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Chapter 8: Problem 16

In the season of \(2004-05,\) the football team of University of South California had the following score differences for its 13 games played. $$\begin{array}{lllllllll}11 & 49 & 32 & 3 & 6 & 38 & 38 & 30 & 8 & 40 & 31 & 5 & 36\end{array}$$ Find (a) the mean score differences: (b) the median score difference.

Short Answer

Expert verified
The mean score difference is 29 and the median score difference is 30.

Step by step solution

01

Calculate the Mean

To calculate the mean, add up all the score differences and then divide by the number of games played. Therefore, the formula for the mean can be expressed as: \[ Mean = \frac{{\Sigma Score Differences}}{{Number of Games}} \]. From the exercise, the total number of games is 13 and the score differences are: 11, 49, 32, 3, 6, 38, 38, 30, 8, 40, 31, 5, 36. Summing up these values gives a total of 377. Thus the mean is \[ Mean = \frac{{377}}{{13}} = 29 \]. The mean score difference is 29.
02

Calculate the Median

The median is the middle value in an ordered set of data. First arrange the score differences in increasing order, which results in: 3, 5, 6, 8, 11, 30, 31, 32, 36, 38, 38, 40, 49. With 13 scores, the median will be the 7th score since it separates the lower and higher half. Therefore, the median score difference is 30.

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Key Concepts

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

Mean Calculation
To find the mean of a set of numbers, we need to sum up all the values and then divide by the number of data points. In the context of the given exercise, where the University of South California's football team's score differences are listed, the mean can help us understand the average score difference across all games.

Here's how you calculate it step-by-step:
  • Add all score differences: 11 + 49 + 32 + 3 + 6 + 38 + 38 + 30 + 8 + 40 + 31 + 5 + 36 = 377
  • Count the number of games played, which is 13 in this case.
  • Divide the total by the number of games: \(Mean = \frac{377}{13} = 29\).
Thus, the mean score difference is 29. This means that, on average, the team had a score difference of 29 points in each game. Analyzing the mean provides us with a central tendency measurement that summarizes the data with a single value.
Median Calculation
The median is another measure of central tendency but focuses on the middle value. It's particularly useful because it is not affected by extreme values, unlike the mean. In the exercise, the task is to find the middle score difference.

To calculate the median, we follow these steps:
  • Order the score differences in ascending order: 3, 5, 6, 8, 11, 30, 31, 32, 36, 38, 38, 40, 49.
  • The dataset contains 13 numbers, which is an odd number, so the median will be the value in the middle position. This means the 7th number in this ordered list.
  • The 7th score is 31, so the median score difference is 31.
The median helps by providing the central score difference that effectively divides the dataset into two equal halves, highlighting the middle value without being influenced by outliers.
Data Analysis
Data analysis refers to the process of inspecting, cleaning, and modeling data to gain insights and support decision-making. In the realm of statistics and sports, analyzing score differences can reveal patterns about a team's performance.

Here are some points to consider:
  • Assessing the mean and median helps us understand both the average performance and whether there is any skewness due to outliers.
  • Suppose a team consistently has wide score differences like these. In that case, it might indicate variability in performance or variability in the quality of opponents.
  • It can be useful to look at other statistics, such as the mode, range, and standard deviation, to understand the distribution of score differences better.
Overall, data analysis in this context can lead to actionable insights, such as identifying patterns that can be used for future game preparations.
Statistical Methods
Statistical methods involve various techniques and tools to analyze data. These methods can be applied to various fields, including sports, to derive meaningful information. In our exercise, we focus on simple methods like the mean and median.

Here's how they fit into the broader scope of statistical methods:
  • Descriptive Statistics: These include calculating the mean and median, which summarize and describe the characteristics of a dataset.
  • Distribution Analysis: Understanding how data is spread. It includes determining the range, variance, and standard deviation.
  • Data Visualization: Often involves graphs or charts that effectively display the data trends and distributions.
Using these methods aids in interpreting data correctly, enhancing our ability to make evidence-based decisions. It's crucial in sports analytics to gauge team and player performance through comprehensive statistical analysis.

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Most popular questions from this chapter

Consider the following measurements of the heat producing capacity of the coal produced by two mines (in millions of calories per ton): $$\begin{array}{lllllll}\text { Mine } & 1: & 8260 & 8130 & 8350 & 8070 & 8340 & \\ \text { Mine 2: } & 7950 & 7890 & 7900 & 8140 & 7920 & 7840\end{array}$$ Can it be concluded that the two population variances are equal?

If the standard deviation of the mean for the sampling distribution of random samples of size 36 from a large or infinite population is \(2,\) how large must the size of the sample become if the standard deviation is to be reduced to \(1.2 ?\)

Two different box-filling machines are used to fill cereal boxes on the assembly line. The critical measurement influenced by these machines is the weight of the product in the machines. Engineers are quite certain that the variance of the weight of product is \(a^{2}=1\) ounce. Experiments are conducted using both machines with sample sizes of 36 each. The sample averages for machine \(A\) and \(B\) are \(X A=4.5\) ounces and \(x_{B}=4.7\) ounces. Engineers seemed surprised that the two sample averages for the filling machines were so different (a) Use the central limit theorem to determine $$P\left(X_{B}-X_{A} \geq 0.2\right)$$ under the condition that \(\mu_{A}=\mu_{B}\). (b) Do the aforementioned experiments seem to, in any way, strongly support a conjecture that the two population means for the two machines are different? Explain using your answer in (a).

The concentration of an active ingredient in the output of a chemical reaction is strongly influenced by the catalyst that is used in the reaction. It is felt that when catalyst \(A\) is used, the population mean concentration exceeds \(65 \%\). The standard deviation is known to be \(a=5 \%\) A sample of outputs from 30 independent experiments gives the average concentration of \(x_{A}=64.5 \%\). (a) Does this sample information with an average concentration of \(\bar{x}_{A}=64.5 \%\) provide disturbing information that perhaps \(\mu_{A}\) is not \(65 \%,\) but less than \(65 \% ?\) Support your answer with a probability statement. (b) Suppose a similar experiment is done with the use of another catalyst, catalyst \(B\). The standard deviation \(a\) is still assumed to be \(5 \%\) and \(\bar{x}_{B}\) turns out to be \(70 \%\). Comment on whether or not the sample information on catalyst \(B\) seems to give strong information that suggests that \(\mu_{B}\) is truly greater than \(\mu_{A}\). Support your answer by computing $$P\left(\bar{X}_{B}-X_{A} \geq 5.5 \quad \mid \mu_{B}=\mu_{A}\right)$$. (c) Under the condition that \(\mu_{A}=\mu_{B}=65 \%,\) give the approximate distribution of the following quantities (with mean and variance of each). Make use of the central limit theorem.

(a) Find \(t_{0.025}\) when \(v=14\). (b) Find \(-t_{0.10}\) when \(v=10\). (c) Find \(t_{0.995}\) when \(v=7\).
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