When only two treatments are involved, ANOVA and the Student t test (Chapter 11) result in the same conclusions. Also, formula120.mml. As an ex, suppose that 14 randomly selected students were divided into two groups, one consisting of 6 students and the other of 8. One group was taught using a combination of lecture and programmed instruction, the other using a combination of lecture and television. At the end of the course, each group was given a 50-item test. The following is a list of the number correct for each of the two groups. Using analysis of variance techniques, test the null hypothesis, formula121.mml that the two mean test scores are equal.
Lecture and
Programmed
Instruction Lecture and
Television
14 33
11 24
25 36
24 21
13 29
12 26
23
22
(a-1) Complete the ANOVA table. (Round SS, MS and F values to 2 decimal places.)
Source SS df MS F
Factors
Error
Total
(a-2) Use a formula122.mml level of significance. (Round your answer to 2 decimal places.)
(b) Using the t test, compute t. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)
(c) There is in the mean test scores.