Ch 14 (2) Slide 2 2 Randomized Blocks (Two-way) Analysis of
Variance The purpose of designing a randomized block experiment is
to reduce the within- treatments variation thus increasing the
relative amount of between treatment variation. This helps in
detecting differences between the treatment means more easily.
Slide 3 3 Block all the observations with some commonality across
treatments Randomized Blocks Slide 4 4 The sum of square total is
partitioned into three sources of variation Treatments Blocks
Within samples (Error) SS(Total) = SST + SSB + SSE Sum of square
for treatmentsSum of square for blocksSum of square for error
Recall. For the independent samples design we have: SS(Total) = SST
+ SSE Partitioning the total variability Slide 5 5 Calculating the
sums of squares Formula for the calculation of the sums of squares
SST = SSB= Slide 6 6 Calculating the sums of squares Formula for
the calculation of the sums of squares SST = SSB= Slide 7 7 To
perform hypothesis tests for treatments and blocks we need Mean
square for treatments Mean square for blocks Mean square for error
Test statistics for treatments Test statistics for blocks Mean
Squares and Test statistics Slide 8 8 Testing the mean responses
for treatments F > F ,k-1,n-k-b+1 Testing the mean response for
blocks F> F ,b-1,n-k-b+1 The F test Rejection Regions Slide 9 9
[ ] Randomized Blocks ANOVA Slide 10 10 ANOVA Table Slide 11 11
Example 1 A randomized block experiment produced the following
statistics: k=5 b=12 SST=1500 SSB=1000 SS(Total)=3500 a. Test to
determine whether the treatment means differ. (Use =0.01) b. Test
to determine whether the blocks means differ. (Use =0.01) Slide 12
12 Solution 1 ANOVA table a. Rejection region: F>F ,k-1,n-k-b+1
=F 0.01,4,44 3.77 F=16.50 >3.77 There is enough evidence to
conclude that the treatment means differ b. Rejection region:
F>F ,b-1,n-k-b+1 =F 0.01,11,44 2.67 Conclusion: F=4.00>2.67
There is enough evidence to conclude that the block means differ
Slide 13 13 Example 2 As an experiment to understand measurement
error, a statistics professor asks four students to measure the
height of the professor, a male student, and a female student. The
differences (in centimeters) between the correct dimension and the
ones produced by the students are listed here. Can we infer at the
5% significance level that there are differences in the errors
between the subjects being measured? studentProfessorMale Student
Female Student 11.41.51.3 23.12.62.4 32.82.11.5 43.43.62.9 Slide 14
14 Solution 2 H 0 : 1 = 2 = 3 H 1 : At least two means differ
Rejection region: F>F ,k-1,n-k-b+1 =F 0.05,2,6 =5.14 K=3, b=4,
grand mean=2.38 =7.3 Slide 15 Solution 2 15 Slide 16 16 Two-Factor
Analysis of Variance Example 2 Suppose in Example 1, two factors
are to be examined: The effects of the marketing strategy on sales.
Emphasis on convenience Emphasis on quality Emphasis on price The
effects of the selected media on sales. Advertise on TV Advertise
in newspapers Slide 17 17 Two-way ANOVA (two factors) City 1 sales
City3 sales City 5 sales City 2 sales City 4 sales City 6 sales TV
Newspapers ConvenienceQualityPrice Factor A: Marketing strategy
Factor B: Advertising media Slide 18 18 Levels of factor A 123
Level 1 of factor B Level 2 of factor B 123 123123 Level 1and 2 of
factor B Difference between the levels of factor A No difference
between the levels of factor B Difference between the levels of
factor A, and difference between the levels of factor B; no
interaction Levels of factor A No difference between the levels of
factor A. Difference between the levels of factor B Interaction M R
e s a p n o n s e M R e s a p n o n s e M R e s a p n o n s e M R e
s a p n o n s e Interaction Slide 19 19 Interaction Without
interaction With interaction B-B- B-B- B+B+ B+B+ B+B+ B+B+ B-B-
B-B- Slide 20 20 Terminology A complete factorial experiment is an
experiment in which the data for all possible combinations of the
levels of the factors are gathered. This is also known as a two-way
classification. The two factors are usually labeled A & B, with
the number of levels of each factor denoted by a & b
respectively. The number of observations for each combination is
called a replicate, and is denoted by r. For our purposes, the
number of replicates will be the same for each treatment, that is
they are balanced. Slide 21 21 Hypothesis H 0 : Factor A and Factor
B do not interact to affect the mean responses H 1 : Factor A and
Factor B do interact to affect the mean responses H 0 : The means
of the a levels of factor A are equal H 1 : At least two means
differ H 0 : The means of the b levels of factor B are equal H 1 :
At least two means differ Slide 22 22 Sums of squares Slide 23 23
Sums of squares Slide 24 24 Slide 25 25 F tests for the Two-way
ANOVA Test for the difference between the levels of the main
factors A and B F= MS(A) MSE F= MS(B) MSE Rejection region: F >
F ,a-1,n-ab F > F , b-1, n-ab Test for interaction between
factors A and B F= MS(AB) MSE Rejection region: F > F
a-1)(b-1),n-ab SS(A)/(a-1) SS(B)/(b-1) SS(AB)/(a-1)(b-1) SSE/(n-ab)
Slide 26 26 ANOVA Table n = abr Slide 27 27 Example 3 The following
data were generated from a 2 X 2 factorial experiment with 3
replicates. Slide 28 28 Example 3 - continued a. Test at the 5%
significance level to determine whether factors A and B interact.
b. Test at the 5% significance level to determine whether
differences exists between the levels of factor A. c. Test at the
5% significance level to determine whether differences exist
between the levels of factor B. Slide 29 29 Solution 3 Slide 30 30
Solution 3 - continued a F =.31, p-value =.5943. There is not
enough evidence to conclude that factors A and B interact. b F =
1.23, p-value =.2995. There is not enough evidence to conclude that
differences exist between the levels of factor A. c F = 13.00,
p-value =.0069. There is enough evidence to conclude that
differences exist between the levels of factor B. Slide 31 Solution
3 - continued 31 Slide 32 Example 4 The required conditions for a
two-factor ANOVA are that the distribution of the response is
__________________ distributed; the variance for each treatment is
________; and the samples are _______. (a) normally; equal;
independent (b) normally; the same; independent (c) normally;
identical; independent 32 Slide 33 33 Two means are considered
different if the difference between the corresponding sample means
is larger than a critical number. Then, the larger sample mean is
believed to be associated with a larger population mean. Conditions
common to all the methods here: The ANOVA model is the one way
analysis of variance The conditions required to perform the ANOVA
are satisfied. Multiple Comparisons Slide 34 34 Inference about :
Equal variances Recall Construct the t-statistic as follows: Build
a confidence interval Slide 35 35 Fisher Least Significant
Different (LSD) Method This method builds on the equal variances
t-test of the difference between two means. The test statistic is
improved by using MSE rather than s p 2. We can conclude that i and
j differ (at % significance level if > LSD, where Slide 36 36
Experimentwise Type I error rate ( E ) (the effective Type I error)
The Fisher s method may result in an increased probability of
committing a type I error. The experimentwise Type I error rate is
the probability of committing at least one Type I error at
significance level of It is calculated by E = 1-(1 ) C where C is
the number of pairwise comparisons (i.e. C = k(k-1)/2) The
Bonferroni adjustment determines the required Type I error
probability per pairwise comparison ( ), to secure a pre-determined
overall E Slide 37 37 The procedure: Compute the number of pairwise
comparisons (C) [C=k(k-1)/2], where k is the number of populations.
Set = E /C, where E is the true probability of making at least one
Type I error (called experimentwise Type I error). We can conclude
that i and j differ (at /C% significance level if Bonferroni
Adjustment Slide 38 38 Example 1 - continued Rank the effectiveness
of the marketing strategies (based on mean weekly sales). Use the
Fisher s method, and the Bonferroni adjustment method Solution (the
Fisher s method) The sample mean sales were 577.55, 653.0, 608.65.
Then, Fisher and Bonferroni Methods Slide 39 39 Solution (the
Bonferroni adjustment) We calculate C=k(k-1)/2 to be 3(2)/2 = 3. We
set =.05/3 =.0167, thus t.0167 2, 60-3 = 2.467 (Excel). Again, the
significant difference is between 1 and 2. Fisher and Bonferroni
Methods Slide 40 40 The test procedure: Find a critical number as
follows: k = the number of treatments =degrees of freedom = n - k n
g = number of observations per sample (recall, all the sample sizes
are the same) = significance level q (k, ) = a critical value
obtained from the studentized range table Tukey Multiple
Comparisons Slide 41 41 If the sample sizes are not extremely
different, we can use the above procedure with n g calculated as
the harmonic mean of the sample sizes. Repeat this procedure for
each pair of samples. Rank the means if possible. Select a pair of
means. Calculate the difference between the larger and the smaller
mean. If there is sufficient evidence to conclude that max >
min. Tukey Multiple Comparisons Slide 42 42 Which Multiple
Comparison Method to Use If you have identified two or three
pairwise comparison, use the Bonferroni method. If you plan to
compare all possible combinations, use Turkey. If the purpose of
the analysis is to point to areas that should be investigated
further, Fisher s LSD method is indicated. Slide 43 43 Example 5 a.
Use Fishers LSD procedure with =0.05 to determine which population
means differ given the following statistics. b. Repeat part a using
the Bonferroni adjustment. c. Repeat part a using Tukeys multiple
comparison method Slide 44 44 Solution 5 Slide 45 45 Solution 5 -
continued Slide 46 46 Solution 5 - continued Slide 47 Example 6
Which of the following statements about multiple comparison methods
is false? a. They are to be use once the F-test in ANOVA has been
rejected. b. They are used to determine which particular population
means differ. c. There are many different multiple comparison
methods but all yield the same conclusions. d. All of these choices
are true. 47
