Chapter 10: Comparisons Involving Means

1 1 Slide

Slide

© 2009, Econ-2030 Applied Statistics-Dr Tadesse

Chapter 10: Comparisons Involving Means

Introduction to Analysis of Variance

Analysis of Variance: Testing for the Equality of

k Population Means

2 2 Slide

Slide


Analysis of Variance

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

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

H0: 1=2=3=. . . = k

Ha: Not all population means are equal

3 3 Slide

Slide



H0: 1=2=3=. . . = k


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

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.

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

4 4 Slide

Slide


Sampling Distribution of the mean Given H0 is True


1x 3x2x

Sample means are close together because there is only

one sampling distribution when H0 is true.

22x n

5 5 Slide

Slide



Sampling Distribution of the mean given H0 is False

33 1x 2x3x 11 22

Sample means come fromdifferent sampling distributionsand are not as close together

when H0 is false.

6 6 Slide

Slide


For each population, the response variable is normally distributed. For each population, the response variable is normally distributed.

Assumptions for Analysis of Variance

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

The observations must be independent. The observations must be independent.

7 7 Slide

Slide


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

8 8 Slide

Slide


Between-Treatments Estimateof Population Variance

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

2

1

( )

MSTR1

k

j jj

n x x

k

Denominator represents the degrees of freedom associated with SSTR

Numerator is the sum of squares

due to treatmentsand is denoted SSTR

9 9 Slide

Slide


The estimate of 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 Estimateof Population Variance

kn

sn

T

k

jjj

1

2)1(

MSE

Denominator represents the degrees of freedom

associated with SSE

Numerator is the sum of squares

due to errorand is denoted SSE

10 10 Slide

Slide


Comparing the Variance Estimates: The F Test

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.

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

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.

11 11 Slide

Slide


Test for the Equality of k Population Means

F = MSTR/MSE

H0: 1=2=3=. . . = k


Hypotheses

Test Statistic

12 12 Slide

Slide



Rejection Rule

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

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

Critical Value Approach: Reject H0 if F > Fa

13 13 Slide

Slide


Sampling Distribution of MSTR/MSE

Rejection Region

Do Not Reject H0Do Not Reject H0

Reject H0Reject H0

MSTR/MSEMSTR/MSE

Critical ValueCritical ValueFF

Sampling Distributionof MSTR/MSE

a

14 14 Slide

Slide


ANOVA Table

SST is partitionedinto SSTR and

SSE.

SST’s degrees of freedom

(d.f.) are partitioned into

SSTR’s d.f. and SSE’s d.f.

TreatmentErrorTotal

SSTRSSESST

k – 1

nT – k

nT - 1

MSTRMSE

Source ofVariation

Sum ofSquares

Degrees ofFreedom

MeanSquares

MSTR/MSE

F

15 15 Slide

Slide


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.

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

2

1 1

SST ( ) SSTR SSEjnk

ijj i

x x

16 16 Slide

Slide


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.

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.

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.

17 17 Slide

Slide


Example:


Janet Reed would like to know ifthere is any significant difference inthe mean number of hours worked per week for the department managersat her three manufacturing plants(in Buffalo, Pittsburgh, and Detroit).

18 18 Slide

Slide



A simple random sample of fivemanagers from each of the three plantswas taken and the number of hoursworked by each manager for theprevious week is shown on the nextslide. Conduct an F test using a = .05.

19 19 Slide

Slide


12345

4854575462

7363666474

5163615456

Plant 1Buffalo

Plant 2Pittsburgh

Plant 3DetroitObservation

Sample MeanSample Variance

55 68 5726.0 26.5 24.5


20 20 Slide

Slide



H0: 1= 2= 3

Ha: Not all the means are equalwhere: 1 = mean number of hours worked per

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

1. Develop the hypotheses.

21 21 Slide

Slide


2. Specify the level of significance. a = .05


3. Compute the value of the test statistic.

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

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

Mean Square Due to Treatments

22 22 Slide

Slide


3. Compute the value of the test statistic.


MSE = 308/(15 - 3) = 25.667

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

F = MSTR/MSE = 245/25.667 = 9.55

23 23 Slide

Slide


TreatmentErrorTotal

490308798

21214

24525.667

Source ofVariation

Sum ofSquares

Degrees ofFreedom

MeanSquares

9.55

F


ANOVA Table

24 24 Slide

Slide


5. Compare the Test Statistic with the critical Value

Because F = 9.55 > 3.89, we reject H0.

Using the Critical Value Approach

4. Determine the critical value and rejection rule.


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 numeratord.f. and 12 denominator d.f., F.05 = 3.89.

25 25 Slide

Slide



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 < .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, thep-value is less than .01 for F = 9.55.

Using the p –Value Approach

4. Compute the p –value.

Documents

Chapter 10: Comparisons Involving Means