An inner city open space area is suspected to have high levels of lead (Pb) in i
ID: 3226009 • Letter: A
Question
An inner city open space area is suspected to have high levels of lead (Pb) in its surface soil. The sampled region is 400 m^2. The region is gridded (i.e. sampling grid) half meter by half meter. 5 grids are selected at random. After removing rocks and other debris from each sample, they are weighed and three 1 gram subsamples are selected from each primary sample. These 1 gram samples are analyzed and the mass of Pb (in micrograms) each 1 gram sample is determined. Measurement errors are assumed negligible. The data are given in Table 2. Estimate the average concentration of Pb and the total mass of Pb present in this region. Calculate standard errors for each. For comparison with the sampling method applied in problem 3, a) Calculate the sample mean and variance for Pb using simple random sampling. b) From these, calculate the estimates for Inventory () and variance s^2 (). c) Comment on your analysis results with respect to sampling method.Explanation / Answer
Average concentraion of Pb = Average of all samples = (20 + 24 + 22 + 18 + 12 + 12 + 7 + 9 + 15 + 4 + 9 + 9 + 22 + 14 + 3)/15 = 13.33 micrograms
Area of grid = 0.5 * 0.5 = 0.25 sq-m
Average Mass of grid = (807+744+782+437+651)/5 = 684.2 gm
Average mass of Pb in a grid = 684.2 * 13.33 micrograms = 9.12 gm
Total number of grids = 400/0.25 = 1600
Total Mass of Pb = 1600 * 9.12 gm = 14.592 kg
Standard error for sample 1 = Mean of sample 1 /sqrt(total mass of primary sample) = (20+24+22)/3/ sqrt(807)
= 22/sqrt(807) = 0.7744 microgram
Standard error for sample 2 = Mean of sample 2/sqrt(total mass of primary sample) = (18+12+12)/3/ sqrt(744)
= 14/sqrt(744) = 0.5133 microgram
Standard error for sample 3 = Mean of sample 3 /sqrt(total mass of primary sample) = (7+9+15)/3/ sqrt(782)
= 10.33/sqrt(782) = 0.3694 microgram
Standard error for sample 4 = Mean of sample 4 /sqrt(total mass of primary sample) = (4+9+9)/3/ sqrt(437)
= 7.33/sqrt(437) = 0.3506 microgram
Standard error for sample 5 = Mean of sample 5 /sqrt(total mass of primary sample) = (22+14+3)/3/ sqrt(651)
= 13/sqrt(651) = 0.5095 microgram