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Unformatted text preview: 22.16 Pvalue = 0.00 Conclusion Rejection Ho Business Answer According to the sample taken there is overwhelming evidence to conclude that the auditors judgements were better than the acutal Anova: Single Factor SUMMARY Groups Count Sum Average Variance Column 1 7 119 17 5.01 Column 2 7 142.8 20.4 6.26 Column 3 7 175 25 4.01 ANOVA SS df MS F Pvalue F crit 225.68 2 112.84 22.16 3.55 SST = Sum of Squares " treatments " = [n j *(xbar j xdoublebar)^2], j=1,2,3 MST = SST/(k1) , where k=number of treatments SSE = Sum of Squares  Errors for all three samples = Sum[(n j1)*s j ^2], j=1,2,3 MSE = SSE/(nk) where n = n1+n2+n3. (Here, n=20+20+20=60) 91.66 18 5.09 317.34 20...
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This note was uploaded on 07/17/2011 for the course STATS 211 taught by Professor Dunne during the Spring '07 term at University of Dayton.
 Spring '07
 Dunne

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