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:  H 0 :  the means of three or more populations are all equal .  vs  H 1 : the means are NOT all equal .

A Motivating Example T h e a r t i c l e “ C o m p r e s s i o n o f S i n g l e - W a l l Corrugated Shipping Containers Using Fixed and Floating Test Platens” ( J . T e s t i n g 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 k å n i x i x = i = 1 n k ( x i - x ) 2 SSB = å n i i = 1 k 2 SSW = å ( n i - 1) s i i = 1 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 ( F > F o bserved ) p-value is with df 1 = k -1, df 2 = n - k .

Example 1  Fifteen fourth-grade students were randomly assigned to 3 groups to experiment with 3 different methods of 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 Method Sum II III Name Notation x i ¯ Means 64.8 73.8 77.6 s i SDs 16.07 13.95 11.84 n i Sample 5 5 5 n = 15 Sizes

Find Critical Value Given α α = .01 A one-way ANOVA test is always right-tailed Area in the right tail is .01 df 1 = k – 1 = 3 – 1 = 2 df 2 = 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.

Making Decision  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.

Finding p-value From F-table, for df 1 = 2, df 2 = 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 > a r i t h < - r e a d . c s v ( " a r i t h . c s v " , h e a d e r = 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 J 1 , J 2 , . . . , J I denote the I sample sizes and let n = S J i denote the total number of observations. 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) obtained 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 S T A T 3 4 5 : D e s i g n a n d A n a l y s i s o f Experiments STAT 344: Applied Regression Analysis STAT 346: Multivariate Analysis STAT 443: Linear Models (Proof of F Distribution)