Pollutants may reduce the strength of bird bones. We believe that the strength r
ID: 3230690 • Letter: P
Question
Pollutants may reduce the strength of bird bones. We believe that the strength reduction, if present, is due to a change in the bone itself, and not a change in the size of the bone. One measure of bone strength is calcium con- tent. We have an instrument which can measure the total amount of calcium in a 1cm length of bone. Bird bones are essentially thin tubes in shape, so the total amount of calcium will also depend on the diameter of the bone.
Thirty-two chicks are divided at random into four groups. Group 1 is a control group and receives a normal diet. Each other group receives a diet including a different toxin (pesticides related to DDT). At 6 weeks, the chicks are sacrificed and the calcium content (in mg) and diameter (in mm) of the right femur is measured for each chick.
Control P #1 P #2 P #3
C Dia C Dia C Dia C Dia
10.41 2.48 12.10 3.10 10.33 2.57 10.46 2.6
11.82 2.81 10.38 2.61 10.03 2.48 8.64 2.17
11.58 2.73 10.08 2.49 11.13 2.77 10.48 2.64
11.14 2.67 10.71 2.69 8.99 2.30 9.32 2.35
12.05 2.90 9.82 2.43 10.06 2.56 11.54 2.89
10.45 2.45 10.12 2.52 8.73 2.18 9.48 2.38
11.39 2.69 10.16 2.54 10.66 2.65 10.08 2.55
12.5 2.94 10.14 2.55 11.03 2.73 9.12 2.29
Analyze these data with respect to the effect of pesticide on calcium in bones.
Explanation / Answer
Mean of Control group (10.41,11.82,11.58,11.14,12.05,10.45,11.39,12.5) is 11.41
Mean of P#1 (12.10,10.38,10.08,10.71,9.82,10.12,10.16,10.14) is 10.44
Mean of P#2 (10.33,10.03,11.13,8.99,10.06,8.73,10.66,11.03) is 10.12
Mean of P#4 (10.46,8.64,10.48,9.32,11.54,9.48,10.08,9.12) is 9.89
Grand mean = 10.47
Total sum of squares, SST = (10.41-10.47)^2 + (11.82-10.47)^2 + .... + (9.12-10.47)^2 = 29.75
Treatment Sum of Squares (SSTR) = [8* (11.41-10.47)^2 + 8* (10.44-10.47)^2 + 8* (10.12-10.47)^2 + 8* (9.89-10.47)^2] = 10.747
Error Sum of Squares (SSE) = SST - SSTR = 29.75 - 10.747 = 19.003
MST = SST/n-1 = 29.75/(32-1) = 0.9596
MSTR = SSTR/c-1 = 10.747/(4-1) = 3.582
MSE = SSE/N-c = 19.003/(32-4) = 0.6787
F = MSTR/MSE = 3.582/0.6787 = 5.2777
Critical value of F at significance level of 0.05 at df = 3,28 is 2.946
As observed F value > Critical value of F, the effect of pesticide is significant on calcium in bones.