CHAPTER 11 ANALYSIS OF VARIANCE ONEWAY ANALYSIS OF

CHAPTER 11 ANALYSIS OF VARIANCE

ONE-WAY ANALYSIS OF VARIANCE Definition ANOVA H 0 : is a procedure used to test: the means of three or more populations are all equal. vs H 1: the means are NOT all equal.

A motivating example dfd

Assumptions of One-Way ANOVA The following assumptions must hold true to use 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.

Calculating the Value of F Test Statistic The value of the test statistic F for an ANOVA test is calculated as

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 df 2=n-k. with df 1 = k-1,

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 II Name Means Method III Sum Notation 64. 8 73. 8 77. 6 SDs 16. 07 13. 95 11. 84 Sample Sizes 5 5 5 n= 15

ANOVA Table

Find Critical Value Given α α =. 01 A one-way ANOVA test is always righttailed 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

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. 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, df 1 = 2, df 2 = 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 m 1 2 73 m 1 3 51 m 1 4 65 m 1 5 87 m 1 6 55 m 2 > # do anova > arith. aov <- aov ( scores ~ methods, data = arith) > # look at the result > summary (arith. aov) Df methods Residuals 2 Sum Sq Mean Sq F value Pr(>F) 432. 13 216. 07 12 2372. 80 197. 73 1. 0927 0. 3665

Another Computing Formula

Example 2