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    One-way ANOVA between 3 Treatments

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    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.

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    https://brainmass.com/statistics/type-i-and-type-ii-errors/one-way-anova-between-treatments-572451

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    Solution:
    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 ...

    Solution Summary

    The solution gives detailed steps on performing One-way ANOVA between 3 treatments. All formula and calculations are shown and explained.

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