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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. \(\begin{array}{llllllllllllll}\mathrm{U}: & 6.0 & 5.0 & 11.0 & 33.0 & 4.0 & 5.0 & 80.0 & 18.0 & 35.0 & 17.0 & 23.0 \\ \mathrm{~F}: & 4.0 & 14.0 & 11.0 & 9.0 & 9.0 & 8.0 & 4.0 & 20.0 & 5.0 & 8.9 & 21.0 \\ & 9.2 & 3.0 & 2.0 & 0.3 & & & & & & & \end{array}\) 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?

Step by step solution

01

Calculate Sample Mean for Each Sample

First, sum all the values in the urban sample \(U\) dataset and divide by the number of observations.\[\text{Mean for U} = \frac{6 + 5 + 11 + 33 + 4 + 5 + 80 + 18 + 35 + 17 + 23}{11} = \frac{237}{11} \approx 21.55\] Similarly, sum all the values in the farm sample \(F\) dataset and divide by the number of observations.\[\text{Mean for F} = \frac{4 + 14 + 11 + 9 + 9 + 8 + 4 + 20 + 5 + 8.9 + 21 + 9.2 + 3 + 2 + 0.3}{15} = \frac{128.4}{15} \approx 8.56\]

02

Calculate Sample Median for Each Sample

The median is the middle value when data is ordered. For \(U\), arrange and find the middle:Values in \(U\): 4, 5, 5, 6, 11, 17, 18, 23, 33, 35, 80. Median for \(U\) = 17 (6th value in the ordered list)For \(F\), arrange values and find the middle:Values in \(F\): 0.3, 2, 3, 4, 4, 5, 8, 8.9, 9, 9, 9.2, 11, 14, 20, 21.Median for \(F\) (8th value in ordered list) = 8.9.

03

Compare Means and Medians

For \(U\), the mean is much larger than the median (21.55 vs. 17), likely due to the 80, a large outlier that skews the mean.For \(F\), the mean and median are quite similar (8.56 vs. 8.9), indicating a more even distribution with less extreme outliers impacting the mean.

04

Calculate Trimmed Means

Remove the smallest and largest values from each dataset. For \(U\), remove 4 and 80:Values: 5, 5, 6, 11, 17, 18, 23, 33, 35Trimmed mean for \(U\): \[\text{Trimmed Mean for U} = \frac{5 + 5 + 6 + 11 + 17 + 18 + 23 + 33 + 35}{9} = \frac{153}{9} \approx 17.00\]For \(F\), remove 0.3 and 21:Values: 2, 3, 4, 4, 5, 8, 8.9, 9, 9, 9.2, 11, 14, 20Trimmed mean for \(F\):\[\text{Trimmed Mean for F} = \frac{2 + 3 + 4 + 4 + 5 + 8 + 8.9 + 9 + 9 + 9.2 + 11 + 14 + 20}{13} = \frac{106.1}{13} \approx 8.16\]

05

Compare Trimmed Means, Means, and Medians

The trimmed mean for \(U\) is much closer to the median than the original mean was (17 vs. 21.55), suggesting the trimming reduced the skew impact.For \(F\), the trimmed mean (8.16) is still close to both the mean (8.56) and median (8.9), indicating trimming had a smaller effect due to less skew in original data.Trimming percentages: - \(U\) removed 2 out of 11 values ~18.18%,- \(F\) removed 2 out of 15 values ~13.33%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Mean

The sample mean is a basic statistical measure that provides an average of all the values in a dataset. It's calculated by summing all the numbers and then dividing by the count of those numbers.

In this exercise, we looked at concentrations in dust samples from both urban and farm homes. For the urban sample, we add all the values, totaling 237, and divide by 11, which gives us a mean of about 21.55. For the farm sample, the total is 128.4, and when divided by the 15 samples, we get a mean of approximately 8.56.

The difference between means suggests that urban homes have a higher average concentration of endotoxin in dust, underlining the importance of knowing how values range across different environments.

Sample Median

The median provides a 'middle' value in a dataset, offering an alternative to the mean that is less affected by extremely high or low values. To find it, you must first arrange numbers in order.

For the urban sample, the ordered values gave a median of 17, while for the farm sample, it was 8.9. Unlike the mean, the sample median can often resist the influence of outliers, providing a better sense of the data's central tendency when anomalies exist.

Notably, in the urban data, the sample median is less than the mean, highlighting potential data skewing by a high outlier of 80. This stark difference reinforces the need for multiple measures of central tendency in data analysis.

Trimmed Mean

The trimmed mean is a way to calculate the average by excluding a set percentage of the highest and lowest data points. This method reduces the potential impact of extreme outliers.

In this case, both datasets had their smallest and largest values removed: 4 and 80 from the urban set, 0.3 and 21 from the farm set. This gave a trimmed mean of 17.00 for urban homes and 8.16 for farm homes.

These values are much closer to the respective medians than the full means, particularly for urban homes. It removes some of the skewing influence and provides a clearer picture of the typical conditions in the environments.

Data Distribution

Data distribution describes how values are spread within a dataset. It helps identify patterns, trends, and potential anomalies. Typically, distributions fall under symmetric, skewed, or uniform categories.

In examining the urban and farm datasets, the urban data appears skewed due to the high value of 80, pushing the mean upwards but not affecting the median equally. Alternatively, the farm data demonstrates a more uniform distribution, with mean and median values being closer.

Understanding a dataset's distribution is crucial to choosing the appropriate statistical tools and interpreting results effectively without being misled by unusual data points.

Outliers

Outliers are data points that diverge significantly from other observations. They can heavily influence statistical measures like the mean, leading to misleading conclusions if not properly addressed.

In the urban data, the value 80 acts as a prominent outlier, impacting the mean significantly more than the median. The farm data had less impactful outliers but still demonstrated values that could be considered variations.

Identifying and understanding outliers can provide insight into data quirks or errors and is crucial in discerning true trends from anomalies. In practice, excluding or adjusting for outliers via methods like the trimmed mean can offer a more representative analysis.