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

The following is hypothetical datasimilar 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 meandifferences 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.

Please see the attachment for detailed explanation of the solution.

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) = …

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