An analysis of variance experiment produced a portion of the accompanying ANOVA table. Assume normality in the underlyin

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An analysis of variance experiment produced a portion of the accompanying ANOVA table. Assume normality in the underlyin

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An Analysis Of Variance Experiment Produced A Portion Of The Accompanying Anova Table Assume Normality In The Underlyin 1
An Analysis Of Variance Experiment Produced A Portion Of The Accompanying Anova Table Assume Normality In The Underlyin 1 (66.27 KiB) Viewed 36 times
An analysis of variance experiment produced a portion of the accompanying ANOVA table. Assume normality in the underlying populations. (You may find it useful to reference the Ftable.) picture Click here for the Excel Data File a. Specify the competing hypotheses in order to determine whether some differences exist between the population means. Ho: HA = HB = HC = HD: HA: Not all population means are equal. O Ho: MA ≥ HB ≥ HC = HD; HA: Not all population means are equal. OHO: MA ≤ MB ≤ HC ≤ HD; HA: Not all population means are equal. b. Fill in the missing statistics in the ANOVA table. (Round "MS" to 4 decimal places and "F" to 3 decimal places.) ANOVA Source of Variation SS df MS F p-value Between Groups 16.94 Within Groups 87.96 Total 104.90 3 63 66 0.0108
ANOVA Source of Variation 16.94 3 Between Groups Within Groups 87.96 63 Total 104.90 66 c. At the 1% significance level, what is the conclusion to the test? Do not reject Ho; we cannot conclude that some means differ. Reject Ho; we can conclude that some means differ. Reject Ho; we cannot conclude that some means differ. Do not reject Ho; we can conclude that some means differ. SS df MS F p-value 0.0108
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