CHAPTER 11 ANALYSIS OF VARIANCE ONE-WAY ANALYSIS OF VARIANCE ANOVA - - PowerPoint PPT Presentation

CHAPTER 11

ANALYSIS OF VARIANCE

ONE-WAY ANALYSIS OF VARIANCE

ANOVA is a procedure used to test: H0: the means of three or more populations are all equal. vs H1: the means are NOT all equal.

A Motivating Example

The article “Compression of Single-Wall Corrugated Shipping Containers Using Fixed and Floating Test Platens” (J. Testing Eval., 1992: 318–320) describes an experiment in which several different types of boxes were compared with respect to compression strength (lb).

Summary Of Data

Boxplot

Assumptions of One-Way ANOVA



• 1. The populations from which the samples are

drawn are (approximately) normally distributed.

• 2. The populations from which the samples are

drawn have the same variance (or standard deviation).

• 3. The samples drawn from different

populations are random and independent.

The Model for one-way ANOVA

Calculating the Value of F Test Statistic

Test statistic F x = ni xi

i=1 k

å

n SSB = ni

i=1 k

å

(xi - x)2 SSW = (ni -1)

i=1 k

å

si

2

F = MSB MSW = SSB / (k -1) SSW / (n - k)

Difference with notations in Devore and Berk

The textbook uses different notations: SSB = SSTr, MSB = MSTr SSW = SSE, MSW = MSE

ANOVA Table

Rejection Region and p-value

The rejection region for F with significance level α is: where, k is the number of groups, n is the total sample size in all groups. p-value is with df1 = k-1, df2=n-k.

P(F > F observed)

Example 1

Fifteen fourth-grade students were randomly assigned to 3 groups to experiment with 3 different methods

• f teaching arithmetic. At the end of

the semester, the same test was given to all 15 students. The table gives the scores of students in the three groups.

Solution

Calculate the value of the test statistic

• F. Assume that all the required

assumptions mentioned at the beginning hold true.

Computing SSB and SSW

Method I Method II Method III Sum Name Notation

Means

64.8 73.8 77.6 SDs 16.07 13.95 11.84

Sample Sizes

5 5 5 n = 15

¯ xi si ni

Find Critical Value Given α

α = .01 A one-way ANOVA test is always right-tailed Area in the right tail is .01 df1 = k – 1 = 3 – 1 = 2 df2 = n – k = 15 – 3 = 12 The rejection region is F > 6.93

ANOVA Table

Critical value of F for df = (2,12) and α = .01.

The value of the test statistic F = 1.09

 It is less than the critical value of F = 6.93  It falls in the nonrejection region

Hence, we fail to reject the null hypothesis We conclude that we do not have statistical evidence to support that the means are not equal.

Making Decision

Finding p-value

From F-table, for df1 = 2, df2 = 12, P(F > 2.81) = 0.10. 1.09 is smaller than 2.81, so p-value = P(F > 1.09) > 0.1. Using R, the p-value is: > pf (1.09, df1 = 2, df2 = 12, lower=F) [1] 0.3673077

Using R

> arith <- read.csv ("arith.csv", header = T) > # have a look at the data > head(arith) scores methods 1 48 m1 2 73 m1 3 51 m1 4 65 m1 5 87 m1 6 55 m2 > # do anova > arith.aov <- aov ( scores ~ methods, data = arith) > # look at the result > summary (arith.aov) Df Sum Sq Mean Sq F value Pr(>F) methods 2 432.13 216.07 1.0927 0.3665 Residuals 12 2372.80 197.73

Another Computing Formula

Let J1, J2, . . . , JI denote the I sample sizes and let n = S Ji denote the total number of

• bservations. The accompanying box gives

ANOVA formulas and the test procedure.

Example 2

The article “On the Development of a New Approach for the Determination of Yield Strength in Mg-Based Alloys” (Light Metal Age, Oct. 1998: 51–53) presented the fol- lowing data on elastic modulus (GPa)

• btained by a new ultrasonic method for

specimens of a certain alloy produced using three different casting processes.

Data and Summary

F Test Procedure

F Table

Studying more about ANOVA

STAT 345: Design and Analysis of Experiments STAT 344: Applied Regression Analysis STAT 346: Multivariate Analysis STAT 443: Linear Models (Proof of F Distribution)