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The article "Snow Cover and Temperature Relationships in North America and Eurasia" (J. Climate and Applied Meteorology, 1983: 460-469) used statistical techniques to relate the amount of snow cover on each continent to average continental temperature. Data presented there included the following ten observations on October snow cover for Eurasia during the years 1970-1979 (in million \(\mathrm{km}^{2}\) ): \(\begin{array}{llllllllll}6.5 & 12.0 & 14.9 & 10.0 & 10.7 & 7.9 & 21.9 & 12.5 & 14.5 & 9.2\end{array}\) What would you report as a representative, or typical, value of October snow cover for this period, and what prompted your choice?

Short Answer

Expert verified
The typical snow cover is 11.35 million km², chosen as the median to minimize outlier influence.

Step by step solution

01

Understand the Data

We have a set of ten observations representing the October snow cover in Eurasia from 1970 to 1979. The values are 6.5, 12.0, 14.9, 10.0, 10.7, 7.9, 21.9, 12.5, 14.5, 9.2 million square kilometers.
02

Consider Central Tendency Measures

To find a typical value, we need to consider measures of central tendency such as the mean, median, and mode. However, the mode may not be applicable here since all values are unique.
03

Calculate the Mean

The mean is the average value and is calculated by summing all observations and dividing by the number of observations. Sum all numbers: \[ 6.5 + 12.0 + 14.9 + 10.0 + 10.7 + 7.9 + 21.9 + 12.5 + 14.5 + 9.2 = 120.1 \]Divide by the number of observations (10): \[ \text{Mean} = \frac{120.1}{10} = 12.01 \] million square kilometers.
04

Calculate the Median

The median is the middle value when the observations are arranged in increasing order. First, arrange the data in ascending order: \[ 6.5, 7.9, 9.2, 10.0, 10.7, 12.0, 12.5, 14.5, 14.9, 21.9 \].As there are 10 values, the median will be the average of the 5th and 6th values. Add 10.7 and 12.0 and divide by 2: \[ \text{Median} = \frac{10.7 + 12.0}{2} = 11.35 \] million square kilometers.
05

Choose the Representative Value

The mean is 12.01 and the median is 11.35. Both values are fairly close, but the mean can be sensitive to outliers or very large values like 21.9. In this data set, the median might serve as a better typical value because it is less influenced by the extreme value of 21.9.

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Key Concepts

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

Mean Calculation
The mean is a measure of central tendency that provides an average value of a data set. In order to calculate the mean, we sum up all the values in the data set and then divide this sum by the number of observations. This gives us a single number representing the center of the data, often referred to as the 'average'. For the given snow cover data from Eurasia, we first add all observations:
  • 6.5, 12.0, 14.9, 10.0, 10.7, 7.9, 21.9, 12.5, 14.5, and 9.2 million square kilometers
  • The total sum: 120.1 million square kilometers
To find the mean, we then divide this sum by the number of data points, which is 10. Thus, the mean is calculated as follows:\[\text{Mean} = \frac{120.1}{10} = 12.01 \text{ million square kilometers}\]The mean provides a good indication of where the majority of data points lie, especially when the data doesn't have extreme values. However, it's essential to understand that the mean can be skewed by very large or very small data points, which leads us to consider other measures like the median.
Median Calculation
The median is another vital measure of central tendency used in statistical analysis, particularly useful when the data set includes outliers. It is the middle value of a data set when the numbers are arranged in ascending order. The median divides your data into two halves, helping to find that typical middle ground.
For our Eurasian snow cover data, the first step in median calculation is organizing the numbers from smallest to largest:
  • 6.5, 7.9, 9.2, 10.0, 10.7, 12.0, 12.5, 14.5, 14.9, 21.9
Since there are 10 numbers, we find the median by averaging the 5th and 6th numbers:
  • 10.7 is the 5th number
  • 12.0 is the 6th number
Thus, the median is calculated as:\[\text{Median} = \frac{10.7 + 12.0}{2} = 11.35 \text{ million square kilometers}\]The median is less affected by outliers and skewed data, making it a reliable measure of central tendency when the data has significant variability or outlier values, like the extreme 21.9 in this data set.
Statistical Analysis
Statistical analysis involves understanding your data using measures that summarize its most crucial features. Central tendency measures like mean and median are fundamental in this practice. They provide insight into what is 'typical' or 'central' for a data set, each offering unique advantages. When analyzing data sets, considering both the mean and median is beneficial for a comprehensive perspective. The mean utilizes all data points and is great for normally distributed data without outliers. However, it can be misleading if there exist extreme values, as seen with the 21.9 million square kilometers observation. The median, on the other hand, provides a more robust central value when the data includes outliers or when it's skewed.
This makes the median a preferred measure in these situations, aligning with why in our snow cover analysis, the median was ultimately chosen as the representative value. Understanding these nuances in statistical analysis helps make informed decisions in selecting the most appropriate measure for describing the central tendency of your data.

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Most popular questions from this chapter

The value of Young's modulus (GPa) was determined for cast plates consisting of certain intermetallic substrates, resulting in the following sample observations ("Strength and Modulus of a Molybdenum-Coated Ti-25Al-10Nb-3U1Mo Intermetallic," J. of Materials Engr and Performance, 1997: 46-50): \(\begin{array}{lllll}116.4 & 115.9 & 114.6 & 115.2 & 115.8\end{array}\) a. Calculate \(\bar{x}\) and the deviations from the mean. b. Use the deviations calculated in part (a) to obtain the sample variance and the sample standard deviation. c. Calculate \(s^{2}\) by using the computational formula for the numerator \(S_{x x}\) d. Subtract 100 from each observation to obtain a sample of transformed values. Now calculate the sample variance of these transformed values, and compare it to \(s^{2}\) for the original data.

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

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