.5)^2/3.5=3.5. To find the chi analysis for the lime larvae, the calculate was (0-2.9)^2/2.9. When all of these are added up 11.37+3.5+2.9=17.8.DiscussThe results for this experiment did not surprise me at all. I did not expect that on the first day, that there would be a color favor, amongst the birds. I realized that although lime might be a more pleasant color than red or blue, the birds might not find this favoritism. Although we knew that the red was the most palable, the birds would have to test the larvae first before coming to a conclusion as to which one they prefered. According to the data, I failed to reject my hypothesis because the chi sqare analysis that I calculated, 2.09, is smaller than the 5.991 which is the chi analysis given in the book. When the chi analysis is smaller, this shows that , my hypothesis was not refused and that there was no color prefrence chosen by the birds on the first day. Also through the graphs, one can also see frrom the pattern of the percentage line of removed food, vs time, that there was no larvae color preference. By the eighth day I expected that the birds would have realize that only the red was since the most palatable larvae, especially since there was only about twenty five percent lime that was palable. However, through tha graph, One can see that toward the eight day the birds were still feeding on some of the lime larvae. This made it an oblivious indication that mimicry had occured. Also I calculated chi square value was much higher that given in the book, which is about a comparision of 31.8 vs 5.991. Since it is higher, I failed to accept my null hypothesis. This means that the birds up on till that they could still not really tell the difference, or had't really noticed that some of the food was more palatable than others.In addition to the eight day, I had also seperated the time periods to see if there was any effect on temperature on the out come of the feed in each the bi...
