MBA 6640

Quantitative Analysis for

Managers

Chapter 11

Analysis of Variance

Troy Program – MBA 6640

1

Chapter Goals

After completing this chapter, you should be able to:

Recognize situations in which to use analysis of variance

Understand different analysis of variance designs

Perform a one-way and two-way analysis of variance and

interpret the results

Conduct and interpret a Kruskal-Wallis test

Analyze two-factor analysis of variance tests with more than

one observation per cell

One-Way Analysis of Variance

Evaluate the difference among the means of three

or more groups

Examples: Average production for 1st, 2nd, and 3rd shift

Expected mileage for five brands of tires

Assumptions

Populations are normally distributed

Populations have equal variances

Samples are randomly and independently drawn

Hypotheses of One-Way ANOVA

H0 : μ1 = μ2 = μ3 = = μK

All population means are equal

i.e., no variation in means between groups

H1 : μi ≠ μ j

for at least one i, j pair

At least one population mean is different

i.e., there is variation between groups

Does not mean that all population means are different

(some pairs may be the same)

One-Way ANOVA

H0 : μ1 = μ2 = μ3 = = μK

H1 : Not all μi are the same

All Means are the same:

The Null Hypothesis is True

(No variation between

groups)

μ1 = μ2 = μ3

One-Way ANOVA

(continued)

H0 : μ1 = μ2 = μ3 = = μK

H1 : Not all μi are the same

At least one mean is different:

The Null Hypothesis is NOT true

(Variation is present between groups)

or

μ1 = μ2 ≠ μ3

μ1 ≠ μ2 ≠ μ3

Variability

The variability of the data is key factor to test the

equality of means

In each case below, the means may look different, but a

large variation within groups in B makes the evidence

that the means are different weak

A

B

A

B

Group

C

Small variation within groups

A

B

Group

C

Large variation within groups

Partitioning the Variation

Total variation can be split into two parts:

SST = SSW + SSG

SST = Total Sum of Squares

Total Variation = the aggregate dispersion of the individual

data values across the various groups

SSW = Sum of Squares Within Groups

Within-Group Variation = dispersion that exists among the

data values within a particular group

SSG = Sum of Squares Between Groups

Between-Group Variation = dispersion between the group

sample means

Partition of Total Variation

Total Sum of Squares

(SST)

=

Variation due to

random sampling

(SSW)

+

Variation due to

differences

between groups

(SSG)

Total Sum of Squares

SST = SSW + SSG

K

ni

SST = ∑∑ (x ij − x)

Where:

2

i=1 j=1

SST = Total sum of squares

K = number of groups (levels or treatments)

ni = number of observations in group i

xij = jth observation from group i

x = overall sample mean

Total Variation

(continued)

SST = (x11 − x )2 + (X12 − x )2 + ... + (x KnK − x )2

Response, X

x

Group 1

Group 2

Group 3

Within-Group Variation

SST = SSW + SSG

K

ni

SSW = ∑∑ (x ij − x i )2

i =1 j=1

Where:

SSW = Sum of squares within groups

K = number of groups

ni = sample size from group i

Xi = sample mean from group i

Xij = jth observation in group i

Within-Group Variation

(continued)

K

ni

SSW = ∑∑ (x ij − x i )

i =1 j=1

Summing the variation

within each group and then

adding over all groups

2

SSW

MSW =

n −K

Mean Square Within =

SSW/degrees of freedom

μi

Within-Group Variation

(continued)

SSW = (x11 − x1 )2 + (x12 − x1 )2 + ... + (x KnK − x K )2

Response, X

x1

Group 1

Group 2

x2

Group 3

x3

Between-Group Variation

SST = SSW + SSG

K

SSG = ∑ ni ( x i − x )

Where:

2

i=1

SSG = Sum of squares between groups

K = number of groups

ni = sample size from group i

xi = sample mean from group i

x = grand mean (mean of all data values)

Between-Group Variation

(continued)

K

SSG = ∑ ni ( x i − x )

2

i=1

Variation Due to

Differences

Between

Groups

SSG

MSG =

K −1

Mean Square Between Groups

= SSG/degrees of freedom

μi

μj

Between-Group Variation

(continued)

SSG = n1(x1 − x) + n2 (x 2 − x) + ... + nK (x K − x)

2

2

Response, X

x1

Group 1

Group 2

x2

Group 3

x3

x

2

Obtaining the Mean Squares

SST

MST =

n −1

SSW

MSW =

n −K

SSG

MSG =

K −1

One-Way ANOVA Table

Source of

Variation

SS

df

Between

Groups

SSG

K-1

Within

Groups

SSW

n-K

SST =

SSG+SSW

n-1

Total

MS

(Variance)

F ratio

SSG

MSG

MSG =

K - 1 F = MSW

SSW

MSW =

n-K

K = number of groups

n = sum of the sample sizes from all groups

df = degrees of freedom

One-Factor ANOVA

F Test Statistic

H0: μ1= μ2 = … = μK

H1: At least two population means are different

Test statistic

MSG

F=

MSW

MSG is mean squares between variances

MSW is mean squares within variances

Degrees of freedom

df1 = K – 1

(K = number of groups)

df2 = n – K

(n = sum of sample sizes from all groups)

Interpreting the F Statistic

The F statistic is the ratio of the between estimate of variance and

the within estimate of variance

The ratio must always be positive

df1 = K -1 will typically be small

df2 = n - K will typically be large

Decision Rule:

Reject H if

0

F > FK-1,n-K,α

α = .05

0

Do not

reject H0

Reject H0

FK-1,n-K,α

One-Factor ANOVA

F Test Example

You want to see if three

different golf clubs yield

different distances. You

randomly select five

measurements from trials on

an automated driving

machine for each club. At

the .05 significance level, is

there a difference in mean

distance?

Club 1

254

263

241

237

251

Club 2

234

218

235

227

216

Club 3

200

222

197

206

204

One-Factor ANOVA Example:

Scatter Diagram

Club 1

254

263

241

237

251

Club 2

234

218

235

227

216

Club 3

200

222

197

206

204

Distance

270

260

250

240

230

•

••

•

•

220

x1 = 249.2 x 2 = 226.0 x 3 = 205.8

x = 227.0

210

x1

••

•

••

x2

•

••

••

200

190

1

2

Club

x

3

x3

One-Factor ANOVA Example

Computations

Club 1

254

263

241

237

251

Club 2

234

218

235

227

216

Club 3

200

222

197

206

204

x1 = 249.2

n1 = 5

x2 = 226.0

n2 = 5

x3 = 205.8

n3 = 5

x = 227.0

n = 15

K=3

SSG = 5 (249.2 – 227)2 + 5 (226 – 227)2 + 5 (205.8 – 227)2 = 4716.4

SSW = (254 – 249.2)2 + (263 – 249.2)2 +…+ (204 – 205.8)2 = 1119.6

MSG = 4716.4 / (3-1) = 2358.2

MSW = 1119.6 / (15-3) = 93.3

2358.2

F=

= 25.275

93.3

One-Factor ANOVA Example

Solution

Test Statistic:

H0: μ1 = μ2 = μ3

H1: μi not all equal

α = .05

df1= 2

df2 = 12

Critical Value:

F2,12,.05= 3.89

α = .05

0

Do not

reject H0

Reject H0

F2,12,.05 = 3.89

MSA 2358.2

F=

=

= 25.275

MSW

93.3

Decision:

Reject H0 at α = 0.05

Conclusion:

There is evidence that

at least one μi differs

F = 25.275

from the rest

Quantitative Analysis for

Managers

Chapter 11

Analysis of Variance

Troy Program – MBA 6640

1

Chapter Goals

After completing this chapter, you should be able to:

Recognize situations in which to use analysis of variance

Understand different analysis of variance designs

Perform a one-way and two-way analysis of variance and

interpret the results

Conduct and interpret a Kruskal-Wallis test

Analyze two-factor analysis of variance tests with more than

one observation per cell

One-Way Analysis of Variance

Evaluate the difference among the means of three

or more groups

Examples: Average production for 1st, 2nd, and 3rd shift

Expected mileage for five brands of tires

Assumptions

Populations are normally distributed

Populations have equal variances

Samples are randomly and independently drawn

Hypotheses of One-Way ANOVA

H0 : μ1 = μ2 = μ3 = = μK

All population means are equal

i.e., no variation in means between groups

H1 : μi ≠ μ j

for at least one i, j pair

At least one population mean is different

i.e., there is variation between groups

Does not mean that all population means are different

(some pairs may be the same)

One-Way ANOVA

H0 : μ1 = μ2 = μ3 = = μK

H1 : Not all μi are the same

All Means are the same:

The Null Hypothesis is True

(No variation between

groups)

μ1 = μ2 = μ3

One-Way ANOVA

(continued)

H0 : μ1 = μ2 = μ3 = = μK

H1 : Not all μi are the same

At least one mean is different:

The Null Hypothesis is NOT true

(Variation is present between groups)

or

μ1 = μ2 ≠ μ3

μ1 ≠ μ2 ≠ μ3

Variability

The variability of the data is key factor to test the

equality of means

In each case below, the means may look different, but a

large variation within groups in B makes the evidence

that the means are different weak

A

B

A

B

Group

C

Small variation within groups

A

B

Group

C

Large variation within groups

Partitioning the Variation

Total variation can be split into two parts:

SST = SSW + SSG

SST = Total Sum of Squares

Total Variation = the aggregate dispersion of the individual

data values across the various groups

SSW = Sum of Squares Within Groups

Within-Group Variation = dispersion that exists among the

data values within a particular group

SSG = Sum of Squares Between Groups

Between-Group Variation = dispersion between the group

sample means

Partition of Total Variation

Total Sum of Squares

(SST)

=

Variation due to

random sampling

(SSW)

+

Variation due to

differences

between groups

(SSG)

Total Sum of Squares

SST = SSW + SSG

K

ni

SST = ∑∑ (x ij − x)

Where:

2

i=1 j=1

SST = Total sum of squares

K = number of groups (levels or treatments)

ni = number of observations in group i

xij = jth observation from group i

x = overall sample mean

Total Variation

(continued)

SST = (x11 − x )2 + (X12 − x )2 + ... + (x KnK − x )2

Response, X

x

Group 1

Group 2

Group 3

Within-Group Variation

SST = SSW + SSG

K

ni

SSW = ∑∑ (x ij − x i )2

i =1 j=1

Where:

SSW = Sum of squares within groups

K = number of groups

ni = sample size from group i

Xi = sample mean from group i

Xij = jth observation in group i

Within-Group Variation

(continued)

K

ni

SSW = ∑∑ (x ij − x i )

i =1 j=1

Summing the variation

within each group and then

adding over all groups

2

SSW

MSW =

n −K

Mean Square Within =

SSW/degrees of freedom

μi

Within-Group Variation

(continued)

SSW = (x11 − x1 )2 + (x12 − x1 )2 + ... + (x KnK − x K )2

Response, X

x1

Group 1

Group 2

x2

Group 3

x3

Between-Group Variation

SST = SSW + SSG

K

SSG = ∑ ni ( x i − x )

Where:

2

i=1

SSG = Sum of squares between groups

K = number of groups

ni = sample size from group i

xi = sample mean from group i

x = grand mean (mean of all data values)

Between-Group Variation

(continued)

K

SSG = ∑ ni ( x i − x )

2

i=1

Variation Due to

Differences

Between

Groups

SSG

MSG =

K −1

Mean Square Between Groups

= SSG/degrees of freedom

μi

μj

Between-Group Variation

(continued)

SSG = n1(x1 − x) + n2 (x 2 − x) + ... + nK (x K − x)

2

2

Response, X

x1

Group 1

Group 2

x2

Group 3

x3

x

2

Obtaining the Mean Squares

SST

MST =

n −1

SSW

MSW =

n −K

SSG

MSG =

K −1

One-Way ANOVA Table

Source of

Variation

SS

df

Between

Groups

SSG

K-1

Within

Groups

SSW

n-K

SST =

SSG+SSW

n-1

Total

MS

(Variance)

F ratio

SSG

MSG

MSG =

K - 1 F = MSW

SSW

MSW =

n-K

K = number of groups

n = sum of the sample sizes from all groups

df = degrees of freedom

One-Factor ANOVA

F Test Statistic

H0: μ1= μ2 = … = μK

H1: At least two population means are different

Test statistic

MSG

F=

MSW

MSG is mean squares between variances

MSW is mean squares within variances

Degrees of freedom

df1 = K – 1

(K = number of groups)

df2 = n – K

(n = sum of sample sizes from all groups)

Interpreting the F Statistic

The F statistic is the ratio of the between estimate of variance and

the within estimate of variance

The ratio must always be positive

df1 = K -1 will typically be small

df2 = n - K will typically be large

Decision Rule:

Reject H if

0

F > FK-1,n-K,α

α = .05

0

Do not

reject H0

Reject H0

FK-1,n-K,α

One-Factor ANOVA

F Test Example

You want to see if three

different golf clubs yield

different distances. You

randomly select five

measurements from trials on

an automated driving

machine for each club. At

the .05 significance level, is

there a difference in mean

distance?

Club 1

254

263

241

237

251

Club 2

234

218

235

227

216

Club 3

200

222

197

206

204

One-Factor ANOVA Example:

Scatter Diagram

Club 1

254

263

241

237

251

Club 2

234

218

235

227

216

Club 3

200

222

197

206

204

Distance

270

260

250

240

230

•

••

•

•

220

x1 = 249.2 x 2 = 226.0 x 3 = 205.8

x = 227.0

210

x1

••

•

••

x2

•

••

••

200

190

1

2

Club

x

3

x3

One-Factor ANOVA Example

Computations

Club 1

254

263

241

237

251

Club 2

234

218

235

227

216

Club 3

200

222

197

206

204

x1 = 249.2

n1 = 5

x2 = 226.0

n2 = 5

x3 = 205.8

n3 = 5

x = 227.0

n = 15

K=3

SSG = 5 (249.2 – 227)2 + 5 (226 – 227)2 + 5 (205.8 – 227)2 = 4716.4

SSW = (254 – 249.2)2 + (263 – 249.2)2 +…+ (204 – 205.8)2 = 1119.6

MSG = 4716.4 / (3-1) = 2358.2

MSW = 1119.6 / (15-3) = 93.3

2358.2

F=

= 25.275

93.3

One-Factor ANOVA Example

Solution

Test Statistic:

H0: μ1 = μ2 = μ3

H1: μi not all equal

α = .05

df1= 2

df2 = 12

Critical Value:

F2,12,.05= 3.89

α = .05

0

Do not

reject H0

Reject H0

F2,12,.05 = 3.89

MSA 2358.2

F=

=

= 25.275

MSW

93.3

Decision:

Reject H0 at α = 0.05

Conclusion:

There is evidence that

at least one μi differs

F = 25.275

from the rest

## Giáo trình nghiên cứu Marketing- Chương 11

## Giáo trình đại số 11

## Hình học xạ ảnh 11

## Tài liệu ôn tập 11: tự luyện thi đại học số 07

## Đề tự luyện thi đại học sô 11

## Slide bài giảng kinh doanh quốc tế - Chương 11

## Bộ đề luyện thi ĐH-CĐ môn Toán P2 - Đề 11

## Đề thi liên thông đại học ngân hàng ngày 5/11/2011

## Ôn thi ĐH môn hóa - Đề 11

## Đề thi Quản trị học - Đề 11

Tài liệu liên quan