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In a random sample of 519 judges, it was found that 285 were introverts. (See reference in Problem 11.) (a) Let \(p\) represent the proportion of all judges who are introverts. Find a point estimate for \(p\). (b) Find a \(99 \%\) confidence interval for \(p\). Give a brief interpretation of the meaning of the confidence interval you have found. (c) Do you think the conditions \(n p > 5\) and \(n q > 5\) are satisfied in this problem? Explain why this would be an important consideration.

Short Answer

Expert verified
(a) Point estimate: 0.549. (b) 99% confidence interval: [0.493, 0.605]. (c) Conditions are satisfied; it's important for using normal approximation.

Step by step solution

01

Calculate Point Estimate for p

The point estimate for the proportion \( p \) is given by the formula \( \hat{p} = \frac{x}{n} \), where \( x \) is the number of introverts and \( n \) is the total number of judges sampled. Here, \( x = 285 \) and \( n = 519 \). Substitute these values to find \( \hat{p} \).\[ \hat{p} = \frac{285}{519} \approx 0.549 \]
02

Calculate 99% Confidence Interval

To find the 99% confidence interval, we use the formula:\[ \hat{p} \pm Z \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]The critical value \( Z \) for a 99% confidence level is approximately 2.576. Calculate the standard error:\[ \sqrt{\frac{0.549(1 - 0.549)}{519}} \approx 0.022 \]Now, compute the confidence interval:\[ 0.549 \pm 2.576 \times 0.022 \]This gives the interval:\[ 0.549 \pm 0.056 \]Thus, the 99% confidence interval is:\[ [0.493, 0.605] \].
03

Interpret the Confidence Interval

The 99% confidence interval \([0.493, 0.605]\) means that we are 99% confident that the true proportion of all judges who are introverts falls between 49.3% and 60.5%.
04

Check Conditions for Approximation

To use the normal approximation to the binomial distribution, both conditions \( n \hat{p} > 5 \) and \( n \hat{q} > 5 \) must be satisfied. Here, \( \hat{q} = 1 - \hat{p} = 0.451 \).Calculate \( n \hat{p} \) and \( n \hat{q} \):\[ n \hat{p} = 519 \times 0.549 \approx 285 \]\[ n \hat{q} = 519 \times 0.451 \approx 234 \]Both values are greater than 5, which means the conditions are satisfied. Meeting these conditions allows us to use the normal approximation, which is crucial for accurately estimating confidence intervals.

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Key Concepts

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

Point Estimate
A point estimate is a single value that serves as a best guess for an unknown population parameter. It's used in statistical practice to provide a simple and immediate estimate of what something is likely to be. In the given exercise, we determined the point estimate for the proportion of introverted judges. To find this estimate, we use the formula \( \hat{p} = \frac{x}{n} \), where \( x \) is the number of introverts observed, and \( n \) is the sample size. For the judges, \( x = 285 \) and \( n = 519 \). After calculating, the point estimate \( \hat{p} \) comes out to approximately 0.549. This means about 54.9% of the sampled judges are introverts. While a point estimate gives a handy number to work with, it's important to note that this 'point' of data isn't foolproof and can be subject to sampling variation. Hence, while it suggests that around half are introverts, this on its own doesn't indicate reliability or precision.
Normal Approximation
Normal approximation is a convenient method used when approximating the distribution of a sample statistic. For this exercise, we are dealing with a proportion within a large sample size, which aligns well with using normal approximation. A key assumption is that the distribution of a binomial random variable can be approximated successfully by a normal distribution if certain conditions are met. Specifically, these conditions are \( n \hat{p} > 5 \) and \( n \hat{q} > 5 \), ensuring sufficient sample data at each tail of the distribution.In our exercise:
  • \( n \hat{p} = 285 \)
  • \( n \hat{q} = 234 \)
Both values are significantly greater than 5, thus satisfying the conditions for using normal approximation. This approximation simplifies the computation of confidence intervals, turning binomial distribution calculations, which can be complex, into a more straightforward normal distribution task.
Binomial Distribution
The binomial distribution arises when an experiment meets specific criteria, typically known as a 'Bernoulli trial.' It involves fixed numbers of trials, with only two possible outcomes, often referred to as "success" and "failure." The exercise involving judges focuses on whether a judge is an introvert or not, aligning perfectly with a binomial framework.For a binomial distribution, we're generally interested in the number of "successes" in \( n \) trials. In this context, our "success" is finding a judge who is an introvert. Here, the parameters are:
  • \( n = 519 \)
  • "Success" probability \( p \), represented by the proportion of introverts amongst the judges.
One advantage of the binomial distribution is that when you meet certain circumstances, such as a large enough sample size, it can be approximated by a normal distribution. This makes statistical calculations like confidence intervals more manageable.
Standard Error
The standard error (SE) gives us an idea of the variability or 'spread' of our sample statistic from one sample to the next. It's a crucial component in constructing confidence intervals, providing us with an estimate of how much the sample proportion is expected to vary from the true population proportion.In this exercise, the standard error is calculated using the formula:\[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]With \( \hat{p} = 0.549 \) and sample size \( n = 519 \), it computes to approximately 0.022. This small number indicates that the sample proportion is quite close to the population proportion; therefore, our estimate is reasonably precise. The SE plays a pivotal role when we expand our point estimate to a confidence interval, ensuring accuracy in how we unroll our point into a possible range of values for the true proportion of interest. Understanding SE allows for deeper insight into the reliability of our confidence interval predictions.

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Most popular questions from this chapter

Consider \(n=100\) binomial trials with \(r=30\) successes. (a) Is it appropriate to use a normal distribution to approximate the \(\hat{p}\) distribution? (b) Find a \(90 \%\) confidence interval for the population proportion of successes \(p\). (c) Explain the meaning of the confidence interval you computed.

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