Mercury is a persistent and dispersive environmental contaminant found in many ecosystems around the world. When released as an industrial by-product, it often finds its way into aquatic systems where it can have deleterious effects on various avian and aquatic species. The accompanying data on blood mercury concentration (gg/g) for adult females near contaminated rivers in Virginia was read from a graph in the article ?"Mercury Exposure Effects the Reproductive Success of a Free-Living Terrestrial Songbird, the Carolina Wren" (The Auk, 2011: 759-769;? this is a publication of the American Ornithologists' Union). a. Determine the values of the sample mean and sample median and explain why they are different. b. Determine the value of the 10% trimmed mean and compare to the mean and median. c. By how much could the observation .20 be increased without impacting the value of the sample median?

Solution: Step 1: Here an experiment is conducted to find the effect of mercury in reproductive system of adult females. The have given data blood mercury concentration (gg/g) for adult females near contaminated rivers in Virginia. .20 .22 .25 .30 .34 .41 .55 .56 1.42 1.70 1.83 2.20 2.25 3.07 3.25 Step 2: a) We have to find the mean and median of the given data. 15 1) Here the total of all observations is = x= 18.55 i=1 i So the sample mean of the observations= 18.55/ 15 = 1.23666 gg/g 2) In order to find the sample median arrange all the given observation in the ascending order It will be .20 .22 .25 .30 .34 .41 .55 .56 1.42 1.70 1.83 2.20 2.25 3.07 3.25 Since the given data is in ascending order The median of sample observations= 0.56 gg/g The average blood mercury concentration and the median are very different. By using a barplot for the given sample values. We can understand clearly that the difference in mean and median value is due to the positive skewness of given data. Step3 : b) We have to find the 10% trimmed mean of the given data. Here a 1/15 trimmed mean obtained by removing the largest and smallest observations and averaging the remaining 13 observations So the observations will be: .22 .25 .30 .34 .41 .55 .56 1.42 1.70 1.83 2.20 2.25 3.07 The mean of this observations is = 15.10/13 = 1.162 Similarly the 2/15 trimmed mean can be find by removing the next smallest and largest value and averaging the middle 11 observations. .25 .30 .34 .41 .55 .56 1.42 1.70 1.83 2.20 2.25 The mean of this observations is = 12.53/11 = 1.074 Since the average of 1/15 and 2/15 is 0.1 3/15 = 2 = 1/10 =0.1 The 10% trimmed mean will be the average of this two means 1.162+1.074 = 2 = 1.118 gg/g As compared to the original mean the trimmed mean is less and the difference is 0.1186 And as compared to the median the 10% trimmed mean is high. Step 4 : c) Here the median of the data set will remain 0.56 so long as that the 8th ordered observation is 0.56. Hence the value .20 could be increased as high as 0.56 without changing the fact that 8th observations is 0.56.