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Chapter 7: Problem 45

The sampling distribution of \(\hat{p}\) is approximately Normal because (a) there are at least 7500 Division I college athletes. (b) \(n p=225\) and \(n(1-p)=525\) are both at least 10 . (c) a random sample was chosen. (d) the athletes' responses are quantitative. (e) the sampling distribution of \(\hat{p}\) always has this shape.

Short Answer

Expert verified
(b) because it satisfies the normal approximation conditions.

Step by step solution

01

Identify Relevant Information

First, we need to understand what each option is stating. Options \((a)\) mentions a large population size, which is not directly related to the shape of a sampling distribution of \(\hat{p}\). Option \((b)\) mentions conditions \(n p \geq 10\) and \(n(1-p) \geq 10\), which are conditions for using a normal approximation for the binomial distribution. Option \((c)\) emphasizes randomness, a fundamental validity requirement for samples. Option \((d)\) states that the responses are quantitative, which is not relevant since \(\hat{p}\) is a proportion. Option \((e)\) incorrectly suggests that the shape is always normal.
02

Check Conditions for Normality

The conditions \( n p \geq 10 \) and \( n(1-p) \geq 10 \) from option \((b)\) ensure that both the number of successes and failures in the sample are large enough for the Central Limit Theorem to apply. In this case, \( n p = 225 \) and \( n(1-p) = 525 \) satisfy these conditions, making it valid to assume the sampling distribution of \(\hat{p}\) is approximately normal.
03

Eliminate Incorrect Options

Given our understanding, we can eliminate options not directly related to the normality of \(\hat{p}\). Options \((a)\), \((d)\), and \((e)\) can be dismissed as they do not address the relevant sampling distribution conditions. Option \((c)\), while important in general for sampling, focuses more on the method of sample selection rather than the distribution shape.
04

Conclusion

The correct answer is option \((b)\). The conditions \(n p = 225\) and \(n(1-p) = 525\) are both at least 10, satisfying the requirements for the sampling distribution of \(\hat{p}\) to be approximately normal.

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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 is a fundamental concept in statistics. It tells us that when we have a large enough random sample, the sampling distribution of the sample mean (or proportion, like \( \hat{p} \)) will be approximately normal, regardless of the shape of the population distribution. This means that as our sample size increases, the more normal the distribution will look.

Key elements include:
  • The need for a sufficiently large sample size. In practical terms, for a distribution of proportions (like in your example where \( n p = 225 \) and \( n(1-p) = 525 \)), both \( n p \) and \( n(1-p) \) should be at least 10 to apply the theorem effectively.
  • Allows for normal distribution approximations, which are helpful for creating confidence intervals and hypothesis tests.
The beauty of the Central Limit Theorem is in its wide applicability, making it a cornerstone for many statistical methods and providing a pathway to manage complex data with ease.
Normal Approximation
Normal approximation is a technique used when dealing with binomial distributions, especially when we have large sample sizes. It involves using the normal distribution to estimate the probabilities of a binomial distribution.

Consider these facts:
  • It simplifies calculations. Instead of working with a more complex binomial probability, we can use the z-score and normal distribution tables.
  • The approximation is valid under certain conditions. Generally, \( n p \) and \( n(1-p) \) must both be greater than or equal to 10.
Normal approximation is particularly useful in hypothesis testing and when constructing confidence intervals for binomially distributed data, thereby supporting data analysts in making informed decisions.
Random Sampling
Random sampling is a key component of reliable statistical analysis. It refers to the method of selecting a sample from a larger population in such a way that every individual has an equal chance of being chosen.

Here's why it matters:
  • Ensures unbiased representation of the population. This means the sample should accurately reflect the diversity of the whole group.
  • Enables generalizations about the population. Findings from random samples can often be applied to the population with reasonable accuracy.
  • Facilitates the assumption of independent observations, crucial for applying many statistical methods effectively.
Random sampling is foundational for making valid and reliable conclusions in statistics. Without it, statistical results could be skewed or biased.
Binomial Distribution
The binomial distribution is a discrete probability distribution. It models the number of successes in a series of independent experiments, each with the same probability of success. This is essential for understanding situations where there are two possible outcomes, like "success" or "failure."

Consider these characteristics:
  • It requires a fixed number of trials. Each trial can either result in success or failure.
  • The probability of success remains constant across trials.
  • The trials are independent, meaning the outcome of one does not affect another.
The binomial distribution is invaluable in numerous fields, including finance and quality control, where predicting the likelihood of a series of events can drive strategic decisions.

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Most popular questions from this chapter

Cereal A company's cereal boxes advertise 9.65 ounces of cereal. In fact, the amount of cereal in a randomly selected box follows a Normal distribution with mean \(\mu=9.70\) ounces and standard deviation \(\sigma=0.03\) ounces. (a) What is the probability that a randomly selected box of the cereal contains less than 9.65 ounces of cereal? Show your work. (b) Now take an SRS of 5 boxes. What is the probability that the mean amount of cereal \(\bar{x}\) in these boxes is 9.65 ounces or less? Show your work.

Dead battery? A car company has found that the lifetime of its batteries varies from car to car according to a Normal distribution with mean \(\mu=48\) months and standard deviation \(\sigma=8.2\) months. The company installs a new brand of battery on an SRS of 8 cars. (a) If the new brand has the same lifetime distribution as the previous type of battery, describe the sampling distribution of the mean lifetime \(\bar{x}\). (b) The average life of the batteries on these 8 cars turns out to be \(\bar{x}=42.2\) months. Find the probability that the sample mean lifetime is 42.2 months or less if the lifetime distribution is unchanged. What conclusion would you draw?

$$ \begin{array}{lccc} \hline \text { Highest education } & \text { Total population } & \text { In labor force } & \text { Employed } \\ \text { Didn't finish high } & 27,669 & 12,470 & 11,408 \\ \text { school } & & & \\ \begin{array}{c} \text { High school but no } \\ \text { college } \end{array} & 59,860 & 37,834 & 35,857 \\ \begin{array}{c} \text { Less than bachelor's } \\ \text { degree } \end{array} & 47,556 & 34,439 & 32,977 \\ \text { College graduate } & 51,582 & 40,390 & 39,293 \\ \hline \end{array} $$ Unemployment (5.3) If you know that a randomly chosen person 25 years of age or older is a college graduate, what is the probability that he or she is in the labor force? Show your work.

Airport security The Transportation Security Administration (TSA) is responsible for airport safety. On some flights, TSA officers randomly select passengers for an extra security check before boarding. One such flight had 76 passengers -12 in first class and 64 in coach class. TSA officers selected an SRS of 10 passengers for screening. Let \(\hat{p}\) be the proportion of first- class passengers in the sample. (a) Is the \(10 \%\) condition met in this case? Justify your answer. (b) Is the Large Counts condition met in this case? Justify your answer.

Do you go to church? The Gallup Poll asked a random sample of 1785 adults whether they attended church during the past week. Let \(\hat{p}\) be the proportion of people in the sample who attended church. A newspaper report claims that \(40 \%\) of all U.S. adults went to church last week. Suppose this claim is true. (a) What is the mean of the sampling distribution of \(\hat{p} ?\) Why? (b) Find the standard deviation of the sampling distribution of \(\hat{p}\). Check to see if the \(10 \%\) condition is met. (c) Is the sampling distribution of \(\hat{p}\) approximately Normal? Check to see if the Large Counts condition is met. (d) Of the poll respondents, \(44 \%\) said they did attend church last week. Find the probability of obtaining a sample of 1785 adults in which \(44 \%\) or more say they attended church last week if the newspaper report's claim is true. Does this poll give convincing evidence against the claim? Explain.
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