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Random variability in baseball A baseball player in the major leagues who plays regularly will have about 500 atbats (that is, about 500 times he can be the hitter in a game) during a season. Suppose a player has a 0.300 probability of getting a hit in an at-bat. His batting average at the end of the season is the number of hits divided by the number of at-bats. This batting average is a sample proportion, so it has a sampling distribution describing where it is likely to fall. a. Describe the shape, mean, and standard deviation of the sampling distribution of the player's batting average after a season of 500 at-bats. b. Explain why a batting average of 0.320 or of 0.280 would not be especially unusual for this player's yearend batting average. (That is, you should not conclude that someone with a batting average of 0.320 one year is necessarily a better hitter than a player with a batting average of \(0.280 .)\)

Short Answer

Expert verified
The sampling distribution is approximately normal with mean 0.300 and standard deviation 0.020. Batting averages of 0.280 and 0.320 are within one standard deviation from the mean, thus not unusual.

Step by step solution

01

Understand the Sampling Distribution Concept

When we talk about the sampling distribution of a player's batting average, we are considering the distribution of the proportion of hits over many seasons, assuming the player has 500 at-bats each season. This distribution can be described by the Central Limit Theorem, which implies the sampling distribution of the sample proportion is approximately normal if the sample size is large enough, as it is in this case.
02

Calculate the Mean of the Sampling Distribution

The mean of the sampling distribution is the probability of success in obtaining a hit in a single at-bat, which is given as 0.300. Thus, the mean (or expected value) of the player's batting average after 500 at-bats is 0.300.
03

Calculate the Standard Deviation of the Sampling Distribution

The standard deviation (also called the standard error) of the sampling distribution of a sample proportion is calculated using the formula \( \sqrt{\frac{p(1-p)}{n}} \), where \( p \) is the probability of obtaining a hit (0.300) and \( n \) is the number of trials (500). So, the standard deviation is \( \sqrt{\frac{0.300 \times 0.700}{500}} \approx 0.020 \).
04

Evaluate Why Batting Averages of 0.280 or 0.320 Are Not Unusual

To understand if 0.280 or 0.320 are unusual batting averages, we can look at how these values relate to the expected mean and standard deviation. An average of 0.280 is \( \frac{0.280 - 0.300}{0.020} \approx -1 \) standard deviations away from the mean, and an average of 0.320 is \( \frac{0.320 - 0.300}{0.020} \approx 1 \) standard deviations away from the mean. Since both these values lie within one standard deviation from the mean in a normal distribution, they are not considered unusual outcomes.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental concept in statistics. It tells us that when we take a large number of random samples from a population, like our baseball player's at-bats, the sampling distribution of the sample means will be approximately normally distributed, regardless of the original population's distribution. In the context of our baseball player, since we have 500 at-bats, this is considered a sufficiently large sample size.

Thanks to the CLT, we know that the player's batting average over a season will resemble a bell curve or normal distribution. This understanding helps us predict where most batting averages will fall and identify what might be considered normal or unusual for the player's performance, such as getting averages close to 0.300.
Sample Proportion
The sample proportion is similar to the batting average in baseball. It is the ratio of successful outcomes to the total number of trials. In our player's scenario, the sample proportion is the player's batting average, calculated as the number of hits divided by the number of at-bats in a season.

This proportion tells us the likelihood of the player getting a hit. In our exercise, we consider his season-long batting average as a sample proportion from the entire population of all potential at-bat outcomes.
  • If a player has a 0.300 probability of hitting, his sample proportion is expected to be 0.300 over a large number of at-bats.
The sample proportion is not just a single number but is supported by the sampling distribution which provides information on variability.
Standard Deviation
Standard deviation here refers to the spread of the player's batting average around the mean across many seasons. It offers insight into how much the batting averages vary when a season's 500 at-bats are repeatedly considered.

In statistical terms, the formula for the standard deviation of a sample proportion is \(\sqrt{\frac{p(1-p)}{n}} \) where \(p\) is the probability of success (0.300 in our context) and \(n\) is the number of at-bats (500).
  • Using the formula gives us a standard deviation (or standard error) of approximately 0.020.
  • This standard deviation allows us to understand how much the actual season batting average might typically deviate from our expected average of 0.300.
Having this value is particularly useful when evaluating unusual batting averages.
Batting Average
A batting average represents a baseball player's ability to hit the ball successfully. Specifically, it is calculated as the number of hits divided by the number of at-bats. In statistical terms, that makes it a sample proportion. The importance of batting average is especially seen in its use for comparing players and understanding performance trends; however, it comes with a random variability.

For example, the relationship between the average and standard deviation tells us whether some batting averages are typical or unusual. A player's batting average might vary from season to season due to natural variation, even if their batting skill remains constant.
  • In a normal distribution, batting averages within one standard deviation of the mean (e.g., 0.280 to 0.320 in our case) are typically expected and not unusual.
  • This understanding prevents judging a player solely on a single season's performance without considering natural statistical fluctuations.
Thus, baseball fans and statisticians alike need to interpret batting averages with context in mind.

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Most popular questions from this chapter

Purpose of sampling distribution You'd like to estimate the proportion of all students in your school who are fluent in more than one language. You poll a random sample of 50 students and get a sample proportion of 0.12. Explain why the standard deviation of the sampling distribution of the sample proportion gives you useful information to help gauge how close this sample proportion is to the unknown population proportion.

Finite populations The formula \(\sigma / \sqrt{n}\) for the standard deviation of \(\bar{x}\) actually is an approximation that treats the population size as infinitely large relative to the sample size \(n\). The exact formula for a finite population size \(N\) is Standard deviation \(=\sqrt{\frac{N-n}{N-1}} \frac{\sigma}{\sqrt{n}}\) The term \(\sqrt{(N-n) /(N-1)}\) is called the finite population correction. a. When \(n=300\) students are selected from a college student body of size \(\mathrm{N}=30,000\), show that the standard deviation equals \(0.995 \sigma / \sqrt{n}\). (When \(n\) is small compared to the population size \(\mathrm{N},\) the approximate formula works very well.) b. If \(n=\mathrm{N}\) (that is, we sample the entire population), show that the standard deviation equals \(0 .\) In other words, no sampling error occurs, since \(\bar{x}=\mu\) in that case.

Relative frequency of heads Construct the sampling distribution of the sample proportion of heads obtained in the experiment of flipping a balanced coin: a. Once. (Hint: The possible sample proportion values are \(0,\) if the flip is a tail, and \(1,\) if the flip is a head. What are their probabilities?) b. Twice. (Hint: The possible samples are \(\\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH},\) \(\mathrm{TT}\\}\), so the possible sample proportion values are 0 , \(0.50,\) and \(1.0,\) corresponding to \(0,1,\) or 2 heads. What are their probabilities?) c. Three times. (Hint: There are 8 possible samples.) d. Refer to parts a-c. Sketch the sampling distributions and describe how the shape is changing as the number of flips \(\mathrm{n}\) increases.

Exit poll CNN conducted an exit poll of 1751 voters in the 2010 Senatorial election in New York between Charles Schumer and Jay Townsend. It is possible that all 1751 voters sampled happened to be Charles Schumer supporters. Investigate how surprising this would be, if actually \(65 \%\) of the population voted for Schumer, by a. Finding the probability that all 1751 people voted for Schumer. (Hint: Use the binomial distribution.) b. Finding the number of standard deviations that a sample proportion of 1.0 for 1751 voters falls from the population proportion of \(0.65 .\)

What good is a standard deviation? Explain how the standard deviation of the sampling distribution of a sample proportion gives you useful information to help gauge how close a sample proportion falls to the unknown population proportion.

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