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    ANOVA (Tukey HSD)

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    The following is hypothetical data similar to the actual research results about birds. The numbers represent relative brain size for the individual birds in each sample.
    NON-MIGRATING SHORT DISTANCE MIGRANTS LONG DISTANCE MIGRANTS
    18 6 4 N=18
    13 11 9 G=180
    19 7 5 EX2=2150
    12 9 6
    16 8 5
    12 13 7

    M=15 M=9 M=6
    T=90 T=54 T=36
    SS=48 SS=34 SS=16

    USE AN ANOVA WITH o=.05 to determine whether there are any significant mean differences among the three groups of birds. (use two decimal places)

    SOURCE SS df MS F F-critical
    BETWEEN ______ _____ _____ _____ _______
    WITHIN ______ _____ _____
    TOTAL ______ ______

    F distribution
    numerator degrees of freedom=6
    denominator degrees of freedom=16

    conclusion
    * fail to reject the null hypothesis; there are significant differences among the three groups of birds
    * reject the null hypothesis; there are significant differences among the three groups of birds
    * reject the null hypothesis; there are no significant differences among the three groups of birds
    * fail to reject the null hypothesis; there are no significant differences among the three groups of birds.

    n2=

    the results show significant differences in the 3 groups of birds______________.

    use the Tukey HSD posttest to determine which groups are significantly different (use 2 decimal places)
    q=________

    non-migrating vs. short distance migrants:
    * not enough information
    *no significant mean difference
    * significant mean difference

    non-migrating vs. long distance migrants
    * no significant mean difference
    * not enough information
    * significant mean difference

    short distance migrants vs. long distance migrants
    * no significant mean difference
    * not enough information
    * significant mean difference.

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    https://brainmass.com/statistics/hypothesis-testing/anova-tukey-hsd-585992

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    Solution
    The entries in the ANOVA table are computed as follows:
    SS(Total) = ∑X2 - G2/N = 2150 - (180^2)/18
    = 2150 - 1800 = 350
    SS(Between) = ∑(T2/n) - G2/N = [(90^2)/6 + (54^2)/6 + (36^2)/6] - [(180^2)/18]
    = 2052 - 1800 = 252
    SS(Within) = SS Total - SS Between = 350 - 252 = 98
    df(Total) = N - 1 = 18 - 1 = 17
    df(Between) = k - 1 = 3 - 1 = 2
    df(Within) = N - k = 18 - 3 = 15
    MS (Between) = SS(Between)/ df(Between) = 252/2 = 126
    MS(Within) = ...

    Solution Summary

    This solution explains the one-way ANOVA test procedure and the pair-wise comparison using Tukey HSD.

    $2.49

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