One of the primary advantages of a repeated-measures design, compared to an independent-measures design, is that it reduces the overall variability by removing variance caused by individual differences. The following data are from a research study comparing three treatment conditions. (see attached file)
1. Assume that the data are form an independent-measures study using three separate samples, each with n = 6 participants. Ignore the column of P totals and use an independent-measures ANOVA with α = .05 to test the significance of the mean differences.
2. Now assume that the data are from a repeated-measures study using the same sample of n = 6 participants in all three treatment conditions. Use a repeated-measures ANOVA with α = .05 to test the significance of the mean differences.
3. Explain why the two analyses lead to different conclusions.
1) We are testing Ho: u1=u2=u3 vs Ha: at least two of u1, u2 and u3 are equal
SStotal=800-108^2/18=152, SSwithin=42+28+34=104, SSbetween=24^2/6+36^2/6+48^2/6-108^2/18=48, DFbetween=3-1=2, DFwithin=18-3=15, MSwithin=104/15=6.933, MSbetween=48/2=24. So F test ...
The solution gives detailed steps on performing One-way ANOVA between 3 treatments. All formula and calculations are shown and explained.
Given the following sample information, test the hypothesis that the treatment means are
equal at the .05 significance level.
Treatment 1 Treatment 2 Treatment 3
8 3 3
11 2 4
10 1 5
a. State the null hypothesis and the alternate hypothesis.
b. What is the decision rule?
c. Compute SST, SSE, and SS total.
d. Complete an ANOVA table.
e. State your decision regarding the null hypothesis.
f. If H0 is rejected, can we conclude