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. U: 6.0 5.0 11.0 33.0 4.0 5.0 80.0 18.0 35.0 17.0 23.0 F: 4.0 14.0 11.0 9.0 9.0 8.0 4.0 20.0 5.0 8.9 21.0 9.2 3.0 2.0 0.3 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?

a).

mean(U)  = 21.54545

mean(F)  = 8.56

Compare:

The mean of U is higher (as expected) compare to the mean of F.

b).

Median(U) =  17

Median(F) = 8.9

The median of U is higher (as expected) compare to the mean of F.

b ii). The median is not affected by outliers. There is evidence of outlier(s) in the data sets (see attached diagram). This is one of the possible reason why median could be less than the mean. Hence, the median of the Urban sample is different from the mean of the Urban sample because the is possible outliers in the data set.

c).

Trimmed mean of U = 16.14286

Trimmed mean of F = 7.566667

The corresponding trimmed mean percentages = 20%.

d).

The trimmed mean is smaller compare to the mean obtained directly from the data sets. It is also smaller to the median (though close to the median than the initial mean).

Step-by-step explanation:

By mean, we mean the sum of all observation divided by the number of observations or sample size (n).

By median, we mean the middle value when the data set/observations is sorted in ascending or descending order.

For replication, kindly use the R codes below:

U = c(6.0, 5.0, 11.0 ,33.0, 4.0, 5.0, 80.0, 18.0, 35.0, 17.0, 23.0)

F = c(4.0, 14.0, 11.0, 9.0, 9.0, 8.0, 4.0, 20.0, 5.0, 8.9, 21.0, 9.2, 3.0, 2.0, 0.3)

mean(U)

mean(F)

median(U)

median(F)

mean(U, trim = 0.2)

mean(F, trim = 0.2)

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that is the sum

step-by-step explanation:

there are no other answers without knowing the variables values