Harley (2003) examined effects on growth of the red algae Mazzaellaparksii, 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?