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For Exercises 27 to 30, check whether each of the conditions is met for calculating a confidence interval for the population proportion p. AIDS and risk factors In the National AIDS Behavioral Surveys sample of 2673 adult heterosexuals, 0.2% had both received a blood transfusion and had a sexual partner from a group at high risk of AIDS. We want to estimate the proportion p in the population who share these two risk factors.

Short Answer

Expert verified
Both conditions for calculating the confidence interval are met.

Step by step solution

01

Identify the sample proportion

Calculate the sample proportion \(\hat{p}\). Given that 0.2% of the sample of 2673 individuals have both risk factors, we compute \(\hat{p} = \frac{0.2}{100} = 0.002\).
02

Calculate the number of individuals with risk factors

Determine the number of individuals in the sample who have both risk factors. This is given by \(2673 \times 0.002 = 5.346\). Round this to the nearest whole number, so approximately 5 individuals.
03

Check the sample size condition

For a confidence interval for a population proportion to be valid, the sample size \(n\) should satisfy the condition that both \(n\hat{p} \geq 5\) and \(n(1-\hat{p}) \geq 5\). In this case, we have \(n\hat{p} = 5\) and \(n(1-\hat{p}) = 2668\), both conditions are met.
04

Justify validation conditions

Since both conditions for \(n\hat{p}\) and \(n(1-\hat{p})\) are satisfied and \(n\) is large, the sampling distribution of \(\hat{p}\) is approximately normal, allowing for the calculation of a confidence interval.

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Key Concepts

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

Sample Proportion
Calculating and understanding the sample proportion is the foundation of creating a confidence interval for a population proportion. When we're given data, like the fact that 0.2% of 2673 individuals have both risk factors for AIDS, we use it to find the sample proportion, represented as \( \hat{p} \). This is found by dividing the percentage by 100, so 0.2% becomes 0.002. Think of this as the fraction of the sample that exhibits the characteristic we're interested in. Remember:
  • The sample proportion \( \hat{p} \) tells us about the sample, not the entire population.
  • A higher sample size often gives a more reliable estimate of the true population proportion.
Getting the sample proportion right is essential because it is the basis upon which further calculations are built.
Population Proportion
The population proportion, denoted by \( p \), represents the true proportion of the entire group we're interested in. In this example, it would be the actual percentage of all adult heterosexuals who have both risk factors for AIDS. In practice, we rarely know \( p \) precisely. Instead, we estimate it using our sample proportion \( \hat{p} \). It’s like trying to guess the taste of an entire dish by sampling just a spoonful. The key here is the concept of estimation – we use sample data to make educated guesses about the larger population. This is crucial for situations where directly gathering data from the whole population is impractical.
Normal Approximation
Normal approximation plays a vital role in finding confidence intervals. This involves assuming that the distribution of the sample proportion follows a normal distribution. This assumption allows us to use statistical techniques to calculate confidence intervals, which estimate where the true population proportion might lie. For a normal approximation to hold, our sample size should be large enough, and it should meet specific conditions (you’ll see how these are checked in the Sample Size Condition section). Put simply, the concept of normal approximation simplifies calculations by allowing us to use the properties of the normal distribution. This makes it much easier to make accurate estimations about the population.
Sample Size Condition
Ensuring that your sample size is adequate is critical for making valid statistical inferences. The sample size condition demands that both \( n\hat{p} \geq 5 \) and \( n(1-\hat{p}) \geq 5 \). These conditions ensure that the sample proportion approximately follows a normal distribution. In our example:
  • \( n\hat{p} = 5 \) confirms that the number of individuals with the characteristic is sufficient.
  • \( n(1-\hat{p}) = 2668 \) confirms that the number of individuals without the characteristic is also sufficient.
Meeting these conditions gives us confidence that we can reasonably approximate the population's characteristics based on our sample. It is a way to check that our sample is robust enough to provide reliable insights into the population, making the analysis outcomes meaningful and trustworthy.

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Most popular questions from this chapter

Multiple choice: Select the best answer for Exercises 21 to 24. In a poll, I. Some people refused to answer questions. II. People without telephones could not be in the sample. III. Some people never answered the phone in several calls. Which of these sources is included in the \(\pm 2 \%\) margin of error announced for the poll? (a) I only (c) III only (e) None of these (b) II only (d) I, II, and III

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How common is SAT coaching? A random sample of students who took the SAT college entrance examination twice found that 427 of the respondents had paid for coaching courses and that the remaining 2733 had not.\(^{14}\) Construct and interpret a 99\(\%\) confidence interval for the proportion of coaching among students who retake the SAT. Follow the four-step process.

Gambling and the NCAA Gambling is an issue of great concern to those involved in college athletics. Because of this concern, the National Collegiate Athletic Association (NCAA) surveyed randomly selected student-athletes concerning their gambling-related behaviors. \(^{17}\) Of the 5594 Division I male athletes in the survey, 3547 reported participation in some gambling behavior. This includes playing cards, betting on games of skill, buying lottery tickets, betting on sports, and similar activities. A report of this study cited a 1% margin of error. (a) The confidence level was not stated in the report. Use what you have learned to find the confidence level, assuming that the NCAA took an SRS. (b) The study was designed to protect the anonymity of the student-athletes who responded. As a result, it was not possible to calculate the number of students who were asked to respond but did not. How does this fact affect the way that you interpret the results?

Explaining confidence \(A 95 \%\) confidence interval for the mean body mass index (BMI) of young American women is \(26.8 \pm 0.6 .\) Discuss whether each of the following explanations is correct. (a) We are confident that 95% of all young women have BMI between 26.2 and 27.4. (b) We are 95% confident that future samples of young women will have mean BMI between 26.2 and 27.4. (c) Any value from 26.2 to 27.4 is believable as the true mean BMI of young American women. (d) In 95% of all possible samples, the population mean BMI will be between 26.2 and 27.4. (e) The mean BMI of young American women cannot be 28.

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