Question

12.2.8. Do the following data appear to violate the as- sumptions underlying the analysis of variance? Explain. Treatment A B C D 16 17 16 17 4 12 2 26 26 22 23 24 8 9 TL 8

Answer 1

Ho: There is no mean difference between the treatments. H.: There is a mean difference between the treatments. EXCEL (Data Analysis) output: Anova: Single Factor SUMMARY Groups Count Sum Average Variance 66 16.5 0.333333 44 11 118.6667 95 23.75 2.916667 4369 2 df ANOVA Source of Variation Between Groups Within Groups Total SS 523.1875 371.75 894.9375 MS F P -value F crit 3 174.3958 5.629455 0.012081 3.490295 12 30.97917 15 The value of test statistic is F = 5.63. P-value = 0.0121.

Conclusion: Use a significance level, a = 0.05. Here, p-value is 0.0121, which is less than the value of a -0.05. That is, p-value (=0.0121) sal=0.05). Therefore, by the rejection rule, reject the null hypothesis H. Interpretation: There is sufficient evidence at level of significance a = 0.05. There is a mean difference between the treatments.