/*! 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] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 22

Find the coefficient of variation for each of the two samples; then compare the variation. (The same data were used in Section 3-I.) 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? $$\begin{array}{lcccccccccc} \text { Collection Contractor Was Brinks } & 1.3 & 1.5 & 1.3 & 1.5 & 1.4 & 1.7 & 1.8 & 1.7 & 1.7 & 1.6 \\ \text { Collection Contractor Was Not Brinks } & 2.2 & 1.9 & 1.5 & 1.6 & 1.5 & 1.7 & 1.9 & 1.6 & 1.6 & 1.8 \end{array}$$

Short Answer

Expert verified
The coefficient of variation is 11.48% for Brinks and 12.72% for Not Brinks. Not Brinks has slightly higher variability.

Step by step solution

01

- Understand the Coefficient of Variation

The coefficient of variation (CV) is a standardized measure of dispersion of a probability distribution or frequency distribution. It is defined as the ratio of the standard deviation to the mean, often expressed as a percentage. The formula is given by: \[ CV = \frac{\sigma}{\mu} \times 100\% \] where \( \sigma \) is the standard deviation and \( \mu \) is the mean.
02

- Calculate the Mean for Each Sample

First, find the mean (\( \mu \)) for each sample. For Brinks: \[ \mu_1 = \frac{1.3 + 1.5 + 1.3 + 1.5 + 1.4 + 1.7 + 1.8 + 1.7 + 1.7 + 1.6}{10} = \frac{15.5}{10} = 1.55 \]For Not Brinks: \[ \mu_2 = \frac{2.2 + 1.9 + 1.5 + 1.6 + 1.5 + 1.7 + 1.9 + 1.6 + 1.6 + 1.8}{10} = \frac{17.3}{10} = 1.73 \]
03

- Calculate the Standard Deviation for Each Sample

Next, find the standard deviation (\( \sigma \)) for each sample. For Brinks: \[ \sigma_1 = \sqrt{\frac{(1.3 - 1.55)^2 + (1.5 - 1.55)^2 + \ (1.3 - 1.55)^2 + (1.5 - 1.55)^2 + (1.4 - 1.55)^2 + (1.7 - 1.55)^2 + (1.8 - 1.55)^2 + (1.7 - 1.55)^2 + (1.7 - 1.55)^2 + (1.6 - 1.55)^2}{10 - 1}} \]\[ \sigma_1 = \sqrt{\frac{0.0625 + 0.0025 + 0.0625 + 0.0025 + 0.0225 + 0.0225 + 0.0625 + 0.0225 + 0.0225 + 0.0025}{9}} = \sqrt{\frac{0.285}{9}} = \sqrt{0.03167} \approx 0.178 \]For Not Brinks: \[ \sigma_2 = \sqrt{\frac{(2.2 - 1.73)^2 + (1.9 - 1.73)^2 + (1.5 - 1.73)^2 + (1.6 - 1.73)^2 + (1.5 - 1.73)^2 + (1.7 - 1.73)^2 + (1.9 - 1.73)^2 + (1.6 - 1.73)^2 + (1.6 - 1.73)^2 + (1.8 - 1.73)^2}{10 - 1}} \]\[ \sigma_2 = \sqrt{\frac{0.2209 + 0.0289 + 0.0529 + 0.0084 + 0.0529 + 0.0009 + 0.0289 + 0.0169 + 0.0169 + 0.0049}{9}} = \sqrt{\frac{0.4335}{9}} = \sqrt{0.04817} \approx 0.219 \]
04

- Calculate the Coefficient of Variation for Each Sample

Now, use the formula for CV to calculate it for both samples. For Brinks: \[ CV_1 = \frac{\sigma_1}{\mu_1} \times 100 \% = \frac{0.178}{1.55} \times 100 \% \approx 11.48 \% \]For Not Brinks: \[ CV_2 = \frac{\sigma_2}{\mu_2} \times 100 \% = \frac{0.219}{1.73} \times 100 \% \approx 12.72 \% \]
05

- Compare the Coefficients of Variation

Compare the calculated CVs: \[ CV_1 (\text{Brinks}) = 11.48 \% CV_2 (\text{Not Brinks}) = 12.72 \% \]Since the coefficient of variation for Not Brinks is slightly higher than that for Brinks, it indicates a slightly higher variability in the amounts collected by contractors who were not Brinks.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

statistical analysis
Statistical analysis involves collecting and examining data to uncover patterns and trends. It utilizes various techniques to make sense of data collected from different sources.
In the case of our exercise, we are interested in comparing the variation in amounts collected from parking meters between two different contractors.
This comparison helps us understand if there are significant differences in how consistent the collection amounts are. We use calculations such as the mean, standard deviation, and coefficient of variation to achieve this.
These metrics provide insights into the central tendency and the dispersion of the data sets, helping us deduce whether the variability in collections is more pronounced for Brinks or the other contractor.
descriptive statistics
Descriptive statistics summarize and describe the main features of a data set. In simpler terms, it provides a way to understand the essential characteristics of the data without delving into every individual data point.
For our exercise, the key descriptive statistics we used are the mean and the standard deviation.
  • The mean, or average, represents the central point of the data set.
  • The standard deviation measures the spread of the data points around this central point.
These two metrics allowed us to calculate the coefficient of variation.
The mean for Brinks and Not Brinks gives us an overall idea of the average collection amounts. Meanwhile, the standard deviation tells us how much the collection amounts deviate from the mean.
With these calculations, we are set to understand how the data for each contractor behaves and how consistent each one's collection amounts are over time.
comparing variability
When comparing variability, we look at how spread out the data is within each group. This is where the coefficient of variation becomes very useful.
The coefficient of variation (CV) is a standardized way to measure and compare the degree of variation between different data sets.
By definition, the CV is the ratio of the standard deviation to the mean, and it is expressed as a percentage. This percentage allows us to compare datasets on a relative basis irrespective of their units or scales.
In our exercise, we found that:
  • Brinks has a CV of approximately 11.48%.
  • Not Brinks has a CV of approximately 12.72%.
These results show that the Not Brinks data set has a slightly higher variability compared to the Brinks data set.
This means that the amounts collected by Not Brinks vary more from their average compared to the amounts collected by Brinks.
Understanding and comparing variability helps in making informed decisions and understanding the consistency of processes or outcomes in different scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the mean and median for each of the two samples, then compare the two sets of results. Waiting times (in seconds) of customers at the Madison Savings Bank are recorded with two configurations: single customer line; individual customer lines. Carefully examine the data to determine whether there is a difference between the two data sets that is not apparent from a comparison of the measures of center. If so, what is it? $$\begin{array}{llllllllll}\text { single Line } & 390 & 396 & 402 & 408 & 426 & 438 & 444 & 462 & 462 & 462 \\\\\text { Individual Lines } & 252 & 324 & 348 & 372 & 402 & 462 & 462 & 510 & 558 & 600\end{array}$$

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-I, where we found measures of center. Here we find measures of variation.) Then answer the given questions. Listed below in dollars are the annual costs of tuition and fees at the 10 most expensive colleges in the United States for a recent year (based on data from \(U . S .\) News \& World Report). The colleges listed in order are Columbia, Vassar, Trinity, George Washington, Carnegie Mellon, Wesleyan, Tulane, Bucknell, Oberlin, and Union. What does this "top 10" list tell us about the variation among costs for the population of all U.S. college tuition? $$\begin{array}{rlllllllll} 49,138 & 47,890 & 47,510 & 47,343 & 46,962 & 46,944 & 46,930 & 46,902 & 46,870 & 46,785 \end{array}$$

Let a population consist of the values 9 cigarettes, 10 cigarettes, and 20 cigarettes smoked in a day (based on data from the California Health Interview Survey). Assume that samples of two values are randomly selected with replacement from this population. (That is, a selected value is replaced before the second selection is made.) a. Find the variance \(\sigma^{2}\) of the population \(\\{9 \text { cigarettes, } 10 \text { cigarettes, } 20 \text { cigarettes }\\}\) b. After listing the nine different possible samples of two values selected with replacement, find the sample variance \(s^{2}\) (which includes division by \(n-1\) ) for each of them; then find the mean of the nine sample variances \(s^{2}\) c. For each of the nine different possible samples of two values selected with replacement, find the variance by treating each sample as if it is a population (using the formula for population variance, which includes division by \(n\) ); then find the mean of those nine population variances. d. Which approach results in values that are better estimates of \(\sigma^{2}:\) part (b) or part (c)? Why? When computing variances of samples, should you use division by \(n\) or \(n-1 ?\) e. The preceding parts show that \(s^{2}\) is an unbiased estimator of \(\sigma^{2}\). Is \(s\) an unbiased estimator of \(\sigma ?\) Explain.

In what sense are the mean, median, mode, and midrange measures of "center"?

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 selling prices (dollars) of TVs that are 60 inches or larger and rated as a "best buy" by Consumer Reports magazine. Are the resulting statistics representative of the population of all TVs that are 60 inches and larger? If you decide to buy one of these TVs, what statistic is most relevant, other than the measures of central tendency? $$\begin{array}{rrrrrrrrrr}1800 & 1500 & 1200 & 1500 & 1400 & 1600 & 1500 & 950 & 1600 & 1150 & 1500 & 1750\end{array}$$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.