.. ). In our case we added up 1.1+ .0024 + .99=2.0924. The results for eight, the red larvae deloped was , while the obsereved removed was 19. The expected removed were 75*19/200=7.1. The chi square analysis for the red larvae deployed was (19-7.1)^2/7.1=19.9.

For day 8 the blue larvae developed was 70, while the observed removed was 0. The expected removed were 70*19/200=6.7. THe chi-square analysis for the blue larvae was (0-6.7)^2/6.7=6.7. For day 8 of the lime larvae, deployed was 55, while the observed removed was 0. The expected removed were 55*19/200=5.2. THe chi square analyis for the lime was (0-5.2)^2/5.2=5.2. When all of these are added up, the total chi square analysis was 19.9+6.7+5.2=31.8 For the time period in which the number of prey was removed, you calculated the chi anaylsis for the AM and period PM of they eight.

For the time perid of AM we would take the number, of larvae deployed, times the number removed divided by the total number deployed for the AM time period. Then we would add them up to get the chi analysis. For the expected removed for red larvae, the calculated was 36*9/100=3.24. For the expected removed number of blue, we calculated 35*9/100=3.15. For expected removed number of lime larvae, we calculated 26*9/100=2.34.

To calculate the chi analysis for red larvae we would take the number removed minus the expected , then square it and divide by the number expected. That is (9-3.24)^2/3.24= 10.24. To find the chi analysis for the blue larvae you would calculate, (0-3.15)^2/3.15=3.15. To calculate the chi square analysis for the lime larvae you would calculate (0-2.43)^2/2.34=2.34. When all of these are added up 10.24+3.15+2.43=15.82 For the time period of PM, wwe would also take the number of larvae deployed, time the number removed removed divived by the total number deployed for the PM time period. Then we would add it up to find the chi analysis for the PM period. For the expected removed for red larvae, the calculated was 36(10)/100=3.6.

For the expected removed for blue 35(10)/100=3.5. For the expected removed red larvae, the calculated was 29(10)/100=2.9. To calculate the chi analysis was the red larvae, you take the number removed minus the expected, then square it and divide by the numberexpected. That is (10-3.6)^2/100=11.37. To find the chi analysis for the blue larvae, you would calculate, (0-3.5)^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. Discuss The 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 birds removed the food. After graphing the data, I failed to accept my null hypothesis.

I thought that the diffent time periods of the day would affect when the birds would usually remove the larvae. I imagined that birds would come out first thig in the morning and then go bacl to their nest. However through the one could see that the results in both days were very similar. In addition when I calculated the chi square analysis for both periods, they were 17.7 for PM and 15.83 for Am. These numbers were both larger than the given value in the book of 3.84, which is why I failed to acceot my null hypothesis.

The results of this lab were sometimes a little difficult to interpret due to various mistakes. One large and costly mistake was missing data. Several groups did not report their data, which resulted in the graphs and calculations being somwhat unusual and often difficult to understand. There were also often large gaps in the graphs. The problem of negative numbers in the data threw off the graphs even more than it already was.

Also, although not proven, it is possible that othe organisms or enviromental conditions could have interfiered with the data collection. The factor of the wind must be taken into account for emoving the prey. If the prey was removed, it doesn’t mean that the predators had eaten it. The predator also could it in his or her mouth and spit it right back out. This response would be very important to mimicry , because the preadator or prey is realizing what is palatable and what isn’t.

The negative numbers that appeared in this experiment could have been the cause of experimental error, due to the fact that there were two extra larvae in each petri dish. There were quite a few mitakes in this data but at least we successfuly obtained viable data. According to the graphs, the birds preyed upon the red larvae the most. The red larvea was 100% palatable and 0% unpalatable. I expected the red larvae to be preyed upon the most, but what shocked me was the amount of lime that was still being preyed on while the birds approched the eighth day.

I thought that since there was only 25% of the lime larvae that was palable the would have picked it up right away. Even the blue larvae was removed about as much as s the lime was another indication that mimicry had occured. The blue was only 25% unpalatable there fore it should have yeilded much higher results than the lime. Some times as was a little confused as to whether my accepting or rejecting a null hypothesis was correct. This is because the chi square analysis isn’t a procedure thatalways ccurately reject or support the null hypothesis.

The chi square analysis is most likely a test of randomness, rather than a test for supporting or rejecting the null hypothesis. This is also why we say that we fail to rejact it, because, just because we fail to reject it doesn’t mean that it is correct, therefore we cannot accept it. This is also why it is good to have graphs, in an experiment like this one, so that they can back up your results and hypothesis, because it is giving you a visual sense of what is going on. Bibliography Charlene Ngong Science.