Chapter 11 - Lecture 1 Single Factor ANOVA

Transcription

1 Chapter 11 - Lecture 1 Single Factor ANOVA April 7th, 2010

2 Means Variance Sum of Squares

3 Review In Chapter 9 we have seen how to make hypothesis testing for one population mean. In Chapter 10 we have seen how to make hypothesis testing for two samples mean. What comes next?

4 Analysis of Variance In this Chapter we will learn how to make tests when we have 3 or more populations and we want to test equality of their means. Let s assume we have I populations. We are testing H 0 : µ 1 = µ 2 =... = µ I vs H A : noth 0 There are two assumptions: The populations are normally distributed with mean µi. The variances are equal, that is σ 2 1 =... = σi 2 To verify equality of variances, there is a formal test called the Levene test, that we will see later. The problem with Levene test, is that you have to do the hypothesis test and then do 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.

5 Before writing down all the formulas, I will give you an example that we will follow to create the theory. I am teaching on four different sections of stat 200 and I am giving them a test. I want to test if the average on the test for all classes is 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 Class 4: 100, 90, 50, 90, 70

6 Sample Mean and Grand Mean Means Variance Sum of Squares With X ij we denote the j th observation in the i th sample. J j=1 X ij Sample mean: X i. = J I J Grand mean: X.. = i=1 j=1 IJ X ij

7 Sample Variance Outline Means Variance Sum of Squares Sample variance: S 2 i = J (X ij X i. ) 2 j=1 J 1 Now that we have the sample variances we can check the condition of equality of variances.

8 Means Variance Sum of Squares Idea We are testing H 0 : µ 1 = µ 2 =... = µ I vs H A : noth 0 In order for the null hypothesis to be true, we need the sample observations x ij to be as close to the sample mean x i. as possible and at the same time the sample means x i. to be as close to the grand mean x.. as possible. So it is logical to assume that we will compare two measures of variation to see if they are close.

9 Error Sum of Square - SSE Means Variance Sum of Squares SSE = I J ( X ij X i. ) 2 = (J 1) I i=1 j=1 i=1 S 2 i Distribution of SSE: SSE σ 2 χ 2 I (J 1)

10 Means Variance Sum of Squares Sum of Square of Group (or Treatment) - SSG or SSTr I SSTr = J ( X i. X.. ) 2 i=1 Distribution of SSTr: (if H 0 is true) SSTr σ 2 χ 2 I 1

11 Means Variance Sum of Squares Mean Square Error (MSE) and Mean Square Treatment (MSTr) MSE = SSE I (J 1) MSTr = SSTr I 1

12 Important results Outline Means Variance Sum of Squares The distribution of SSE tell us that ( ) SSE E = E(MSE) = σ 2 I (J 1) The distribution of SSTr tell us that if H 0 is true: ( ) SSTr E = E(MSTr) = σ 2 I 1) What happens to E(MSTr) if H 0 is not true?

13 F - test So it is logical to say that a good way to make the test is to use the following test statistic. F = MSTr MSE What is the distribution of the test statistic above and why?

14 ANOVA Table All the previous results can be summarized in the following Table: Table: ANOVA TABLE Source of Sum of Mean variation df Squares Squares F Treatments I 1 SSTr MSTr MSTr/MSE Error I (J 1) SSE MSE Total IJ 1 SSTo

15 Fundamental ANOVA Identity SSTo = SSTr + SSE SSTo = I J (x ij x.. ) 2 i=1 j=1

16 Levene To test equality of variances using the Levene you need to calculate x ij x i.. The new values you get is your new dataset. Perform an ANOVA test on these new values.

17 Section 11.1 page 550 1, 2, 3, 4, 5, 6, 7, 8, 9, 10