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Chapter 1: Problem 31

In Superbowl XXXVII, Michael Pittman of Tampa Bay rushed (ran with the football) 17 times on first down, and the results were the following gains in yards: \(\begin{array}{lllllllll}23 & 1 & 4 & 1 & 6 & 5 & 9 & 6 & 2 \\ -1 & 3 & 2 & 0 & 2 & 24 & 1 & 1 & \end{array}\) a. Determine the value of the sample mean. b. Determine the value of the sample median. Why is it so different from the mean? c. Calculate a trimmed mean by deleting the smallest and largest observations. What is the corresponding trimming percentage? How does the value of this \(\bar{x}_{r r}\) compare to the mean and median?

Short Answer

Expert verified
a. Mean: 5.18 b. Median: 2 (Mean is higher due to extreme values) c. Trimmed Mean: 4.4; 11.76% trimming

Step by step solution

01

Calculate the Sample Mean

To find the sample mean, sum all the data points and then divide by the number of data points. The data points are:\(23, 1, 4, 1, 6, 5, 9, 6, 2, -1, 3, 2, 0, 2, 24, 1, 1\).Firstly, sum these values: \(23 + 1 + 4 + 1 + 6 + 5 + 9 + 6 + 2 - 1 + 3 + 2 + 0 + 2 + 24 + 1 + 1 = 88\).Next, divide the total by the number of observations (17):\(\text{Mean} = \frac{88}{17} \approx 5.18\).
02

Find the Sample Median

To find the median, first list the numbers in ascending order:\(-1, 0, 1, 1, 1, 1, 2, 2, 2, 3, 4, 5, 6, 6, 9, 23, 24\).With 17 observations, the median is the 9th value:Thus, the Median is 2.
03

Discuss Difference Between Mean and Median

The mean (\(5.18\)) is larger than the median (2) because the mean is affected by extreme values, such as 24 and 23, whereas the median is not. The sizeable values raise the mean, while the median remains centrally located among sorted values.
04

Calculate Trimmed Mean

To find the trimmed mean, remove the smallest and largest data points: -1 and 24.Remaining observations: \(0, 1, 1, 1, 1, 2, 2, 2, 3, 4, 5, 6, 6, 9, 23\).Sum these:\(0 + 1 + 1 + 1 + 1 + 2 + 2 + 2 + 3 + 4 + 5 + 6 + 6 + 9 + 23 = 66\).Total observations now: 15 (since 2 are removed).Trimmed mean: \(\text{Trimmed Mean} = \frac{66}{15} = 4.4\).The trimming percentage is \(\frac{2}{17} \times 100\% \approx 11.76\%\).The trimmed mean (4.4) is less than the mean and closer to the median due to reduced influence of extreme values.

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Key Concepts

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

Sample Mean
The sample mean, often referred to as the arithmetic mean, is a measure that sums up all the data values and divides by the count of values. This gives us the average value in the data set. In this exercise, Michael Pittman's 17 rushing attempts resulted in yard gains that we summed (88 yards) and divided by the number of attempts (17). Thus, the sample mean is calculated as: \[\text{Sample Mean} = \frac{88}{17} \approx 5.18\]
  • The mean provides a central value that represents the data.
  • It is sensitive to all data points, especially outliers.
The sample mean of 5.18 yards suggests that, on average, Pittman gained approximately 5.18 yards per rush on first down during the game.
Sample Median
The sample median is the middle value of a data set arranged in ascending order. It divides the data into two equal halves, offering a measure of central tendency that is less affected by outliers. In Pittman鈥檚 data, when arranged from lowest to highest value, the 9th value in the sequence is 2. Thus, the median is:
  • Median: 2
  • It is less influenced by extreme values compared to the mean.
  • Represents a central point that is not skewed by anomalies.
This difference with the sample mean demonstrates how one outstanding or low value can shift the mean, while the median stays firmly representative of the majority of the data.
Trimmed Mean
The trimmed mean is a method to calculate an average that reduces the influence of extreme values or outliers by excluding a fixed percentage of the smallest and largest data points. By removing these outliers, we can achieve a more robust mean that reflects the central tendency without the skewing effect of extremes.In Pittman's case:
  • Remove the smallest (-1) and largest (24) values.
  • The remaining values were summed to get 66, with 15 observations left.
  • Trimmed Mean: \( \frac{66}{15} = 4.4 \)
This 11.76% trimming yields a mean (\(4.4\)) that lies closer to the median and less than the untrimmed sample mean, highlighting its immunity to outliers.
Outliers Impact
Outliers are data points significantly distant from others in the dataset. They can skew certain measures of central tendency like the mean but have negligible effect on the median. Examining Pittman's rushing yards, gains like -1 and 24 clearly stand out as outliers.
  • Outliers inflate or deflate the sample mean, making it an unreliable measure for central tendency if present.
  • The presence of 23 and 24 increased the calculated mean from potential values around the median.
  • Modern statistical analysis often involves methods like trimming or using the median to manage outlier impacts.
Understanding outliers is crucial in descriptive statistics to ensure comprehensive data analysis and interpretation.
Descriptive Statistics
Descriptive statistics aim to summarize or describe characteristics of a dataset. They offer insights into the dataset's main features through numerical or graphical representations. Key tools include:
  • Mean: Provides a measure of central tendency, affected by outliers.
  • Median: Centre point of the dataset, resilient against outliers.
  • Range: Difference between the maximum and minimum values, highlighting data spread.
  • Variance and Standard Deviation: Measures of data dispersion or variability.
Using descriptive statistics, such as sample mean and median, one can better understand patterns and tendencies in datasets like Pittman's rushing attempts, assisting in identifying strategies or performance evaluations.

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