Single Factor Analysis of Variance (ANOVA) - Lecture 1, Study notes of Construction

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Chapter 11 - Lecture 1 Single Factor ANOVA

Yuan Huang

April 5, 2013

Chapter 9 : hypothesis testing for one population mean.

Chapter 10: hypothesis testing for two population means.

What comes next?

Examples of multiple population comparisons:

• (^) Compare safety factors for multiple makers of cars.
• (^) Compare effect of the same drug in different doses.
• Compare survival time for different treatments for cancers.
• (^) Compare the yield of crop with different pesticides.

Father of Modern statistics - Sir R.A. Fisher

• ANOVA 1918, 1919 (Rothamsted Experimental Station (

studies in crop variation)).

Example : An experiment to study the effect of sunshine and water

in growing corns. Suppose the sunshine can be: intense, weak,

bare; amount of water can be: much, little.

• 2 × 3 = 6 populations;
• (^) Factors: sunshine and water (two factors);
• Levels of factors: intense, weak and bare for sunshine; much

and little for water.

When the response is numerical and factor is categorical (finite

number of levels), we could apply the analysis of variance ANOVA.

Depending on how many factors you have in your study (to label

your population), comparisons can be classified as:

• Single-factor study (One-way ANOVA);
• (^) Two-factor study (Two-way ANOVA).

In this lecture, we are introducing One-way ANOVA.

There are three assumptions:

• (^) Each sample has the same size, denoted by J - the data is

balanced.

• (^) The populations are normally distributed with mean μ i.
• Equal variance: σ

2

1

=... = σ

2

I

Note: To verify equality of variances, there is a formal test called

the Levene test. A rule of thumb that one can use is that the

largest standard deviation is not larger than two times the smaller.

Example

Let me first give some numbers to help understand the notations

will be introduced.

• Assume I am teaching on three different sections of stat 200

and I am giving them a test. I want to test if the true

averages on the test for all classes are equal. I am selecting a

sample of 5 students from each class.

• Class 1: 70, 50, 100, 100, 70
• (^) Class 2: 60, 85, 65, 100, 30
• Class 3: 80, 50, 90, 75, 85

Variance

Sample Variance

Sample variance:

S

2

i

J ∑

j=

(X

ij

X

i.

2

J − 1

Idea

Hypothesis:

H

0

: μ 1

=... = μ I

vs H 1

: at least two μ

′

i

s differ.

• In order for the null hypothesis to be true, we expect the

sample means ¯xi. to be as close to the grand mean ¯x.. as

possible.

• (^) How to define close?

Sum of Squares

Error Sum of Square SSE

Definition: error sum of squares SSE:

SSE =

I ∑

i=

J ∑

j=

( ¯X

ij

X

i.

2

= (J − 1)

I ∑

i=

S

2

i

• (^) Distribution of SSE:

SSE

σ

2

∼ χ

2

I (J−1)

Sum of Squares

Treatment sum of square SSTr

Definition: treatment sum of squares SSTr:

SSTr = J

I ∑

i=

( ¯X

i.

X

..

2

• (^) Distribution of SSTr: (If H 0 is true)

SSTr

σ

2

∼ χ

2

I − 1