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Percent of Free Throws Made Usually, in sports, we expect top athletes to get better over time. We expect future athletes to run faster, jump higher, throw farther. One thing has remained remarkably constant, however. The percent of free throws made by basketball players has stayed almost exactly the same for 50 years. \(^{5}\) For college basketball players, the percent is about \(69 \%,\) while for players in the NBA (National Basketball Association) it is about \(75 \%\). (The percent in each group is also very similar between male and female basketball players.) In each case below, find the mean and standard deviation of the distribution of sample proportions of free throws made if we take random samples of the given size. (a) Samples of 100 free throw shots in college basketball (b) Samples of 1000 free throw shots in college basketball (c) Samples of 100 free throw shots in the \(\mathrm{NBA}\) (d) Samples of 1000 free throw shots in the \(\mathrm{NBA}\)

Short Answer

Expert verified
(a) Mean and standard deviation of 100 free throws in college basketball are 0.69 and 0.04620 respectively. (b) Mean and standard deviation of 1000 free throws in college basketball are 0.69 and 0.01462 respectively. (c) Mean and standard deviation of 100 free throws in the NBA are 0.75 and 0.043301 respectively. (d) Mean and standard deviation of 1000 free throws in the NBA are 0.75 and 0.013693 respectively.

Step by step solution

01

Calculation of mean and standard deviation for college basketball with 100 free throws

Using given % of successful throws in college basketball, mean = \(p = 0.69\). Now, using the formula for standard deviation, \(\sigma = \sqrt{p(1-p)/n} = \sqrt{0.69(1-0.69)/100} = 0.04620\)
02

Calculation of mean and standard deviation for college basketball with 1000 free throws

Using the same percentage of successful throws in college basketball, mean = \(p = 0.69\). Now, the standard deviation is calculated as \(\sigma = \sqrt{p(1-p)/n} = \sqrt{0.69(1-0.69)/1000} = 0.01462\)
03

Calculation of mean and standard deviation for NBA with 100 free throws

Using given % of successful throws in NBA, mean = \(p = 0.75\). Now, the standard deviation is calculated as \(\sigma = \sqrt{p(1-p)/n} = \sqrt{0.75(1-0.75)/100} = 0.043301\)
04

Calculation of mean and standard deviation for NBA with 1000 free throws

Using the same percentage of successful throws in NBA, mean = \(p = 0.75\). Now, the standard deviation is calculated as \(\sigma = \sqrt{p(1-p)/n} = \sqrt{0.75(1-0.75)/1000} = 0.013693\)

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Key Concepts

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

Sample Proportions
In statistics, sample proportions are used to estimate population parameters. When we talk about the proportion of successful free throws, we're discussing how often players make their throws within a sample size. For example, if we have a group of 100 shots, and 69 are successful in college basketball, the sample proportion is 0.69, or 69%.

Sample proportions are easy to calculate. Simply divide the number of successful outcomes by the total number of trials. In our case, it’s the successful throws divided by the total throws. Understanding this concept helps us make generalizations about larger groups based on the behavior of smaller, manageable groups.

When we compare different sample sizes, we might notice changes in variability and accuracy. A larger sample size tends to better represent the true proportion of the whole group, as demonstrated by comparing samples of 100 to those of 1000 in basketball. It's crucial in making these estimations as precise as possible.
Standard Deviation
Standard deviation is a measure that tells us how much variation exists from the mean or expected value. When dealing with proportions, like the percentage of successful free throws, standard deviation helps us understand how consistent performance is across multiple trials or samples.

The formula for computing standard deviation in this context is \( \sigma = \sqrt{p(1-p)/n} \), where \( p \) is the sample proportion and \( n \) is the number of trials in the sample. This formula gives us insight into how much a player’s performance might fluctuate within any given number of shots.

For smaller sample sizes, such as 100 throws, standard deviation is larger, indicating more variability. As we increase the sample size to 1000, standard deviation decreases, which suggests performance becomes more consistent—aligning closer to the true mean. This concept is vital for interpreting how sample sizes impact data stability and reliability.
Mean Calculation
The mean in statistics is often referred to as the average. It serves as a central point around which the data is distributed. In the context of basketball free throws, the mean proportion tells us the typical success rate expected, based on historical data or sample observations.

To find the mean of the sample proportion, we assume it equals the recorded proportion of successful shots. For example, in college basketball, with a 69% success rate, the mean of our sample proportion is simply \( p = 0.69 \). For NBA players, this mean is \( p = 0.75 \).

The mean is a critical component when comparing different datasets because it provides a benchmark to evaluate if a player's performance is improving, declining, or staying consistent over time. Understanding how to calculate and interpret the mean helps us understand broader trends in athletic performance, making informed decisions based on statistical data.

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Most popular questions from this chapter

A young statistics professor decided to give a quiz in class every week. He was not sure if the quiz should occur at the beginning of class when the students are fresh or at the end of class when they've gotten warmed up with some statistical thinking. Since he was teaching two sections of the same course that performed equally well on past quizzes, he decided to do an experiment. He randomly chose the first class to take the quiz during the second half of the class period (Late) and the other class took the same quiz at the beginning of their hour (Early). He put all of the grades into a data table and ran an analysis to give the results shown below. Use the information from the computer output to give the details of a test to see whether the mean grade depends on the timing of the quiz. (You should not do any computations. State the hypotheses based on the output, read the p-value off the output, and state the conclusion in context.) $$ \begin{aligned} &\text { Two-Sample T-Test and Cl }\\\ &\begin{array}{lrrrr} \text { Sample } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \text { Late } & 32 & 22.56 & 5.13 & 0.91 \\ \text { Early } & 30 & 19.73 & 6.61 & 1.2 \end{array} \end{aligned} $$ Difference \(=\mathrm{mu}\) (Late) \(-\mathrm{mu}\) (Early) Estimate for difference: 2.83 \(95 \%\) Cl for difference: (-0.20,5.86) T-Test of difference \(=0\) (vs not \(=\) ): T-Value \(=1.87\) P-Value \(=0.066 \quad \mathrm{DF}=54\)

Standard Error from a Formula and a Bootstrap Distribution In Exercises 6.19 to \(6.22,\) use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of lie detector trials in which the technology misses a lie, with \(n=48\) and \(\hat{p}=0.354\)

IQ tests scale the scores so that the mean IQ score is \(\mu=100\) and standard deviation is \(\sigma=15\). Suppose that 30 fourth graders in one class are given such an IQ test that is appropriate for their grade level. If the students are really a random sample of all fourth graders, what is the chance that the average IQ score for the class is above \(105 ?\)

Involve scores from the high school graduating class of 2010 on the SAT (Scholastic Aptitude Test). The distribution of sample means \(\bar{x}_{N}-\bar{x}_{E},\) where \(\bar{x}_{N}\) represents the mean Mathematics score for a sample of 100 people for whom the native language is not English and \(\bar{x}_{E}\) represents the mean Mathematics score for a sample of 100 people whose native language is English, is centered at 10 with a standard deviation of 17.41 . Give notation and define the quantity we are estimating with these sample differences. In the population of all students taking the test, who scored higher on average, non-native English speakers or native English speakers? Standard Error from a Formula and a Bootstrap Distribution In Exercises 6.224 and \(6.225,\) use StatKey or other technology to generate a bootstrap distribution of sample differences in means and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample standard deviations as estimates of the population standard deviations.

Standard Error from a Formula and a Bootstrap Distribution In Exercises 6.19 to \(6.22,\) use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of survey respondents who say exercise is important, with \(n=1000\) and \(\hat{p}=0.753\)

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