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Chapter 1: Problem 34

The article "Snow Cover and Temperature Relationships in North America and Eurasia" (J. Climate Appl. Meteorol., 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 October snow cover is 12.01 million km虏, the mean value.

Step by step solution

01

Understanding the Data

We have a set of ten observations representing October snow cover in Eurasia over the years 1970-1979. The values are 6.5, 12.0, 14.9, 10.0, 10.7, 7.9, 21.9, 12.5, 14.5, and 9.2 million km虏.
02

Choosing a Measure of Central Tendency

To find the typical value, we need to decide which measure of central tendency (mean, median, or mode) is most suitable. The mean provides an average, while the median gives the central value of the ordered dataset. The mode, which is the most frequent value, is not applicable here as all values are unique.
03

Calculating the Mean

The mean is calculated by summing all values and dividing by the number of observations. Mean = \( \frac{6.5 + 12.0 + 14.9 + 10.0 + 10.7 + 7.9 + 21.9 + 12.5 + 14.5 + 9.2}{10} \).
04

Performing the Sum

Add up all the values: 6.5 + 12.0 + 14.9 + 10.0 + 10.7 + 7.9 + 21.9 + 12.5 + 14.5 + 9.2 = 120.1
05

Dividing by Number of Observations

Divide the total sum by the number of observations: \( \frac{120.1}{10} = 12.01 \).
06

Evaluating the Median

To find the median, order the data set and identify the central number(s). Ordered data: 6.5, 7.9, 9.2, 10.0, 10.7, 12.0, 12.5, 14.5, 14.9, 21.9. The median is the average of the 5th and 6th numbers: \( \frac{10.7 + 12.0}{2} = 11.35 \).
07

Conclusion

Both mean (12.01) and median (11.35) provide insights into the typical snow cover. In this case, the mean is slightly higher due to the influence of an extreme value (21.9). However, both values are relatively close, indicating either could represent the typical snow cover.

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Key Concepts

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

Central Tendency Measures
Central tendency measures are essential tools in statistical analysis. They help us understand the typical or most representative value from a dataset.

The key measures are:
  • **Mean**: The average of all data points.
  • **Median**: The middle value when data is ordered from smallest to largest.
  • **Mode**: The most frequently occurring value in the dataset.
Choosing between these depends on the dataset and the goal of the analysis. The mean takes into account all data points, providing a comprehensive view. However, it might be influenced by extreme values. The median, on the other hand, represents the center of data, reducing the effect of outliers. When data has many repetitions, the mode can be useful to identify the most common value.
Mean Calculation
To calculate the mean, you need to sum all the values in your data set and divide by the number of observations. This method gives you the arithmetic average. In our dataset, the values were:
  • 6.5
  • 12.0
  • 14.9
  • 10.0
  • 10.7
  • 7.9
  • 21.9
  • 12.5
  • 14.5
  • 9.2
First, we add all values together:\[6.5 + 12.0 + 14.9 + 10.0 + 10.7 + 7.9 + 21.9 + 12.5 + 14.5 + 9.2 = 120.1\]Next, divide by the number of data points:\[\frac{120.1}{10} = 12.01\]Thus, the mean snow cover for this data is 12.01 million km虏.
Median Calculation
The median gives the central point of a dataset by arranging values in numerical order. For our data:
  • 6.5
  • 7.9
  • 9.2
  • 10.0
  • 10.7
  • 12.0
  • 12.5
  • 14.5
  • 14.9
  • 21.9
Since there are 10 values, which is even, the median is found by taking the average of the 5th and 6th numbers:\[\frac{10.7 + 12.0}{2} = 11.35\]Therefore, the median snow cover is 11.35 million km虏. This central value is less influenced by extreme variations in the data.
Outliers Influence
Outliers are data points significantly different from others in the dataset. They can skew results, especially impacting measures like the mean.

In our example, the data point 21.9 is substantially higher than the others. This outlier increases the mean:
  • The mean (12.01 million km虏) is pulled up by the outlier.
  • Conversely, the median (11.35 million km虏) stays stable as it's determined by central order, not magnitude.
When data contains outliers, the median often provides a better sense of central tendency as it minimizes the influence of extreme values.

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Most popular questions from this chapter

The amount of flow through a solenoid valve in an automobile's pollution- control system is an important characteristic. An experiment was carried out to study how flow rate depended on three factors: armature length, spring load, and bobbin depth. Two different levels (low and high) of each factor were chosen, and a single observation on flow was made for each combination of levels. a. The resulting data set consisted of how many observations? b. Does this study involve sampling an existing population or a conceptual population?

Many people who believe they may be suffering from the flu visit emergency rooms, where they are subjected to long waits and may expose others or themselves be exposed to various diseases. The article "Drive-Through Medicine: A Novel Proposal for the Rapid Evaluation of Patients During an Influenza Pandemic" (Ann. Emerg. Med., 2010: 268-273 described an experiment to see whether patients could be evaluated while remaining in their vehicles. The following total processing times (min) for a sample of 38 individuals were read from a graph that appeared in the cited article: \(\begin{array}{llllllll}9 & 16 & 16 & 17 & 19 & 20 & 20 & 20 \\ 23 & 23 & 23 & 23 & 24 & 24 & 24 & 24 \\ 25 & 25 & 26 & 26 & 27 & 27 & 28 & 28 \\ 29 & 29 & 29 & 30 & 32 & 33 & 33 & 34 \\ 37 & 43 & 44 & 46 & 48 & 53 & & \end{array}\) a. Calculate several different measures of center and compare them. b. Are there any outliers in this sample? Any extreme outliers? c. Construct a boxplot and comment on any interesting features.

At the beginning of the 2007 baseball season each American League team had nine starting position players (this includes the designated hitter but not the pitcher). Here are the salaries for the New York Yankees and the Cleveland Indians in thousands of dollars: \(\begin{array}{llllll}\text { Yankees: } & 12000 & 600 & 491 & 22709 & 21600 \\\ & 13000 & 13000 & 15000 & 23429 & \\ \text { Indians: } & 3200 & 3750 & 396 & 383 & 1000 \\ & 3750 & 917 & 3000 & 4050 & \end{array}\) Construct a comparative boxplot and comment on interesting features. Compare the salaries of the two teams. The Indians won more games than the Yankees in the regular season and defeated the Yankees in the playoffs.

A sample of \(n=10\) automobiles was selected, and each was subjected to a 5 -mph crash test. Denoting a car with no visible damage by \(S\) (for success) and a car with such damage by \(\mathrm{F}\), results were as follows: \(\begin{array}{llllllllll}\mathrm{S} & \mathrm{S} & \mathrm{F} & \mathrm{S} & \mathrm{S} & \mathrm{S} & \mathrm{F} & \mathrm{F} & \mathrm{S} & \mathrm{S}\end{array}\) a. What is the value of the sample proportion of successes \(x / n\) ? b. Replace each \(S\) with a 1 and each \(F\) with a 0 . Then calculate \(\bar{x}\) for this numerically coded sample. How does \(x\) compare to \(x / n\) ? c. Suppose it is decided to include 15 more cars in the experiment. How many of these would have to be \(S\) 's to give \(x / n=.80\) for the entire sample of 25 cars?

A sample of 77 individuals working at a particular office was selected and the noise level (dBA) experienced by each one was determined, yielding the following data ("Acceptable Noise Levels for Construction Site Offices, Build. Serv. Engr. Res. Technol., 2009: 87-94). \(\begin{array}{lllllllll}55.3 & 55.3 & 55.3 & 55.9 & 55.9 & 55.9 & 55.9 & 56.1 & 56.1 \\ 56.1 & 56.1 & 56.1 & 56.1 & 56.8 & 56.8 & 57.0 & 57.0 & 57.0 \\\ 57.8 & 57.8 & 57.8 & 57.9 & 57.9 & 57.9 & 58.8 & 58.8 & 58.8 \\ 59.8 & 59.8 & 59.8 & 62.2 & 62.2 & 63.8 & 63.8 & 63.8 & 63.9 \\ 53.9 & 63.9 & 64.7 & 64.7 & 64.7 & 65.1 & 65.1 & 65.1 & 65.3 \\ 65.3 & 65.3 & 65.3 & 67.4 & 67.4 & 67.4 & 67.4 & 68.7 & 68.7 \\ 58.7 & 68.7 & 69.0 & 70.4 & 70.4 & 71.2 & 71.2 & 71.2 & 73.0 \\ 73.0 & 73.1 & 73.1 & 74.6 & 74.6 & 74.6 & 74.6 & 79.3 & 79.3 \\ 79.3 & 79.3 & 83.0 & 83.0 & 83.0 & & & & \end{array}\) Use various techniques discussed in this chapter to organize, summarize, and describe the data.
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