Mathematics Homework Help
Mathematics Homework Help. Need statistics help to use a two-factor ANOVA to analyze these data and interpret your results.
Harley (2003) examined effects on growth of the red algae Mazzaella parksii, which forms lawn-like
growths in the intertidal region. He focused on two factors: (1) location and
(2) herbivore abundance. The two locations were in areas along the shoreline below
where this algae normally occurs (Low zone) and in areas along the shoreline
where this algae does normally grows (Mid zone). For herbivore abundance, he
had areas where limpets (Lottia
species) were removed (Absent) and areas where limpets were not removed
(Present). Harley then measured the area covered by Mazzaella (in cm2) for each replicate plot, and obtained
the following data (NOTE: 0.5 has been added to each value to remove any zeroes
from the data):
Height |
Herbivores |
Area |
Height |
Herbivores |
Area |
low |
absent |
88.46 |
mid |
absent |
0.50 |
low |
absent |
1188.02 |
mid |
absent |
622.54 |
low |
absent |
2178.41 |
mid |
absent |
2168.20 |
low |
absent |
276.96 |
mid |
absent |
0.50 |
low |
absent |
594.26 |
mid |
absent |
26.42 |
low |
absent |
1470.77 |
mid |
absent |
38.20 |
low |
absent |
1140.90 |
mid |
absent |
2204.33 |
low |
absent |
3732.72 |
mid |
absent |
524.36 |
low |
absent |
1257.14 |
mid |
absent |
377.49 |
low |
absent |
2132.08 |
mid |
absent |
1378.88 |
low |
absent |
1254.00 |
mid |
absent |
201.56 |
low |
absent |
627.25 |
mid |
absent |
9.92 |
low |
absent |
0.50 |
mid |
absent |
467.03 |
low |
absent |
1689.11 |
mid |
absent |
276.96 |
low |
absent |
1307.41 |
mid |
absent |
2073.96 |
low |
absent |
1746.44 |
mid |
absent |
2297.80 |
low |
present |
143.44 |
mid |
present |
0.50 |
low |
present |
209.42 |
mid |
present |
1037.23 |
low |
present |
0.50 |
mid |
present |
1230.44 |
low |
present |
64.12 |
mid |
present |
307.59 |
low |
present |
0.50 |
mid |
present |
38.98 |
low |
present |
0.50 |
mid |
present |
170.15 |
low |
present |
763.91 |
mid |
present |
657.09 |
low |
present |
1847.76 |
mid |
present |
1332.54 |
low |
present |
9.92 |
mid |
present |
1407.94 |
low |
present |
69.62 |
mid |
present |
2095.95 |
low |
present |
0.50 |
mid |
present |
401.05 |
low |
present |
16.99 |
mid |
present |
99.46 |
low |
present |
0.50 |
mid |
present |
7.57 |
low |
present |
1294.05 |
mid |
present |
1610.57 |
low |
present |
12.28 |
mid |
present |
1665.55 |
low |
present |
7.57 |
mid |
present |
2012.69 |
A)
Use a two-factor ANOVA to analyze these data and
interpret your results. The data are not currently set up correctly for doing a
two-factor ANOVA in Excel, so you will likely need to rearrange them!
B)
Test these data to see if (1) they are normally
distributed and (2) the groups have equal variances.
C)
If either assumption is violated, perform a data
transformation to best eliminate the problem (you may or may not be able to completely
eliminate any violations!). Then perform a second two-factor ANOVA, this time
using the transformed data. How do your F-values and P-values differ if you
compare the results of the two ANOVAs?