conduct an impartial test. She randomly selected 20 golf

professionals and then randomly assigned four golfers to each of five brands. Each golfer drove

the assigned brand of ball. The driving distances, in yards, are displayed in the following table.

Brand 1

286

276

281

274

Brand 2

279

277

284

288

Brand 3

270

262

277

280

Brand 4

284

271

269

275

Brand 5

281

293

276

292

Preliminary data analyses indicate that the independent samples come from normal populations with

equal standard deviations. Do the data provide sufficient evidence to conclude that a difference

exists in mean weekly earnings among nonsupervisory workers in the five industries? Perform the

required hypothesis test using ¦Á = 0.05. (Note: T1 = 1,117, T2 = 1,128 , T3 = 1,089 , T4 = 1,099,

5

T5 = 1,142 , and

4

¡Æ¡Æ y

i=1 j =1

2

ij

= 1,555,185 .)

Let the subscripts 1, 2, 3, 4, and 5 refer to Brand 1, Brand 2, Brand 3, Brand 4, and Brand 5 respectively.

Each sample size is 4, and the total number of data pieces is 20.

Step 0 :

Step 1 :

Check Assumptions

Hypotheses

H0 : ?1 = ?2 = ?3 = ? 4 = ?5 (The mean driving distances are equal.)

Ha : Not all of the means are equal.

Step 2 :

Significance Level

Step 3 :

Critical Value and Rejection Region

Step 4 :

Reject the null hypothesis if F ¡Ý 3.06 (P-value ¡Ü 0.05).

Construct the One-way ANOVA Table

¦Á = 0.05

F¦Á ,( df 1 =k ?1,df 2 =N ? k) = F0.05,( df 1 =5?1,df 2 = 20? 5) = F0.05,( df 1 =4,df 2 =15) = 3.06

5

T = ¡Æ Ti = 5,575

i=1

5

4

SSTo = ¡Æ ¡Æ y ?

i=1 j =1

2

2

ij

(T )

2

N

= 1,555,185 ?

(5,575)

20

2

= 1,153.75

2

2

2

2

2

? (T )

T2 )

T3 )

T4 )

T5 ) ? (T )

(

(

(

(

1

SSTr = ?

+

+

+

+

??

n2

n3

n4

n5 ?

N

? n1

? (1,117) 2 (1,128)2 (1,089) 2 (1,099) 2 (1,142)2 ? (5,575)2

??

=?

+

+

+

+

= 458.5

4

4

4

4 ?

20

? 4

SSE = SSTo ? SSTr = 1,153.75 ? 485.5 = 695.25

Source

Treatments

Error

Total

Step 5 :

Step 6 :

df

4

15

19

SS

458.50

695.25

1,153.75

MS = SS/df

114.625

46.35

F-statistic

2.4730

p-value

0.05 < p-value < 0.10

Decision

Since 2.4730 < 3.06 (p-value > 0.05), fail to reject the null hypothesis.

State conclusion in words

At the ¦Á = 0.05 level of significance, there is not enough evidence to conclude that the

mean driving distances of the brands of golf balls differ.

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