he 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=31.8For 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=15.82For 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...
