Question

into aquatic systems where it can have deleterious effects Mercury is a persistent and dispersive environmental contaminant found n many ecosystems around the world. When released as an industrial by-product, iIt often finds its way ecosystems a on verious avian and aquatic species. The accompanying data on blood mercury concentration (Hg/g) for adult near contaminated rivers in a state was read from a graph in an article. 0.21 0.23 0.25 0.31 0.35 0.41 o.54 0.57 1.43 1.71 1.84 2.21 2.26 3.0e 3.24 (a) Determine the values o, the sample mean and sample median Hint Σ _1864 Round your answers to three decr al aces Explain why they are different O There is heavy positive skewness in the data. O There is heavy negative skewness in the data O They aren't different since the data is symmetrie. O The sample mean and sample median are never equal O There are an odd number of data values (b) Determine the value of thn 10% trimmed mean, (Round your answer to three decimal places.) Compare to the mean and median O It is the same as the mean. O It is between the values of the mean and median O It is below both the mean and median. O It is above both the mean and the median It is the same as the median. (c) By how much could the observation 0.21 be increased without impacting the value of the sample median

Answer 1

here from the data values,

a) mean = ( $\sum$ x) / n = 1.243

median = ((n+1)/2)^th value = 8^th ordered value = 0.57

since mean > median, there is heavy positive skewness in the data

b) 10% trimmed mean can be calculated by eliminating the smallest 10% and largest 10% of the sample then fine the average of the remaining samples

so, 15*10% = 1.5 = ~2 so remove 2 values from top and bottom (each time one value)

mean(trim1) = (0.23+0.25+........+3.08)/13 = 1.168

mean(trim 2) = (0.25+0.31+.....2.26)/11 = 1.08

therefore, trimmed mean = (1.168 + 1.08)/2 = 1.124

trimmed mean is in between the values of mean and median

c) here,

sample median = 0.57

so, 0.21/0.57 = 0.37

the observation 0.21 can be increased by 0.37 and then the median value is not affected