Exposure to microbial products, especially endotoxin, may have an impact on vulnerability to allergic diseases. The article “Dust Sampling Methods for Endotoxin—An Essential, But Underestimated Issue” (Indoor Air, 2006: 20–27) considered various issues associated with determining endotoxin concentration. The following data on concentration (EU/mg) in settled dust for one sample of urban homes and another of farm homes was kindly supplied by the authors of the cited article. a. Determine the sample mean for each sample. How do they compare? b. Determine the sample median for each sample. How do they compare? Why is the median for the urban sample so different from the mean for that sample? c. Calculate the trimmed mean for each sample by deleting the smallest and largest observation. What are the corresponding trimming percentages? How do the values of these trimmed means compare to the corresponding means and medians?

Answer Step 1 of 6 a)For sample U arrange the values in ascending order 4, 5, 5, 6, 11, 17, 18, 23, 33, 35, 80 Mean=X /n=237/11=21.55 For sample F arrange the values in ascending order 0.3, 2, 3, 4, 4, 5, 8, 8.9, 9, 9,9.2, 11, 14, 20, 21 Mean=X /n=128.4/15=8.56 Mean of sample U > Mean of sample F Step 2 of 6 n+1 th b) Median for U=( 2 ) value =( 11+1 ) value th =6 value =17 Median for F=( n21) value 15+1 th =( 2 ) value =8 value =8.9 Median of sample U > Median of sample F Step 3 of 6 The median for the urban sample =17 The mean for the urban sample=21.55 Median is the positional average that is the middle value of the given values Here mean is the average of all the given values In that the last value 80 is very far from the previous value 35 So there is more gap between 35 and 80 Because of that mean value is more than median So, the median for the urban sample so different from the mean for that sample