1. i) Two way ANOVA without replication
CHAPTER 4
Analysis of Variance
One-way ANOVA
Two-way ANOVA
i) Two way ANOVA without replication
ii) Two way ANOVA with replication

2. Introduction
In this chapter, expand the idea of hypothesis tests. We describe a test of variances and then a test that simultaneously compares several means to determine if they came from equal populations.
The simultaneous comparison of several population means called analysis of variance (ANOVA).
Using F-test where test whether two samples are populations having equal variances and compare several population means simultaneously.

3. ANOVA assumptions:
The population follows the normal distribution.
The populations have equal standard deviation.
The populations are independent.

4. Key Concepts ANOVA is analysis of variance.
ANOVA can be used to analyze the data obtained from experimental or observational studies.
A factor is a variable that the experimenter has selected for investigation.
A level is the intensity setting of a factor.
A treatment is a specific combination of a factor levels.
Experimental units are the objects of interest in the experiment.
Variation between treatment groups captures the effect of the treatment.
Variation within treatment groups represents random error not explained by the experimental treatments.

5. Example 4.1
Suppose that the experimenter began by randomly selecting 20 men and 20 women for the experiment. These two groups were then randomly divided into ten each for the experimental and control groups. What are the factors, levels and treatments in this experiment?
Solution :
Now there are two factors of interest to the experimental and control groups and each factor has two levels:
“Gender” at two levels: men and women
“Meal” at two levels: breakfast and no breakfast
In this more complex experiment, there are four treatments, one for each specific combination of factor levels: men without breakfast, men with breakfast, women without breakfast and women with breakfast.

6. One-way ANOVA
The one-way analysis of variance specifically allows us to compare several groups of observations whether or not their population mean are equal. One way ANOVA is also known as Completely Randomized Design (CRD).
This design only involves one factor.
The application of one way ANOVA requires that the following assumptions hold true:
(i) The populations from which the samples are drawn are
(approximately) normally distributed.
(ii) The populations from which the samples are drawn have the same variance.
(iii) The samples drawn from different populations are random and independent.

7. Each observation may be written as: Or alternatively written as: Where

8. The is the total of all observations from the treatment, while
is the grand total of all N observations.
Then the hypothesis can be written as:
Note: This hypothesis is use for model
Treatment 1 2 … i k .
Total

9. For model the hypothesis is as follows: The computations for an analysis of variance problem are usually summarized in tabular form as shown in table below. This table is referred to as the ANOVA table.
Source of Variation
Sum of Squares
Degree of freedom
Mean Square
F Calculated
Treatment
(Between levels)
SSTR
k - 1
Error
(within levels)
SSE
N - k
Total
SST
N - 1

10. where We reject if and conclude that some of the data is due to differences in the treatment levels.

11. Example 4.2 Three different types of acid can be used in a particular chemical process. The resulting yield (in %) from several batches using the different types of acid are given below: Test whether or not the three populations appear to have equal means using = 0.05.
Acid A B C 93 95 76 97 77 74 87 84

12. Solution: 1. Construct the table of calculation: 2
Solution: 1. Construct the table of calculation: 2. Set up the hypothesis:
Acid A B C 93 95 76 97 77 74 87 84

13. 3. Construct ANOVA table:

14. Treatment (Between levels)
Source of Variation
Sum of Squares
Degrees of Freedom
Mean Square
F Calculated
Treatment (Between levels)
3 – 1 = 2
Error (within levels)
9 – 3 = 6
Total
9 – 1 = 8
4. At = 0.05, from the statistical table for f distribution, we have
5. Since , thus we failed to reject
and conclude that there is no difference for mean in the three types of acid at significance at = 0.05

15. Anova: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
Column 1 3
262
Column 2
279 93 28
Column 3
237 79 19
ANOVA
Source of Variation SS df MS F
P-value
F crit
Between Groups 2
Within Groups 6
Total 8

16. Output from Excel Compare calculated values from example to Excel output:
Groups
Count
Sum
Average
Variance
Column 1 3
262
87.333
Column 2
279 93 28
Column 3
237 79 19
ANOVA
Source of Variation SS df MS F
P-value
F crit
Between Groups
298 2
148.78
2.4614
0.1657
5.143
Within Groups
363 6
60.444
Total
660 8
The test statistic
The critical
bound
The p-value

17. Exercise 4.1: Four catalyst that may affect the concentration of one component in a three-component liquid mixture are being investigated. The following concentrations are obtained. Compute a one-way analysis of variance for this experiment and test the hypothesis at 0.05 level of significance and state your conclusion concerning the effect of catalyst on the concentration of one component in three-component liquid mixture. 1 2 3 4
58.2
56.3
50.1
52.9
57.2
54.5
54.2
49.9
58.4
57.0
55.4
50.0
55.8
55.3
51.7
54.9

18. Exercise 4.2
The following is sample information. Test the hypothesis that the treatment means are equal. Use (Answer: Reject )
Treatment 1
Treatment 2
Treatment 3 8 3 6 2 4 10 5 9