When only two treatments are involved, ANOVA and the Student t test (Chapter 11) result in the same conclusions. Also, . As an example, 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, that the two mean test scores are equal.
Lecture and Program Instruction Values: 12,15,26,22,12,13
Lecture and Television: 37,23,33,24,27,29,25,25
I have figured out Anova Table
Factor SS-430.72, Factor DF1, Factor MS 430.72
Error SS-342.21, Error DF 12, Error MS 28.52
Total SS=772.93
Total DF 13
.01 level of significance - Compute the test statistic F is 15.10
What I need step by step is the manual way to figure out using the t-test to compute.