Answer to Question #120485 in Statistics and Probability for Joseph Se

Question

A study compared five different methods for teaching descriptive statistics. The five methods were traditional lecture and discussion, programmed textbook instruction, programmed text with lectures, computer instruction, and computer instruction with lectures. 45 students were randomly assigned, 9 to each method. After completing the course, students took a 1-hour exam. We are interested in finding out if the average test scores are different for the different teaching methods.

If the original significance level for the ANOVA was 0.05, what should be the adjusted significance level for the pairwise tests to compare all pairs of means to each other?

A. 0.05
B. 0.25
C. 0.01
D. 0.5
E. 0.005

Answer 1

Recognize that the null hypothesis in ANOVA sets all means equal to each other, and the alternative hypothesis suggest that at least one mean is different.

The group degrees of freedom is number of levels (categories) minus 1 (k−1=5−1=4

k−1=5−1=4) and the error degrees of freedom is the sample size minus the number of levels (n−k=45−5=40

n−k=45−5=40).

H₀: μ₁=μ₂=...=μ_k

H_A: at least one mean is different

 p-value of the test is 0.0168

The p-value is low so we reject the null hypothesis.

The conclusion of the test is that at least two group means are significantly different from each other.

$K=k(k−1)2. K=(5−4)/2, α∗=0.05/10=0.005$

answer: E. 0.005