business statics final (2hour duration)

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Chapter 13, Part A Analysis of Variance and Experimental Design ■  Introduction to Analysis of Variance ■  Analysis of Variance: Testing for the Equality of k Population Means ■  Multiple Comparison Procedures

Introduction to Analysis of Variance

Analysis of Variance (ANOVA) can be used to test for the equality of three or more population means.

Data obtained from observational or experimental studies can be used for the analysis.

We want to use the sample results to test the following hypotheses:

H0: µ1 = µ2 = µ3 = . . . = µk

Ha: Not all population means are equal

Introduction to Analysis of Variance

H0: µ1 = µ2 = µ3 = . . . = µk

H1: Not all population means are equal

If H0 is rejected, we cannot conclude that all population means are different.

Rejecting H0 means that at least two population means have different values.

■  Sampling Distribution of Given H0 is True x

Introduction to Analysis of Variance

µ 1x 3x2x

Sample means are close together because there is only

one sampling distribution when H0 is true.

2 2 x n

σ σ =

Introduction to Analysis of Variance

■  Sampling Distribution of Given H0 is False x

µ3 1x 2x3x µ1 µ2

Sample means come from different sampling distributions

and are not as close together when H0 is false.

For each population, the response variable is normally distributed.

Assumptions for Analysis of Variance

The variance of the response variable, denoted σ 2, is the same for all of the populations.

The observations must be independent.

Analysis of Variance: Testing for the Equality of k Population Means

■  Between-Treatments Estimate of Population Variance ■  Within-Treatments Estimate of Population Variance ■  Comparing the Variance Estimates: The F Test ■  ANOVA Table

Between-Treatments Estimate of Population Variance

■  A between-treatment estimate of σ 2 is called the mean square treatment and is denoted MSTR.

( ) MSTR

j j j

n x x

k =

−

= −

∑

Denominator represents the degrees of freedom associated with SSTR

Numerator is the sum of squares

due to treatments and is denoted SSTR

■  The estimate of σ 2 based on the variation of the sample observations within each sample is called the mean square error and is denoted by MSE.

Within-Samples Estimate of Population Variance

j jj

−

= ∑ =1

2)1( MSE

Denominator represents the degrees of freedom

associated with SSE

Numerator is the sum of squares

due to error and is denoted SSE

Comparing the Variance Estimates: The F Test

n  If the null hypothesis is true and the ANOVA assumptions are valid, the sampling distribution of MSTR/MSE is an F distribution with MSTR d.f. equal to k - 1 and MSE d.f. equal to nT - k.

n  If the means of the k populations are not equal, the value of MSTR/MSE will be inflated because MSTR overestimates σ 2.

n  Hence, we will reject H0 if the resulting value of MSTR/MSE appears to be too large to have been selected at random from the appropriate F distribution.

Test for the Equality of k Population Means

F = MSTR/MSE

H0: µ1 = µ2 = µ3 = . . . = µk H1: Not all population means are equal

■  Hypotheses

■  Test Statistic

Test for the Equality of k Population Means

■  Rejection Rule

where the value of Fα is based on an F distribution with k - 1 numerator d.f. and nT - k denominator d.f.

Reject H0 if p-value < α p-value Approach:

Critical Value Approach: Reject H0 if F > Fα

Sampling Distribution of MSTR/MSE

■  Rejection Region

Do Not Reject H0

Reject H0

MSTR/MSE

Critical Value Fα

Sampling Distribution of MSTR/MSE

ANOVA Table

SST is partitioned into SSTR and SSE.

SST’s degrees of freedom (d.f.) are partitioned into

SSTR’s d.f. and SSE’s d.f.

Treatment Error Total

SSTR SSE SST

k – 1 nT – k nT - 1

MSTR MSE

Source of Variation

Sum of Squares

Degrees of Freedom

Mean Square

MSTR/MSE

ANOVA Table

SST divided by its degrees of freedom nT – 1 is the overall sample variance that would be obtained if we treated the entire set of observations as one data set.

With the entire data set as one sample, the formula for computing the total sum of squares, SST, is:

1 1

SST ( ) SSTR SSE jnk

ij j i

x x = =

= − = +∑∑

ANOVA Table

ANOVA can be viewed as the process of partitioning the total sum of squares and the degrees of freedom into their corresponding sources: treatments and error.

Dividing the sum of squares by the appropriate degrees of freedom provides the variance estimates and the F value used to test the hypothesis of equal population means.

■  Example: Kako Manufacturing

Test for the Equality of k Population Means

Susi Kako would like to know if there is any significant difference in the mean number of hours worked per week for the department managers at her three manufacturing plants (in Budapest, Balaton, and Bratislava).

■  Example: Kako Manufacturing

Test for the Equality of k Population Means

A simple random sample of five managers from each of the three plants was taken and the number of hours worked by each manager for the previous week is shown on the next slide. Conduct an F test using α = 0.05.

1 2 3 4 5

48 54 57 54 62

73 63 66 64 74

51 63 61 54 56

Plant 1 Budapest

Plant 2 Balaton

Plant 3 Bratislava Observation

Sample Mean Sample Variance

55 68 57 26.0 26.5 24.5

Test for the Equality of k Population Means

H0: µ 1 = µ 2 = µ 3 H1: Not all the means are equal

where: µ 1 = mean number of hours worked per

week by the managers at Plant 1 µ 2 = mean number of hours worked per week by the managers at Plant 2 µ 3 = mean number of hours worked per week by the managers at Plant 3

1. Develop the hypotheses.

n  p -Value and Critical Value Approaches

2. Specify the level of significance. α = 0.05

Test for the Equality of k Population Means

n  p -Value and Critical Value Approaches

3. Compute the value of the test statistic.

MSTR = 490/(3 - 1) = 245 SSTR = 5(55 - 60)2 + 5(68 - 60)2 + 5(57 - 60)2 = 490

= (55 + 68 + 57)/3 = 60 x (Sample sizes are all equal.)

Mean Square Due to Treatments

3. Compute the value of the test statistic.

Test for the Equality of k Population Means

MSE = 308/(15 - 3) = 25.667

SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308 Mean Square Due to Error

(continued)

F = MSTR/MSE = 245/25.667 = 9.55

n  p -Value and Critical Value Approaches

Treatment Error Total

490 308 798

2 12 14

245 25.667

Source of Variation

Sum of Squares

Degrees of Freedom

Mean Square

9.55

Test for the Equality of k Population Means

n  ANOVA Table

Test for the Equality of k Population Means

5. Determine whether to reject H0.

We have sufficient evidence to conclude that the mean number of hours worked per week by department managers is not the same at all 3 plant.

The p-value < 0.05, so we reject H0.

With 2 numerator d.f. and 12 denominator d.f., the p-value is .01 for F = 6.93. Therefore, the p-value is less than 0.01 for F = 9.55.

n  p –Value Approach

4. Compute the p –value.

5. Determine whether to reject H0.

Because F = 9.55 > 3.89, we reject H0.

n  Critical Value Approach

4. Determine the critical value and rejection rule.

Reject H0 if F > 3.89

Test for the Equality of k Population Means

We have sufficient evidence to conclude that the mean number of hours worked per week by department managers is not the same at all 3 plant.

Based on an F distribution with 2 numerator d.f. and 12 denominator d.f., F.05 = 3.89.

Multiple Comparison Procedures

■  Suppose that analysis of variance has provided statistical evidence to reject the null hypothesis of equal population means.

■  Fisher’s least significant difference (LSD) procedure can be used to determine where the differences occur.

Fisher’s LSD Procedure

■  Test Statistic

1 1MSE( )

i j

x x t

n n

− =

■  Hypotheses µ µ−0 : i jH µ µ≠: a i jH

Fisher’s LSD Procedure

where the value of ta/2 is based on a t distribution with nT - k degrees of freedom.

■  Rejection Rule

Reject H0 if p-value < α

p-value Approach:

Critical Value Approach:

Reject H0 if t < -ta/2 or t > ta/2

Fisher’s LSD Procedure Based on the Test Statistic xi - xj

■  Test Statistic

_ _

/2 1 1LSD MSE( )

i j t n nα= +

where

−i jx x

Reject H0 if > LSD −i jx x

■  Hypotheses

■  Rejection Rule

µ µ−0 : i jH µ µ≠: a i jH

Fisher’s LSD Procedure Based on the Test Statistic xi - xj

■  Example: Kako Manufacturing

Recall that Susi Kako wants to know if there is any significant difference in the mean number of hours worked per week for the department managers at her three manufacturing plants.

Analysis of variance has provided statistical evidence to reject the null hypothesis of equal population means. Fisher’s least significant difference (LSD) procedure can be used to determine where the differences occur.

Fisher’s LSD Procedure Based on the Test Statistic xi - xj

For α = 0.05 and nT - k = 15 – 3 = 12 degrees of freedom, t.025 = 2.179

LSD = + =2 179 25 667 15 1 5 6 98. . ( ) .LSD = + =2 179 25 667

1 5 1 5 6 98. . ( ) .

/2 1 1LSD MSE( )

i j t n nα= +

MSE value was computed earlier

Fisher’s LSD Procedure Based on the Test Statistic xi - xj

■  LSD for Plants 1 and 2

•  Conclusion

•  Test Statistic −1 2x x = |55 - 68| = 13

Reject H0 if > 6.98 −1 2x x •  Rejection Rule

µ µ−0 1 2: H µ µ≠1 2: aH

•  Hypotheses (A)

The mean number of hours worked at Plant 1 is not equal to the mean number worked at Plant 2.

■  LSD for Plants 1 and 3

Fisher’s LSD Procedure Based on the Test Statistic xi - xj

•  Conclusion

•  Test Statistic −1 3x x = |55 - 57| = 2

Reject H0 if > 6.98 −1 3x x •  Rejection Rule

µ µ−0 1 3: H µ µ≠1 3: aH

•  Hypotheses (B)

There is no significant difference between the mean number of hours worked at Plant 1 and the mean number of hours worked at Plant 3.

■  LSD for Plants 2 and 3

Fisher’s LSD Procedure Based on the Test Statistic xi - xj

•  Conclusion

•  Test Statistic −2 3x x = |68 - 57| = 11

Reject H0 if > 6.98 −2 3x x •  Rejection Rule

µ µ−0 2 3: H µ µ≠2 3: aH

•  Hypotheses (C)

The mean number of hours worked at Plant 2 is not equal to the mean number worked at Plant 3.

Type I Error Rates

■  The experimentwise Type I error rate gets larger for problems with more populations (larger k).

αEW = 1 – (1 – α)(k – 1)!

■  The comparisonwise Type I error rate α indicates the level of significance associated with a single pairwise comparison.

■  The experimentwise Type I error rate αEW is the probability of making a Type I error on at least one of the (k – 1)! pairwise comparisons.

End of Chapter 13, Part A