1. Statistics
Analysis of Variance

2. Contents Statistics in practice Introduction to Analysis of Variance
Analysis of Variance: Testing for the Equality of k Population Means
Multiple Comparison Procedures

3. STATISTICS in PRACTICE
Burke Marketing Services, Inc., is one of the most experienced market research firms in the industry.
In one study, a firm retained Burke to evaluate potential new versions of a children’s dry cereal. Analysis of variance was the statistical method used to study the data obtained from the taste tests.

4. STATISTICS in PRACTICE (Continue)
The experimental design employed by Burke and the subsequent analysis of variance were helpful in making a product design recommendation.

5. Introduction to Analysis of Variance (ANOVA)
Analysis of Variance (ANOVA) can be used to test for the equality of three or more populations.
Data obtained from observational or experimental studies can be used for the analysis.

6. Introduction to Analysis of Variance (ANOVA)
We want to use the sample results to test the following hypotheses:
H0: 1=2=3= = k
Ha: Not all population means are equal

7. Introduction to Analysis of Variance (ANOVA)
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.

8. Introduction to Analysis of Variance (ANOVA)
Sample means are close together
because there is only
one sampling distribution
when H0 is true. 
Sampling Distribution of Given H0 is True

9. Introduction to Analysis of Variance (ANOVA)
Sample means come from
different sampling distributions
and are not as close together
when H0 is false. 3 1 2
Sampling Distribution of Given H0 is False

10. Introduction to Analysis of Variance (ANOVA)
For each population, the response variable is normally distributed.
The variance of the response variable, denoted  2, is the same for all of the populations.
The observations must be independent.
TA will teach how to valid. use MINITAB to test if these three assumptions are valid.

11. Introduction to Analysis of Variance (ANOVA)
Terminology:
Treatment
Between Treatment
Within Treatment
TA will teach how to valid. use MINITAB to test if these three assumptions are valid.

12. 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

13. Analysis of Variance: Testing for the Equality of k Population Means
Analysis of variance can be used to test for the equality of k population means.
The hypotheses tested is
H0:
Ha: Not all population means are equal
where mean of the jth population.

14. Analysis of Variance: Testing for the Equality of k Population Means
Sample data
= value of observation i for treatment j
= number of observations for treatment j
= sample mean for treatment j
= sample variance for treatment j
= sample standard deviation for treatment j

15. Analysis of Variance: Testing for the Equality of k Population Means
statisitcs
The sample mean for treatment j
The sample variance for treatment j

16. Analysis of Variance: Testing for the Equality of k Population Means
The overall sample mean
where nT = n1 + n nk
If the size of each sample is n, nT = kn then

17. Analysis of Variance: Testing for the Equality of k Population Means
Between-Treatments Estimate of Population Variance
The sum of squares due to treatments (SSTR)

18. Analysis of Variance: Testing for the Equality of k Population Means
Between-Treatments Estimate of Population Variance
The mean square due to treatments (MSTR)

19. Analysis of Variance: Testing for the Equality of k Population Means
Within-Treatments Estimate of Population Variance
The sum of squares due to error (SSE
The mean square due to error (MSE)

20. Between-Treatments Estimate of Population Variance
A between-treatment estimate of  2 is called the mean square treatment and is denoted MSTR.
Numerator is the
sum of squares
due to treatments
and is denoted SSTR
Denominator represents
the degrees of freedom
associated with SSTR

21. Within-Samples Estimate of Population Variance
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.
Numerator is the
sum of squares
due to error
and is denoted SSE
Denominator represents
the degrees of freedom
associated with SSE

22. 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  σ2.

23. Test for the Equality of k Population Means
Hypotheses
H0: 1=2=3= = k
Ha: Not all population means are equal
Test Statistic
F = MSTR/MSE

24. Reject H0 if p-value < a
Test for the Equality of k Population Means
Rejection Rule:
p-value Approach:
Critical Value Approach:
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 < a
Reject H0 if F > Fa

25. Sampling Distribution
Sampling Distribution of MSTR/MSE
Rejection Region
Do Not Reject H0
Reject H0
MSTR/MSE
Critical Value F
Sampling Distribution
of MSTR/MSE a

26. ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean
Treatment
Error
Total
SSTR
SSE
SST
k – 1
nT – k
nT - 1
MSTR
MSE
MSTR/MSE
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.

27. 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:

28. 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.

29. Test for the Equality of k Population Means-- Example
National Computer Products, Inc. (NCP), manufactures printers and fax machines at plants located in Atlanta, Dallas, and Seattle.
Objective: To measure how much employees at these plants know about total quality management.
A random sample of six employees was selected from each plant and given a quality awareness examination.

30. Test for the Equality of k Population Means--Example
Data
Let = mean examination score for population 1
= mean examination score for population 2
= mean examination score for population 3

31. Test for the Equality of k Population Means-- Example
Hypotheses
H0: = =
Ha: Not all population means are equal
In this example
1. dependent or response variable : examination score
2. independent variable or factor : plant location
3. levels of the factor or treatments : the values of a factor selected for investigation, in the NCP example the three treatments or three population are Atlanta, Dallas, and Seattle.

32. Test for the Equality of k Population Means-- Example
Three assumptions:
1. For each population, the response variable is normally distributed. The examination scores (response variable) must be normally distributed at each plant.
2. The variance of the response variable, , is the same for all of the populations. The variance of examination scores must be the same for all three plants.
3. The observations must be independent. The examination score for each employee must be independent of the examination score for any other employee.

33. Test for the Equality of k Population Means-- Example
ANOVA Table
p-value = < α = .05. We reject H0.

34. Test for the Equality of k Population Means-- Example
Example: Reed Manufacturing
Janet Reed 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 Buffalo, Pittsburgh, and Detroit).

35. Test for the Equality of k Population Means-- Example
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 = .05.

36. Test for the Equality of k Population Means-- Example1 2 3 4 5 48 54 57 62 73 63 66 64 74 51 61 56
Plant 1
Buffalo
Plant 2
Pittsburgh
Plant 3
Detroit
Observation
Sample Mean
Sample Variance

37. Test for the Equality of k Population Means-- Example
p -Value and Critical Value Approaches
1. Develop the hypotheses.
H0:  1= 2= 3
Ha: Not all the means are equal
where:
 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

38. Test for the Equality of k Population Means-- Example
2. Specify the level of significance.
a = .05
3. Compute the value of the test statistic.
Mean Square Due to Treatments
(Sample sizes are all equal.)
= ( )/3 = 60
SSTR = 5( )2 + 5( )2 + 5( )2 = 490
MSTR = 490/(3 - 1) =
MSE = 308/(15 - 3) =
SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308
Mean Square Due to Error
F = MSTR/MSE = 245/ =

39. Test for the Equality of k Population Means-- Example
ANOVA Table
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Mean
Squares F
Treatment
Error
Total
490
308
798 2 12 14
245
25.667
9.55

40. Test for the Equality of k Population Means-- Example
p -Value Approaches
4. Compute the p –value.
With 2 numerator d.f. and 12 denominator d.f.,the p-value is .01 for F = Therefore, the p-value is less than .01 for F = 9.55.
5. Determine whether to reject H0.
The p-value < .05, so we 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.

41. Test for the Equality of k Population Means-- Example
Critical Value Approaches
4. Determine the critical value and rejection rule.
Based on an F distribution with 2 numerator
d.f. and 12 denominator d.f., F.05 = 3.89.
Reject H0 if F > 3.89
5. Determine whether to reject H0.
Because F = 9.55 > 3.89, we 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.

42. Test for the Equality of k Population Means
Summary

43. 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.

44. Fisher’s LSD Procedure
Hypotheses
Test Statistic

45. Fisher’s LSD Procedure
Rejection Rule
p-value Approach:
Reject H0 if p-value < a
Critical Value Approach:
Reject H0 if t < -ta/2 or t > ta/2
where the value of ta/2 is based on a
t distribution with nT - k degrees of freedom.

46. Fisher’s LSD Procedure Based on the Test Statistic xi - xj
Hypotheses
Test Statistic
Rejection Rule
Reject H0 if > LSD

47. Fisher’s LSD Procedure Based on the Test Statistic xi – xj -- Example
Example: Reed Manufacturing
Recall that Janet Reed 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.

48. Fisher’s LSD Procedure Based on the Test Statistic xi – xj -- Example
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.

49. Fisher’s LSD Procedure Based on the Test Statistic xi – xj -- Example
For  = .05 and nT - k = 15 – 3 = 12
degrees of freedom, t.025 = 2.179
MSE value was
computed earlier

50. Fisher’s LSD Procedure Based on the Test Statistic xi – xj -- Example
LSD for Plants 1 and 2
Hypotheses (A)
Rejection Rule
Reject H0 if > 6.98
Test Statistic
= | | = 13
Conclusion:
The mean number of hours worked at Plant 1 is
not equal to the mean number worked at Plant 2.

51. Fisher’s LSD Procedure Based on the Test Statistic xi – xj -- Example
LSD for Plants 1 and 3
Hypotheses (B)
Rejection Rule
Reject H0 if > 6.98
Test Statistic
= | | = 2
Conclusion:
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.

52. Fisher’s LSD Procedure Based on the Test Statistic xi – xj -- Example
LSD for Plants 2 and 3
Hypotheses (C)
Rejection Rule
Reject H0 if > 6.98
Test Statistic
= | | = 11
Conclusion:
The mean number of hours worked at Plant 2 is
not equal to the mean number worked at Plant 3.

53. Type I Error Rates
The experimentwise Type I error rate aEW is the probability of making a Type I error on at least one of the (k – 1)! pairwise comparisons.
The comparisonwise Type I error rate a indicates the level of significance associated with a single pairwise comparison.
aEW = 1 – (1 – a)(k – 1)!

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