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Chapter 2: Problem 109

Consider these two sets of data: $$\begin{array}{llllll} \hline \text { Set 1 } & 46 & 55 & 50 & 47 & 52 \\ \text { Set 2 } & 30 & 55 & 65 & 47 & 53 \\ \hline \end{array}$$ Both sets have the same mean, \(50 .\) Compare these measures for both sets: \(\Sigma(x-\bar{x}), \operatorname{SS}(x),\) and range. Comment on the meaning of these comparisons.

Short Answer

Expert verified
The sum of differences from the mean \(\Sigma(x-\bar{x})\) is 0 for both sets. The sum of squares of differences from the mean \(\operatorname{SS}(x)\) is 54 for Set 1 and 668 for Set 2. The range is 9 for Set 1 and 35 for Set 2. So, Set 2 is more variable than Set 1, despite having the same mean.

Step by step solution

01

Calculate Sum of Differences

The sum of the differences \(\Sigma(x-\bar{x})\) represents the cumulative difference of each data point from the mean. The mean for both sets is 50. For each set, subtract each data point from the mean and then sum up these differences. This results in a sum of 0 for both sets, because the sum of the differences from the mean always equals 0 in any data set.
02

Calculate Sum of Squares

The sum of the squared differences \(\operatorname{SS}(x)\) is calculated by squaring each of the differences you computed in the previous step, and then summing these squared differences. For Set 1, this is \(16+25+0+9+4=54\). For Set 2, this is \(400+25+225+9+9=668\). This value shows how much variation or dispersion there is from the mean.
03

Calculate Range

The range is the difference between the highest and the lowest values in the data set. For Set 1, this is \(55-46=9\). For Set 2, this is \(65-30=35\). This shows the spread of data in the data set.
04

Comment on Comparisons

For both sets, \(\Sigma(x-\bar{x})\) is 0, which is always the case for any data set. The SS(x) values and ranges are different for the two sets. Set 1 is less dispersed around the mean (SS(x)=54, range=9) while Set 2 is more dispersed (SS(x)=668, range=35). Despite having the same mean, Set 2 has a larger variability compared to Set 1.

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Key Concepts

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

Mean
The mean, often called the average, provides a central value for a data set. To find the mean, you sum up all the data points and divide by the number of points. For both Set 1 and Set 2, the mean is 50. This means that if you divide the sum of the data points in each set by five (since each set has five points), you'll get 50. This single value offers a quick snapshot of the overall level of the data. However, the mean doesn't tell the whole story about the data set, because it doesn't show how spread out the data points are.
Sum of Squares
The Sum of Squares (SS) gauge variation in a data set. It measures how far each data point differs from the mean. To calculate it, follow these steps:
  • Subtract the mean from each data point to get the differences.
  • Square each of these differences to avoid negative values.
  • Add all the squared differences together.
For Set 1, the SS equals 54, meaning the data points do not vary much from the mean. For Set 2, the SS is 668, which shows much greater variation. The higher the SS, the more spread out about the mean the data points are.
Range
The range helps you understand the spread of the data by looking at only the extreme values. Calculating the range is simple: find the highest and lowest values in the set, then subtract the lowest from the highest. For Set 1, the range is 9, which indicates relatively close numbers. Set 2, however, has a range of 35, signaling a broader spread of values. Thus, the range is a handy tool for a basic understanding of data spread, but it doesn't consider every data point, unlike other measures such as the sum of squares.
Data Dispersion
Data dispersion highlights how spread out the numbers in a data set are. This can be measured using the sum of squares or the range. Less dispersion means the values are closely knit around the mean, as in Set 1. Greater dispersion indicates a wider distribution around the mean, like in Set 2. Dispersion matters because it affects data reliability; high dispersion might suggest that the mean isn't a good representation of the data.
Data Comparison
Data comparison involves looking at how different data sets behave. Even if they share the same mean, like Set 1 and Set 2, their spread can be vastly different. By analyzing the sum of squares, range, and dispersion, even subtle differences between data sets become clear. In our example, Set 2's greater range and sum of squares highlight it as more variable than Set 1. This means that even though on average they 'look' the same, their data behavior is quite different.

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Most popular questions from this chapter

Explain why it is possible to find the mean for the data of a quantitative variable, but not for a qualitative variable.

A survey of 100 resort club managers on their annual salaries resulted in the following frequency distribution: $$\begin{array}{lccccc} \hline \text { Annual Salary (s1000s) } & 15-25 & 25-35 & 35-45 & 45-55 & 55-65 \\ \text { No. of Managers } & 12 & 37 & 26 & 19 & 6 \\ \hline \end{array}$$ a. The data value "35" belongs to which class? b. Explain the meaning of "35-45." c. Explain what "class width" is, give its value, and describe three ways that it can be determined. d. Draw a frequency histogram of the annual salaries for resort club managers. Label class boundaries. (Retain these solutions to use in Exercise 2.53 on p. \(61 .\) )

What kinds of financial transactions do you do online? Are you worried about your security? According to Consumer Internet Barometer, the source of a March \(25,2009,\) USA Today Snapshot titled "Security of online accounts," the following transactions and percent of people concerned about their online security were reported. $$\begin{array}{lc} \text { What } & \text { Percent } \\ \hline \text { Bankirg } & 72 \\ \text { Paying bills } & 70 \\ \text { Buying stocks, bords } & 62 \\ \text { Fling toxes } & 62 \\ \hline \end{array}$$ Prepare two bar graphs to depict the percentage data. Scale the vertical axis on the first graph from 50 to \(80 .\) Scale the second graph from 0 to \(100 .\) What is your conclusion concerning how the percentages of the four responses stack up based on the two bar graphs, and what would you recommend, if anything, to improve the presentations?

What property does the distribution need for the median, the midrange, and the midquartile to all be the same value?

According to Chebyshev's theorem, what proportion of a distribution will be within \(k=4\) standard deviations of the mean?
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