/*! 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 31 The standard deviation alone doe... [FREE SOLUTION] | 91Ó°ÊÓ

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The standard deviation alone does not measure relative variation. For example, a standard deviation of \(\$ 1\) would be considered large if it is describing the variability from store to store in the price of an ice cube tray. On the other hand, a standard deviation of \(\$ 1\) would be considered small if it is describing store-to-store variability in the price of a particular brand of freezer. A quantity designed to give a relative measure of variability is the coefficient of variation. Denoted by CV, the coefficient of variation expresses the standard deviation as a percentage of the mean. It is defined by the \(C V=100\left(\frac{s}{\bar{x}}\right)\) formula Consider two samples. Sample 1 gives the actual weight (in ounces) of the contents of cans of pet food labeled as having a net weight of 8 ounces. Sample 2 gives the actual weight (in pounds) of the contents of bags of dry pet food labeled as having a net weight of 50 pounds. The weights for the two samples are \(\begin{array}{lrrrrr}\text { Sample 1 } & 8.3 & 7.1 & 7.6 & 8.1 & 7.6 \\ & 8.3 & 8.2 & 7.7 & 7.7 & 7.5 \\ \text { Sample 2 } & 52.3 & 50.6 & 52.1 & 48.4 & 48.8 \\ & 47.0 & 50.4 & 50.3 & 48.7 & 48.2\end{array}\) a. For each of the given samples, calculate the mean and the standard deviation. b. Compute the coefficient of variation for each sample. Do the results surprise you? Why or why not?

Short Answer

Expert verified
This analysis will provide the mean, standard deviation and coefficient of variation for both sets of samples. These numerical values will provide a quantitative representation of the variability among the sample sets. A higher Coefficient of Variation indicates greater variability in the dataset.

Step by step solution

01

Calculate Mean

Add all values in the sample and then divide by the count of values. Do this for both samples.
02

Calculate Standard Deviation

Subtract the mean from each value, square the results, add those, divide by how many values minus one, then take the square root. Do this for both samples.
03

Calculate Coefficient of Variation

Take the standard deviation, divide it by the mean, and multiply by 100. Do this for both samples.
04

Analyzing Results

Discuss which sample has a higher coefficient of variation, and what this tells us about the spread of the data.

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Key Concepts

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

Standard Deviation
The standard deviation is a statistical measure that captures the extent of spread or variability in a set of data. It tells us how much the individual data points in a sample deviate from the mean (average) of the data set. To calculate it, you need to follow a few steps:
First, subtract the mean from each data point in your sample. Then, square each of these differences. Next, add all of these squared differences together. Divide this sum by one less than the number of data points in the sample (this is called degrees of freedom). Finally, take the square root of this quotient.
This entire process gives you the standard deviation, which can provide a valuable insight into the data's variability. For example, a small standard deviation suggests that the data points are close to the mean, indicating low variability. Conversely, a large standard deviation suggests a wider range of data points.
Mean Calculation
Calculating the mean is often the first step in understanding your data set. The mean gives you the average value of the data, and it's quite straightforward to compute. To find the mean, simply sum up all the data points and then divide by the number of data points.
Let's say you have five data points like in Sample 1: 8.3, 7.1, 7.6, 8.1, and 7.6. Add these numbers up to get a total, and then divide by 5, which is the count of data points. This simple calculation provides you with the mean, which serves as a reference point for further analysis, such as standard deviation and coefficient of variation.
Knowing the mean is crucial since it allows you to compare other statistical measures, providing a foundation for deeper analysis.
Data Variability
Data variability quantifies the extent to which data values in a set diverge from the average (mean) value. It's an essential aspect of statistical analysis because it gives insights into the reliability and consistency of your data.
High variability implies that the data points are widely spread out, and this could signify inconsistency or diverse conditions affecting the observation. On the other hand, low variability suggests that data points are closely clustered around the mean, indicating stability and consistency.
Understanding variability is important when comparing data from different samples or conditions, as it contextualizes the results. For instance, even if two data sets have the same mean, their variability could offer starkly different stories.
Statistical Analysis
Statistical analysis involves using mathematical methods to make sense of data. This includes summarizing the data, identifying patterns or trends, and making conclusions based on the data. Key elements of statistical analysis include measures of central tendency, like the mean, and measures of dispersion, such as variance and standard deviation.
One useful tool in statistical analysis is the coefficient of variation (CV). It is calculated by dividing the standard deviation by the mean and multiplying by 100 to express it as a percentage. The CV helps compare the degree of variability between different samples, even if their means are vastly different.
By using concepts like these, statistical analysis enables researchers to derive meaningful insights, test hypotheses, and make decisions based on data. It's a vital skill in many fields, from biology and economics to engineering and social sciences.

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Most popular questions from this chapter

Fiber content (in grams per serving) and sugar content (in grams per serving) for 18 high fiber cereals (www.consumerreports.com) are shown below. Fiber Content \(\begin{array}{rrrrrrrrr}7 & 10 & 10 & 7 & 8 & 7 & 12 & 12 & 8 \\ 13 & 10 & 8 & 12 & 7 & 14 & 7 & 8 & 8\end{array}\) Sugar Content \(\begin{array}{llllllll}11 & 6 & 14 & 13 & 0 & 18 & 9 & 10\end{array}\) $$ \begin{array}{rrrrrrrrr} 11 & 6 & 14 & 15 & 0 & 18 & 9 & 10 \\ 6 & 10 & 17 & 10 & 10 & 0 & 9 & 5 & 11 \end{array} $$ a. Find the median, quartiles, and interquartile range for the fiber content data set. b. Find the median, quartiles, and interquartile range for the sugar content data set. C. Are there any outliers in the sugar content data set? d. Explain why the minimum value for the fiber content data set and the lower quartile for the fiber content data set are equal. e. Construct a comparative boxplot and use it to comment on the differences and similarities in the fiber and sugar distributions.

Give two sets of five numbers that have the same mean but different standard deviations, and give two sets of five numbers that have the same standard deviation but different means.

Houses in California are expensive, especially on the Central Coast where the air is clear, the ocean is blue, and the scenery is stunning. The median home price in San Luis Obispo County reached a new high in July 2004, soaring to \(\$ 452,272\) from \(\$ 387,120\) in March 2004\. (San Luis Obispo Tribune, April 28, 2004). The article included two quotes from people attempting to explain why the median price had increased. Richard Watkins, chairman of the Central Coast Regional Multiple Listing Services was quoted as saying, "There have been some fairly expensive houses selling, which pulls the median up." Robert Kleinhenz, deputy chief economist for the California Association of Realtors explained the volatility of house prices by stating: "Fewer sales means a relatively small number of very high or very low home prices can more easily skew medians." Are either of these statements correct? For each statement that is incorrect, explain why it is incorrect and propose a new wording that would correct any errors in the statement.

The paper "Answer Changing on MultipleChoice Tests" ( Journal of Experimental Education \([1980]: 18-21)\) reported that for a group of 162 college students, the average number of responses changed from the correct answer to an incorrect answer on a test containing 80 multiple-choice items was 1.4. The corresponding standard deviation was reported to be 1.5 . Based on this mean and standard deviation, what can you tell about the shape of the distribution of the variable number of answers changed from right to wrong? What can you say about the number of students who changed at least six answers from correct to incorrect?

The accompanying data on number of minutes used for cell phone calls in one month was generated to be consistent with summary statistics published in a report of a marketing study of San Diego residents (TeleTruth, March 2009 ): $$ \begin{array}{rrrrrrrrrr} 189 & 0 & 189 & 177 & 106 & 201 & 0 & 212 & 0 & 306 \\ 0 & 0 & 59 & 224 & 0 & 189 & 142 & 83 & 71 & 165 \\ 236 & 0 & 142 & 236 & 130 & & & & & \end{array} $$ a. Would you recommend the mean or the median as a measure of center for this data set? Give a brief explanation of your choice. (Hint: It may help to look at a graphical display of the data.) b. Compute a trimmed mean by deleting the three smallest observations and the three largest observations in the data set and then averaging the remaining 19 observations. What is the trimming percentage for this trimmed mean? c. What trimming percentage would you need to use in order to delete all of the 0 minute values from the data set? Would you recommend a trimmed mean with this trimming percentage? Explain why or why not.

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