Question: Consider the data from Exercise 11.7.1. For these data, SS(species of birch) = 2.19781, SS(flooding) = 2.25751, SS(interaction) = 0.097656, and SS(within) = 0.47438. (a)

Consider the data from Exercise 11.7.1. For these data, SS(species of birch) = 2.19781, SS(flooding) = 2.25751, SS(interaction) = 0.097656, and SS(within) = 0.47438.
(a) Construct the ANOVA table.
(b) How many degrees of freedom are there for the ANOVA F test for interactions?
(c) The P-value for the test is 0.142. If α = 0.05, what is your conclusion regarding the null hypothesis?
(d) What is the value of the F test statistic for testing that species has no effect on ATP concentration?
(e) The P-value for the testing that species has no effect on ATP concentration is 0.000008. If α = 0.01, what is your conclusion regarding this null hypothesis?
(f) Assuming that each of the four populations has the same standard deviation, use the data to calculate an estimate of that standard deviation.

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