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Basic Computation: Confidence Interval for \(p\) Consider \(n=100\) binomial trials with \(r=30\) successes. (a) Check Requirements 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) Interpretation Explain the meaning of the confidence interval you computed.

Short Answer

Expert verified
A normal approximation is appropriate; the 90% confidence interval is (0.2247, 0.3753). The true proportion is likely within this range.

Step by step solution

01

Check Requirements

To determine if it is appropriate to use a normal distribution to approximate the sampling distribution of the sample proportion \( \hat{p} \), we need to check the requirements. These requirements are that both \( np \) and \( n(1-p) \) are greater than 5. Here, \( n = 100 \) and \( r = 30 \), so \( \hat{p} = \frac{r}{n} = \frac{30}{100} = 0.3 \). Using \( \hat{p} \) as an estimate for \( p \), we have: \( np = 100 \times 0.3 = 30 \) and \( n(1 - p) = 100 \times 0.7 = 70 \). Both are greater than 5, so a normal approximation is appropriate.
02

Calculate the Standard Error

The standard error (SE) of the sample proportion \( \hat{p} \) is calculated using the formula: \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.3 \times 0.7}{100}} = \sqrt{\frac{0.21}{100}} = 0.0458\]
03

Find the Z-score for a 90% Confidence Interval

For a 90% confidence interval, the Z-score is based on the 5% in each tail of the normal distribution (since 5% + 5% = 10% leaves 90% in the middle). The Z-score corresponding to this is approximately 1.645. This value is determined from Z-tables or using a calculator.
04

Calculate the Confidence Interval

The formula for the confidence interval is:\[ \hat{p} \pm Z \times SE \]Substitute the values:\[0.3 \pm 1.645 \times 0.0458 \]Calculate the margin of error:\[1.645 \times 0.0458 = 0.0753\]Thus, the confidence interval is:\[0.3 \pm 0.0753 = (0.2247, 0.3753)\]
05

Interpretation of the Confidence Interval

The 90% confidence interval (0.2247, 0.3753) means that we are 90% confident that the true population proportion \( p \) of successes lies between 22.47% and 37.53%. This interval gives us a range of plausible values for the true proportion in the population.

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Key Concepts

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

Binomial Distribution
The binomial distribution is a fundamental concept in statistics. It describes the behavior of a fixed number of trials, each with two possible outcomes. Think of it as flipping a coin a certain number of times and observing how many times heads appear. Each flip is an independent event with the same probability of success, denoted as \( p \). This distribution is utilized when the number of trials, \( n \), is fixed, and the probability of success in each trial is constant.
For example, in the given exercise, the number of trials \( n \) is 100, and the number of successes \( r \) is 30. The sample proportion \( \hat{p} \) of successes is 0.3, derived from \( \hat{p} = \frac{30}{100} \).
This sets the foundation for determining whether we can use other statistical methods, such as normal approximation, to calculate confidence intervals.
  • This concept is key for calculating probabilities in a range of statistics and real-world applications.
  • The behavior of binomial distribution is idealized using assumptions that guide proper understanding.
Standard Error
The standard error (SE) is a crucial measure in statistics, representing how much the sample proportion \( \hat{p} \) is expected to vary from sample to sample. It provides insight into the precision of an estimate of a population parameter. Think of it as an indicator of the reliability of your sample proportion when estimating a population proportion.
In our exercise, the standard error is computed using:
\[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.3 \times 0.7}{100}} = 0.0458 \]
The calculated standard error of 0.0458 indicates a moderate level of variance in the sample proportion given the sample size and proportion.
  • Standard error helps in constructing confidence intervals, providing a range of values within which the true population parameter likely lies.
  • The smaller the standard error, the more precise the estimate.
Sample Proportion
The sample proportion \( \hat{p} \) is a simple yet important concept in statistics. It represents the fraction of successes in a sample, serving as an estimate of the population proportion \( p \). In our case, with a sample size \( n = 100 \) and successes \( r = 30 \), it is calculated as:
\( \hat{p} = \frac{r}{n} = \frac{30}{100} = 0.3 \)
This indicates that 30% of the sample has the characteristic of interest.
The sample proportion serves multiple purposes:
  • It is a point estimate of the unknown population proportion.
  • It is used for calculating the standard error and forming the basis of a confidence interval.
  • It plays a crucial role in hypothesis testing and other inferential statistics.

Understanding sample proportion helps in the interpretation of statistical data, facilitating better decision-making.
Normal Approximation
Normal approximation is a statistical technique employed when a binomial distribution can be closely approximated by a normal distribution. This simplification is especially useful for large sample sizes, as it allows for easier calculation of probabilities and confidence intervals.
In the exercise, the requirements \( np > 5 \) and \( n(1-p) > 5 \) are satisfied, evidenced by \( np = 30 \) and \( n(1-p) = 70 \).
By meeting these criteria, the binomial distribution is approximated using a normal distribution, allowing the calculation of a 90% confidence interval. The normal approximation simplifies calculations, especially when using Z-scores from standard normal distributions.
  • It eases the computational burden for large sample sizes.
  • This technique facilitates quick estimation and interpretation of confidence intervals and other statistical measures.

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Most popular questions from this chapter

Assume that the population of \(x\) values has an approximately normal distribution. Wildlife: Mountain Lions How much do wild mountain lions weigh? The 77 th Annual Report of the New Mexico Department of Game and Fish, edited by Bill Montoya, gave the following information. Adult wild mountain lions 18 months or older) captured and released for the first time in the San Andres Mountains gave the following weights (pounds): \(\begin{array}{llllll}68 & 104 & 128 & 122 & 60 & 64\end{array}\) (a) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x}=91.0\) pounds and \(s \approx 30.7\) pounds. (b) Find a \(75 \%\) confidence interval for the population average weight \(\mu\) of all adult mountain lions in the specified region. (c) Interpretation What does the confidence interval mean in the context of this problem?

Confidence Intetvals: Values of \(\sigma\) A random sample of size 36 is drawn from an \(x\) distribution. The sample mean is \(100 .\) (a) Suppose the \(x\) distribution has \(\sigma=30\). Compute a \(90 \%\) confidence interval for \(\mu\). What is the value of the margin of error? (b) Suppose the \(x\) distribution has \(\sigma=20\). Compute a \(90 \%\) confidence interval for \(\mu\). What is the value of the margin of error? (c) Suppose the \(x\) distribution has \(\sigma=10\). Compute a \(90 \%\) confidence interval for \(\mu\). What is the value of the margin of error? (d) Compare the margins of error for parts (a) through (c). As the standard deviation decreases, does the margin of error decrease? (e) Critical Thinking Compare the lengths of the confidence intervals for parts (a) through (c). As the standard deviation decreases, does the length of a \(90 \%\) confidence interval decrease?

Assume that the population of \(x\) values has an approximately normal distribution. Franchise: Candy Store Do you want to own your own candy store? With some interest in running your own business and a decent credit rating, you can probably get a bank loan on startup costs for franchises such as Candy Express, The Fudge Company, Karmel Corn, and Rocky Mountain Chocolate Factory. Startup costs (in thousands of dollars) for a random sample of candy stores are given below (Source: Entrepreneur Magazine, Vol. 23, No. 10 ). \(\begin{array}{lllllllll}95 & 173 & 129 & 95 & 75 & 94 & 116 & 100 & 85\end{array}\) (a) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x} \approx 106.9\) thousand dollars and \(s \approx 29.4\) thousand dollars. (b) Find a \(90 \%\) confidence interval for the population average startup costs \(\mu\) for candy store franchises. (c) Interpretation What does the confidence interval mean in the context of this problem?

Basic Computation: Confidence Interval for \(\mu_{1}-\mu_{2}\) Consider two independent distributions that are mound-shaped. A random sample of size \(n_{1}=36\) from the first distribution showed \(\bar{x}_{1}=15\), and a random sample of size \(n_{2}=40\) from the second distribution showed \(\bar{x}_{2}=14\) (a) Check Requirements If \(\sigma_{1}\) and \(\sigma_{2}\) are known, what distribution does \(\bar{x}_{1}-\bar{x}_{2}\) follow? Explain. (b) Given \(\sigma_{1}=3\) and \(\sigma_{2}=4\), find a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (c) Check Requirements Suppose \(\sigma_{1}\) and \(\sigma_{2}\) are both unknown, but from the random samples, you know \(s_{1}=3\) and \(s_{2}=4\). What distribution approximates the \(\bar{x}_{1}-\bar{x}_{2}\) distribution? What are the degrees of freedom? Explain. (d) With \(s_{1}=3\) and \(s_{2}=4\), find a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (e) If you have an appropriate calculator or computer software, find a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using degrees of freedom based on Satterthwaite's approximation. (f) Interpretation Based on the confidence intervals you computed, can you be \(95 \%\) confident that \(\mu_{1}\) is larger than \(\mu_{2} ?\) Explain.

Answer true or false. Explain your answer. Every random sample of the same size from a given population will produce exactly the same confidence interval for \(\mu\).

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