Question

When an ANOVA comparing the means of 3 groups indicates that at least one group is different from the others, a common follow-up analysis to determine which group(s) is (are) different is pairwise two-sample t-tests each assessed using i) the pooled standard deviation when calculating the standard error for the difference in means and ii) a Bonferonni-corrected alpha level of 0.0167 to control the type I error rate for the overall inference to 5% .

Answer 1

Answer:

ii) a Bonferonni-corrected alpha level of 0.0167 to control the type I error rate for the overall inference to 5% .

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".  

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"  

The hypothesis for this case are:

Null hypothesis: $\mu_{A}=\mu_{B}=\mu_{C}$

Alternative hypothesis: Not all the means are equal $\mu_{i}\neq \mu_{j}, i,j=A,B,C$

Since we reject the null hypothesis we want to see which method it's the best to determine which group(s) is (are) different is pairwise two-sample t-tests each assessed using.

And on this case the best option is:

ii) a Bonferonni-corrected alpha level of 0.0167 to control the type I error rate for the overall inference to 5% .

The reason is because the Bonferroni correction "compensates for that increase by testing each individual hypothesis at a significance level of $\alpha$ who represent the desired overall alpha level and m is the number of hypotheses". For our case m=3 hypotheses with a desired $\alpha = 0.05$, then the Bonferroni correction would test each individual hypothesis at $0.05/3=0.0167$

One advatange of this method is that "This method not require any assumptions about dependence among the p-values or about how many of the null hypotheses are true" . And is more powerful than the individual paired t tests.