1. CHAPTER 4 Analysis of Variance One-way ANOVA

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. 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.

5. The is the total of all observations from the treatment, while
is the grand total of all N observations.
Treatment 1 2 … i k .
Total

6. Then the hypothesis can be written as: For model : For model :

7. 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

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

9. 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

10. 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

11. 3. Construct ANOVA table:

12. Source of Variation
Sum of Squares
Degrees of Freedom
Mean Square
F Calculated
Treatment
3 – 1 = 2
Error
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

13. 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

14. The following is a partial ANOVA table:
Exercise 4.2:
The following is a partial ANOVA table:
Complete the table and answer the following questions:
Complete the ANOVA table. (red colour is the answer)
Write all the hypothesis
What is your conclusion regarding the hypothesis? Using 5%significance level.
Source
Sum of Squares df
Mean Square F
Treatment
= 320 2
320 / 2 = 160
160 / 20 = 8
Error
9 * 20 =180
11-2=9 20
Total
500 11

15. Exercise 4. 3 The following is sample information
Exercise 4.3 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