rking pens. Disposable gloves, wax paper, and masking tape assisted the making process of making the food easier. A refrigerator was utalized for the storage of prepared petri dishes with larvae. The chi-sqare analysis and graphs, will help analyze how many larvae were removed or remained at the end of each deployment/retrieval period. ResultsThe results of this kind of experiment are typically used to either prove a null hypothesis (H) or approve an alternative hypothesis (H). The chi-square analysis is an effective means to prove or disprove hypothesis. The total # of prey removed for day 1 was 3. The # of red larvae deployed was 64. The observed removed from the red larvae was 2. The expected removed was 64*3/200 =.96. This was calculated frm taking the total amount deployed times the total amount of larvae that was observed as being removed, divided by the total amount of larvae deployed in that day. THe chi-square analysis for the red larvaewas (2-.96)^2/.96 = 1.1. For day one, the blue larvae deployed was 70., while the observed removed was 0. The expected removed was 70*3/200=1.05. The chi-square analysis for the blue was(0-1.05)^2/1.05= .0024. For day one, the lime larvae was 66, while the observed removed was 0. THe expected removed was 66*3/200= .99. The chi-square analysis for the green larvae was (0-.99)^2/.99=.99. In order to find the chi-square analysis for the total # of prey by color (combining AM and PM data) on day #1, you would simply add up all the chi-square numbers calculated and compare it to the (x^2 0.05, 2=5.991). 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 t...
