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ANOVA (Tukey HSD)

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Please help me with this question if possible. Thank you!
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

Solution Preview

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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.

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See Also This Related BrainMass Solution

One ANOVA with eta and Tukey HSD.

Suppose that you are interested in determining the effectiveness of an intensive supervision probation (ISP) program. To examine its effectiveness, you keep track of 12 offenders sentenced to ISP for one year and tabulate the number of new offenses they committed. You also gather two control groups. The first consists of 12 offenders sentenced to routine probation. The second consists of 12 offenders whose cases were dismissed. The following results describe the number of new offenses each offender in each group committed during your one-year follow-up:

Using these data, answer the following questions:
1. Identify the independent and dependent variables.
2. With an alpha of 0.01, test the null hypothesis that the three population means are equal by doing the following: What are the null and alternative hypotheses? Show the calculation and explain how to construct the ANOVA table. Determine the appropriate test statistic and determine whether it is significant.
3. If appropriate, conduct a mean comparison for all pairs of means using Tukey?s HSD test.
4. Calculate the value of Eta-squared, and make a conclusion about the strength of the relationship.

See attached file for full problem description.

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