Question

In an effort to determine the most effective way to teach safety principles to a group of employees, four different methods were tried. Some employees were given programmed instruction booklets and worked through the course at their own pace. Other employees attended lectures. A third group watched a television presentation, and a fourth group we divided into small discussion groups. A high of 10 was possible. A sample of 5 tests were selected from each group. The results were: Sample Programmed Instruction Lecture TV Group Discussion 1 6 8 7 8 2 7 5 9 5 3 6 8 6 6 4 5 6 8 6 5 6 8 5 5 At the 0.01 level, what is the critical value

Answer 1

critical value = 5.29

Explanation:

Given that they are divided into 4 groups and a sample of 5 test was selected

N = 5 * 4 = 20

k = 4

∝ = 0.01

Df for numerator ( SS group )= k - 1 = 3

Df for denominator ( SSE group ) = N - k = 20 - 4 = 16

DF ( degree of freedom )

Next we will use the F table to determine the critical value

Critical value =
`F_(16,3,0.01)` = 5.29