Introduction to Probability and Statistics, 14th Edition by William Mendenhall – Test Bank
MBB.IntroProb13.ch11sec9-10
TRUE/FALSE
1. In ANOVA for an a b factorial experiment, it is necessary to have at least two measurements at each level of each factor in order to analyze any interaction between the factors.
ANS: T PTS: 1
2. The number of cells in a two-way factor ANOVA is equal to a + b 1; where a is the number of levels of factor A and b is the number of levels of factor B.
ANS: F PTS: 1
3. In order to conduct a two-way factor ANOVA with replications, the number of replications r must be the same in each cell.
ANS: T PTS: 1
4. In a two-way factor ANOVA, the smallest number of replications required in any cell is 2, but all cells must have the same number of replications.
ANS: T PTS: 1
5. In a two-way factor ANOVA, the total sum of squares can be partitioned into four parts: the variation due to factor A, the variation due to factor B, the error variation, and the variation due to blocking.
ANS: F PTS: 1
6. In a two-way factor ANOVA, the variances of the populations are assumed to be equal unless the error variation is zero.
ANS: F PTS: 1
7. In a two-way factor ANOVA, if factors A and B do not interact, then neither A nor B can be considered statistically significant.
ANS: F PTS: 1
8. In a two-way factor ANOVA with replications, the null hypothesis for testing whether interaction exists is that no interaction exists, while the alternative hypothesis is that interaction does exist.
ANS: T PTS: 1
9. In a two-way factor ANOVA with replications in which all hypotheses are to be tested at the .05 significance level, if the p-value for interaction is .0257, then we should conclude that no interaction exists between the levels of the two factors.
ANS: F PTS: 1
10. In a two-way factor ANOVA with replications, the reason for separating out the sum of squares due to interaction between factors A and B is to increase the chance of detecting significant differences across levels of factor A and factor B.
ANS: T PTS: 1
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