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Chapter 1: Problem 37

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 median, 11.35 million km虏, is a typical value, as it better represents the data by not being influenced by extreme values.

Step by step solution

01

Arrange the Data

First, list the data in ascending order to make it easy to identify certain statistical measures. Arrange the snow cover data: \(6.5, 7.9, 9.2, 10.0, 10.7, 12.0, 12.5, 14.5, 14.9, 21.9\).
02

Calculate the Mean

The mean is calculated by summing all the values and dividing by the number of observations. \[\text{Mean} = \frac{6.5 + 7.9 + 9.2 + 10.0 + 10.7 + 12.0 + 12.5 + 14.5 + 14.9 + 21.9}{10} = \frac{120.1}{10} = 12.01\] The mean value is \(12.01\, \text{million km}^2.\)
03

Calculate the Median

The median is the middle value in a list of numbers. If the list has an even number of observations, the median is the average of the two middle values. Since there are 10 observations, we find the median between the 5th and 6th values: 10.7 and 12.0.\[\text{Median} = \frac{10.7 + 12.0}{2} = 11.35\] The median is \(11.35\, \text{million km}^2.\)
04

Interpretation and Choice

In choosing a representative value, consider the mean and median values. The mean is affected by the extreme value 21.9, making the mean slightly larger than the central tendency of the data. The median is less affected by extreme values and, in this case, gives a central tendency closer to most of the data points. Therefore, the median (11.35 million \(\text{km}^2\)) is a better representative value.

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Key Concepts

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

Mean
The concept of the mean is one of the most fundamental in statistics. It provides an average or central value for a set of numbers. Calculating the mean involves adding up all of the numbers in a data set and then dividing by the number of items in the set.

To illustrate, let's take the snow cover data from Eurasia between 1970 and 1979: 6.5, 7.9, 9.2, 10.0, 10.7, 12.0, 12.5, 14.5, 14.9, and 21.9 million km虏. Adding these numbers together gives us a total of 120.1. Since there are 10 observations, we divide 120.1 by 10 to get the mean, which is 12.01 million km虏.

The mean is influenced by extremely high or low values. In this example, the particularly high value of 21.9 pulls the mean up, showing us that while the mean gives a sense of the data's average, it can sometimes be misleading if the data contains outliers.
Median
The median is another measure of central tendency, representing the middle point in a data set. Unlike the mean, the median is not skewed by very large or small values.

To find the median, first arrange the data in ascending order. For the Eurasia snow cover data, the ordered list is: 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, a pair lies in the middle position. Thus, we take the average of these two middle numbers (10.7 and 12.0), resulting in a median value of 11.35 million km虏.

The median is often considered a better representation of the central tendency, especially in data sets with outliers, as it gives a more accurate reflection of the "typical" value.
Central Tendency
Central tendency is a critical concept in data analysis, providing a summary measure that describes a whole set of data with a single value that represents the center of the data distribution.

The most common measures of central tendency are the mean and median. When analyzing the Eurasia data, the mean was calculated as 12.01, while the median was 11.35 million km虏. In scenarios with skewed distributions or outliers, like the higher end of this snow data, the median is often preferred as it is more robust.

Understanding central tendency is vital for statistical analysis as it highlights trends and patterns, guiding us to make informed interpretations or decisions based on data.
Data Analysis
Data analysis involves examining data sets to infer conclusions and insights. In our example of snow cover data in Eurasia, we first organized and understood the distribution of the data by arranging it in order, then calculated measures like the mean and median.

Through data analysis, we determined that the median was a better indicator of central tendency than the mean in the presence of an outlier. This practice of evaluating various aspects of data helps us derive meaningful patterns and make well-grounded predictions or decisions.

Good data analysis not only focuses on the numbers but also considers the context and implications of statistical findings, as it aims to provide clarity and insight into raw data.

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Most popular questions from this chapter

The article "The Pedaling Technique of Elite Endurance Cyclists" (Int. J. of Sport Biomechanics, 1991: 29-53) reported the accompanying data on single-leg power at a high workload: \(\begin{array}{lllllll}244 & 191 & 160 & 187 & 180 & 176 & 174 \\ 205 & 211 & 183 & 211 & 180 & 194 & 200\end{array}\) a. Calculate and interpret the sample mean and median. b. Suppose that the first observation had been 204 rather than 244 . How would the mean and median change? c. Calculate a trimmed mean by eliminating the smallest and largest sample observations. What is the corresponding trimming percentage? d. The article also reported values of single-leg power for a low workload. The sample mean for \(n=13\) observations was \(\bar{x}=119.8\) (actually 119.7692), and the 14th observation, somewhat of an outlier, was 159 . What is the value of \(\bar{x}\) for the entire sample?

Fire load \(\left(\mathrm{MJ} / \mathrm{m}^{2}\right)\) is the heat energy that could be released per square meter of floor area by combustion of contents and the structure itself. The article "Fire Loads in Office Buildings" (J. of Structural Engr., 1997: 365-368) gave the following cumulative percentages (read from a graph) for fire loads in a sample of 388 rooms: $$ \begin{array}{lccccc} \text { Value } & 0 & 150 & 300 & 450 & 600 \\ \text { Cumulative \% } & 0 & 19.3 & 37.6 & 62.7 & 77.5 \\ \text { Value } & 750 & 900 & 1050 & 1200 & 1350 \\ \text { Cumulative \% } & 87.2 & 93.8 & 95.7 & 98.6 & 99.1 \\ \text { Value } & 1500 & 1650 & 1800 & 1950 & \\ \text { Cumulative \% } & 99.5 & 99.6 & 99.8 & 100.0 & \end{array} $$ a. Construct a relative frequency histogram and comment on interesting features. b. What proportion of fire loads are less than 600 ? At least \(1200 ?\) c. What proportion of the loads are between 600 and 1200 ?

The article "Can We Really Walk Straight?" (Amer. J. of Physical Anthropology, 1992: 19-27) reported on an experiment in which each of 20 healthy men was asked to walk as straight as possible to a target \(60 \mathrm{~m}\) away at normal speed. Consider the following observations on cadence (number of strides per second): \(\begin{array}{rrrrrrrrrr}.95 & .85 & .92 & .95 & .93 & .86 & 1.00 & .92 & .85 & .81 \\ .78 & .93 & .93 & 1.05 & .93 & 1.06 & 1.06 & .96 & .81 & .96\end{array}\) Use the methods developed in this chapter to summarize the data; include an interpretation or discussion wherever appropriate. [Note: The author of the article used a rather sophisticated statistical analysis to conclude that people cannot walk in a straight line and suggested several explanations for this.]

Consider the following data on active repair time (hours) for a sample of \(n=46\) airborne communications receivers: $$ \begin{array}{rrrrrrrrrrrr} .2 & .3 & .5 & .5 & .5 & .6 & .6 & .7 & .7 & .7 & .8 & .8 \\ .8 & 1.0 & 1.0 & 1.0 & 1.0 & 1.1 & 1.3 & 1.5 & 1.5 & 1.5 & 1.5 & 2.0 \\ 2.0 & 2.2 & 2.5 & 2.7 & 3.0 & 3.0 & 3.3 & 3.3 & 4.0 & 4.0 & 4.5 & 4.7 \\ 5.0 & 5.4 & 5.4 & 7.0 & 7.5 & 8.8 & 9.0 & 10.3 & 22.0 & 24.5 & & \end{array} $$ Construct the following: a. A stem-and-leaf display in which the two largest values are displayed separately in a row labeled HI. b. A histogram based on six class intervals with 0 as the lower limit of the first interval and interval widths of 2 , \(2,2,4,10\), and 10 , respectively.

A transformation of data values by means of some mathematical function, such as \(\sqrt{x}\) or \(1 / x\), can often yield a set of numbers that has "nicer" statistical properties than the original data. In particular, it may be possible to find a function for which the histogram of transformed values is more symmetric (or, even better, more like a bell-shaped curve) than the original data. As an example, the article "Time Lapse Cinematographic Analysis of BerylliumLung Fibroblast Interactions" (Environ. Research, 1983: 34-43) reported the results of experiments designed to study the behavior of certain individual cells that had been exposed to beryllium. An important characteristic of such an individual cell is its interdivision time (IDT). IDTs were determined for a large number of cells both in exposed (treatment) and unexposed (control) conditions. The authors of the article used a logarithmic transformation, that is, transformed value \(=\log\) (original value). Consider the following representative IDT data: $$ \begin{array}{lccccc} \hline \text { IDT } & \log _{10}(\text { IDT }) & \text { IDT } & \log _{10}(\text { IDT }) & \text { IDT } & \log _{10}(\text { IDT }) \\ \hline 28.1 & 1.45 & 60.1 & 1.78 & 21.0 & 1.32 \\ 31.2 & 1.49 & 23.7 & 1.37 & 22.3 & 1.35 \\ 13.7 & 1.14 & 18.6 & 1.27 & 15.5 & 1.19 \\ 46.0 & 1.66 & 21.4 & 1.33 & 36.3 & 1.56 \\ 25.8 & 1.41 & 26.6 & 1.42 & 19.1 & 1.28 \\ 16.8 & 1.23 & 26.2 & 1.42 & 38.4 & 1.58 \\ 34.8 & 1.54 & 32.0 & 1.51 & 72.8 & 1.86 \\ 62.3 & 1.79 & 43.5 & 1.64 & 48.9 & 1.69 \\ 28.0 & 1.45 & 17.4 & 1.24 & 21.4 & 1.33 \\ 17.9 & 1.25 & 38.8 & 1.59 & 20.7 & 1.32 \\ 19.5 & 1.29 & 30.6 & 1.49 & 57.3 & 1.76 \\ 21.1 & 1.32 & 55.6 & 1.75 & 40.9 & 1.61 \\ 31.9 & 1.50 & 25.5 & 1.41 & & \\ 28.9 & 1.46 & 52.1 & 1.72 & & \\ \hline \end{array} $$ Use class intervals \(10-<20,20-<30, \ldots\) to construct a histogram of the original data. Use intervals \(1.1-<1.2\), \(1.2-<1.3, \ldots\) to do the same for the transformed data. What is the effect of the transformation?
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