/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Find the coefficient of variatio... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the coefficient of variation for each of the two samples; then compare the variation. (The same data were used in Section 3-1.) Theft Listed below are amounts (in millions of dollars) collected from parking meters by Brinks and others in New York City during similar time periods. A larger data set was used to convict five Brinks employees of grand larceny. The data were provided by the attorney for New York City, and they are listed on the DASL Website. Do the two samples appear to have different amounts of variation? \(\begin{array}{lllllllllll} \text { Collection Contractor Was Brinks } & 1.3 & 1.5 & 1.3 & 1.5 & 1.4 & 1.7 & 1.8 & 1.7 & 1.7 & 1.6 \\ \text { Collection Contractor Was Not Brinks } & 2.2 & 1.9 & 1.5 & 1.6 & 1.5 & 1.7 & 1.9 & 1.6 & 1.6 & 1.8 \end{array}\)

Short Answer

Expert verified
The coefficients of variation show which sample has more variability.

Step by step solution

01

- Calculate the mean for each sample

The mean for Brinks: \(\bar{x}_1 = \frac{1.3 + 1.5 + 1.3 + 1.5 + 1.4 + 1.7 + 1.8 + 1.7 + 1.7 + 1.6}{10}\) The mean for non-Brinks: \(\bar{x}_2 = \frac{2.2 + 1.9 + 1.5 + 1.6 + 1.5 + 1.7 + 1.9 + 1.6 + 1.6 + 1.8}{10}\)
02

- Calculate the standard deviation for each sample

Standard deviation for Brinks: \(s_1 = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x}_1)^2}{n-1}}\) Standard deviation for non-Brinks:\(s_2 = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x}_2)^2}{n-1}}\)
03

- Calculate the coefficient of variation (CV) for each sample

Coefficient of variation is given by \(CV = \frac{s}{\bar{x}}\times100\)For Brinks: \(CV_1 = \frac{s_1}{\bar{x}_1}\times100\)For non-Brinks: \(CV_2 = \frac{s_2}{\bar{x}_2}\times100\)
04

- Compare the coefficients of variation

Compare \(CV_1\) and \(CV_2\) to determine which sample has more variability.

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Key Concepts

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

Statistics
Statistics is a branch of mathematics that helps us understand and analyze data. For instance, it allows us to make sense of the collections from parking meters in the exercise. Key statistical measures include the mean and standard deviation. These help us summarize and describe data sets.

In this exercise, we calculate and compare the mean and standard deviation of two samples collected by different contractors. By doing this, we understand the spread and average of the collected amounts. This process assists in verifying if there's a noticeable difference in the variation of data between the two contractors.
Sample Variability
Sample variability refers to the extent to which data values in a sample differ from one another. It’s an important concept in statistics, as it gives insight into the consistency of the data within a sample.

We measure variability using the standard deviation. A smaller standard deviation indicates that data points are close to the mean, showing less variability. Conversely, a larger standard deviation means the data points are spread out over a wider range, indicating higher variability.

For example, if we see high variability in the money collected from parking meters by Brinks compared to non-Brinks contractors, it raises questions about the consistency and reliability of their collection methods.
Standard Deviation
The standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how spread out the data points are from the mean. Calculating it involves the following steps:
  • Find the mean (average) of the data set.
  • Subtract the mean from each data point and square the result.
  • Find the average of these squared differences.
  • Take the square root of this average.

In this exercise, we calculate the standard deviation for the collections of Brinks and non-Brinks contractors separately. This gives us a clear understanding of the spread of each collection sample, allowing us to compare their consistency.
Mean Calculation
The mean, often referred to as the average, is a measure that provides a central value for a set of numbers. It is calculated by summing up all the data values and then dividing by the number of values. Here’s how you compute the mean:
  • Add all the numbers in the data set.
  • Divide the sum by the number of data points.

In our example, we've calculated the mean for both Brinks and non-Brinks collections. These means give us a reference point to understand the general level of collected amounts. This average value plays a crucial role in further calculations, like the standard deviation and the coefficient of variation.

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Most popular questions from this chapter

The 20 brain volumes \(\left(\mathrm{cm}^{3}\right)\) from Data Set 8 "IQ and Brain Size" in Appendix B have a mean of \(1126.0 \mathrm{~cm}^{3}\) and a standard deviation of \(124.9 \mathrm{~cm}^{3}\). Use the range rule of thumb to identify the limits separating values that are significantly low or significantly high. For such data, would a brain volume of \(1440 \mathrm{~cm}^{3}\) be significantly high?

Use the following cell phone airport data speeds (Mbps) from Sprint. Find the percentile corresponding to the given data speed. $$\begin{array}{llllllllll} 0.2 & 0.3 & 0.3 & 0.3 & 0.3 & 0.3 & 0.3 & 0.4 & 0.4 & 0.4 \\ 0.5 & 0.5 & 0.5 & 0.5 & 0.5 & 0.6 & 0.6 & 0.7 & 0.8 & 1.0 \\ 1.1 & 1.1 & 1.2 & 1.2 & 1.6 & 1.6 & 2.1 & 2.1 & 2.3 & 2.4 \\ 2.5 & 2.7 & 2.7 & 2.7 & 3.2 & 3.4 & 3.6 & 3.8 & 4.0 & 4.0 \\ 5.0 & 5.6 & 8.2 & 9.6 & 10.6 & 13.0 & 14.1 & 15.1 & 15.2 & 30.4 \end{array}$$ 13.0 Mbps

Consider a value to be significantly low if its score is less than or equal to \(-2\) or consider the value to be significantly high if its \(z\) score is greater than or equal to \(2 .\) In a recent year, scores on the Medical College Admission Test (MCAT) had a mean of \(25.2\) and a standard deviation of \(6.4\). Identify the MCAT scores that are significantly low or significantly high.

Watch out for these little buggers. Each of these exercises involves some feature that is somewhat tricky. Find the (a) mean, \((b)\) median, (c) mode, (d) midrange, and then answer the given question. Listed below are the numbers of Atlantic hurricanes that occurred in each year. The data are listed in order by year, starting with the year 2000 . What important feature of the data is not revealed by any of the measures of center? $$ \begin{array}{llllllllllllll} 8 & 9 & 8 & 7 & 9 & 15 & 5 & 6 & 8 & 4 & 12 & 7 & 8 & 2 \end{array} $$

Find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as "minutes") in your results. (The same data were used in Section 3-1, where we found measures of center. Here we find measures of variation.) Then answer the given questions. Listed below are the numbers of Atlantic hurricanes that occurred in each year. The data are listed in order by year, starting with the year 2000 . What important feature of the data is not revealed by any of the measures of variation? $$ \begin{array}{llllllllllllll} 8 & 9 & 8 & 7 & 9 & 15 & 5 & 6 & 8 & 4 & 12 & 7 & 8 & 2 \end{array} $$

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