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Chapter 2: Problem 80

In the seasons that followed his 2001 record-breaking season, Barry Bonds hit \(46,45,45,5,\) and 26 homers, respectively (www.espn.com). \(^{14}\) Two boxplots, one of Bond's homers through 2001 , and a second including the years 2002-2006, follow. The statistics used to construct these boxplots are given in the table. $$ \begin{array}{lccccccc} \text { Years } & \text { Min } & a_{1} & \text { Median } & a_{3} & \text { IQR } & \text { Max } & n \\ \hline 2001 & 16 & 25.00 & 34.00 & 41.50 & 16.5 & 73 & 16 \\ 2006 & 5 & 25.00 & 34.00 & 45.00 & 20.0 & 73 & 21 \end{array} $$ a. Calculate the upper fences for both of these boxplots. b. Can you explain why the record number of homers is an outlier in the 2001 boxplot, but not in the 2006 boxplot?

Short Answer

Expert verified
Question: Calculate the upper fences for the 2001 and 2006 boxplots, and explain why the record number of homers (73) is considered an outlier in the 2001 boxplot but not in the 2006 boxplot. Answer: The upper fences for the 2001 and 2006 boxplots are 66.25 and 75, respectively. The record number of homers (73) is an outlier in the 2001 boxplot because it is greater than the Upper Fence (66.25), but it is not considered an outlier in the 2006 boxplot because it is not greater than the Upper Fence (75). This indicates that the distribution of homers changed with the inclusion of the additional years (2002-2006), resulting in a higher range of values and making the record number of homers no longer an outlier in the 2006 dataset.

Step by step solution

01

Calculate the Upper Fences

For the 2001 dataset, we have Q3 = 41.5 and IQR = 16.5. Using the formula for Upper Fence, we get: $$Upper Fence_{2001} = 41.5 + 1.5 * 16.5 = 41.5 + 24.75 = 66.25$$ For the 2006 dataset, we have Q3 = 45 and IQR = 20. Using the formula for Upper Fence, we get: $$Upper Fence_{2006} = 45.00 + 1.5 * 20 = 45.00 + 30 = 75$$ Hence, the Upper Fence for 2001 boxplot is 66.25 and for 2006 boxplot is 75.
02

Determine the outlier

Recall that the record number of homers is 73. Now let's compare the Upper Fences with this value: For the 2001 dataset, 73 is greater than the Upper Fence (66.25), so the record number of homers is considered an outlier in the 2001 boxplot. For the 2006 dataset, 73 is not greater than the Upper Fence (75), so the record number of homers is not considered an outlier in the 2006 boxplot. In conclusion: a. The upper fences for 2001 and 2006 boxplots are 66.25 and 75, respectively. b. The record number of homers (73) is an outlier in the 2001 boxplot because it is greater than the Upper Fence (66.25) but not in the 2006 boxplot because it is not greater than the Upper Fence (75). This means that with the inclusion of the additional years (2002-2006), the distribution of homers has changed, resulting in a higher range of values, and thus, the record number of homers is no longer considered as an outlier in the 2006 dataset.

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Key Concepts

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

Upper Fence Calculation
The upper fence in a boxplot serves as a critical cutoff to determine whether a data point is an outlier or not. Understanding how to calculate it is essential. The formula for the upper fence is given as \(Q3 + 1.5 \times IQR\), where \(Q3\) is the third quartile, and \(IQR\) is the interquartile range.

For the 2001 dataset in our exercise:
\(Upper Fence_{2001} = 41.5 + 1.5 \times 16.5 = 66.25\)
And for the 2006 dataset:
\(Upper Fence_{2006} = 45.00 + 1.5 \times 20 = 75\)

This calculation demonstrates how we determine the boundary for identifying outliers within our dataset; any data point lying above this value can significantly affect the interpretation of our data.
Interpreting Boxplots
A boxplot, or box-and-whisker plot, visually summarizes the distribution of a dataset. Important features of boxplots include the median, quartiles, and potentially any outliers. The box represents the middle 50% of the data, showing the interquartile range, while the 'whiskers' extend to the smallest and largest values within the fences.

When interpreting boxplots, look for the positioning of the median, the spread of the quartiles (which tells us about the variability in the data), and the length of the whiskers. If any data points are plotted outside the 'fences', they are potential outliers which can indicate extremes that may warrant further investigation.
Identification of Outliers
Outliers are data points that fall outside the expected range of a dataset, and they can be easily identified using a boxplot. When the data points exceed the upper fence or fall below the lower fence (which is \(Q1 - 1.5 \times IQR\)), they are considered outliers. In our example with Barry Bonds鈥� homers:

In 2001, the number of homers (73) exceeds the upper fence (66.25), qualifying it as an outlier.
In 2006, however, the number of homers (73) does not exceed the upper fence (75), hence it is not an outlier.

Outliers can influence the mean of a dataset significantly and may indicate a need for further analysis to understand why they are different from the rest of the data.
Interquartile Range (IQR)
The Interquartile Range (IQR) is a measure of statistical dispersion and is calculated as the difference between the third quartile (\(Q3\)) and the first quartile (\(Q1\)) of a dataset. It tells us about the spread of the middle 50% of the data.

In the context of our exercise:
The IQR for 2001 is \(Q3 - Q1 = 41.5 - 25 = 16.5\)
The IQR for 2006 is \(Q3 - Q1 = 45.0 - 25 = 20.0\)

The IQR is crucial for determining the upper and lower fences and thereby identifying outliers. Any significant changes in the IQR can alter the calculation of the fences and potentially change which data points are considered outliers.

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Most popular questions from this chapter

The number of Starbucks coffee shops in 18 cities within 20 miles of the University of California, Riverside is shown in the following table (www.starbucks.com). $$ \begin{array}{rrrrr} 16 & 7 & 2 & 6 & 4 \\ 1 & 7 & 1 & 1 & 1 \\ 3 & 2 & 11 & 1 & \\ 5 & 1 & 4 & 12 & \end{array} $$ a. Find the mean, the median, and the mode. b. Compare the median and the mean. What can you say about the shape of this distribution? c. Draw a dotplot for the data. Does this confirm your conclusion about the shape of the distribution from part b?

The length of time required for an automobile driver to respond to a particular emergency situation was recorded for \(n=10\) drivers. The times (in seconds) were \(.5, .8,1.1, .7, .6,\) .9, .7, .8, .7, .8 a. Scan the data and use the procedure in Section 2.5 to find an approximate value for \(s\). Use this value to check your calculations in part b. b. Calculate the sample mean \(\bar{x}\) and the standard deviation \(s\). Compare with part a.

The number of raisins in each of 14 miniboxes (1/2-ounce size) was counted for a generic brand and for Sunmaid brand raisins. The two data sets are shown here: $$ \begin{array}{llll|llll} &&&{\text { Generic Brand }} &&&& \ {\text { Sunmaid }} \\ \hline 25 & 26 & 25 & 28 & 25 & 29 & 24 & 24 \\ 26 & 28 & 28 & 27 & 28 & 24 & 28 & 22 \\ 26 & 27 & 24 & 25 & 25 & 28 & 30 & 27 \\ 26 & 26 & & & 28 & 24 & & \end{array} $$ a. What are the mean and standard deviation for the generic brand? b. What are the mean and standard deviation for the Sunmaid brand? c. Compare the centers and variabilities of the two brands using the results of parts a and b.

A pharmaceutical company wishes to know whether an experimental drug being tested in its laboratories has any effect on systolic blood pressure. Fifteen randomly selected subjects were given the drug, and their systolic blood pressures (in millimeters) are recorded. $$ \begin{array}{lll} 172 & 148 & 123 \\ 140 & 108 & 152 \\ 123 & 129 & 133 \\ 130 & 137 & 128 \\ 115 & 161 & 142 \end{array} $$ a. Guess the value of \(s\) using the range approximation. b. Calculate \(\bar{x}\) and \(s\) for the 15 blood pressures. c. Find two values, \(a\) and \(b\), such that at least \(75 \%\) of the measurements fall between \(a\) and \(b\).

A strain of longstemmed roses has an approximate normal distribution with a mean stem length of 15 inches and standard deviation of 2.5 inches. a. If one accepts as "long-stemmed roses" only those roses with a stem length greater than 12.5 inches, what percentage of such roses would be unacceptable? b. What percentage of these roses would have a stem length between 12.5 and 20 inches?
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