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For all these problems, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding. In order to use a normal distribution to compute confidence intervals for \(p\), what conditions on \(n p\) and \(n q\) need to be satisfied?

Short Answer

Expert verified
Both \( n p \) and \( n q \) should be at least 5 to use the normal distribution for \( p \)'s confidence intervals.

Step by step solution

01

Understand the definition

To use the normal distribution to compute confidence intervals for a population proportion \( p \), it's important to ensure that the sampling distribution of the sample proportion \( \hat{p} \) is approximately normal. This is possible by using the central limit theorem.
02

Examine the conditions for normal approximation

The conditions that help ensure that the sampling distribution is approximately normal are that both \( n p \) and \( n q \) must be sufficiently large, where \( q = 1 - p \). Generally, the rule of thumb is that both \( n p \) and \( n q \) should be greater than or equal to 5.
03

Verify the conditions

To directly apply the normal approximation when computing confidence intervals, verify that \( n p \geq 5 \) and \( n (1-p) \geq 5 \). If these conditions are met, the approximation is considered valid.

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Key Concepts

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

Confidence Intervals
Confidence intervals are a statistical tool that helps us estimate the range in which a population parameter, such as a population proportion, is likely to fall. They offer a way to express the uncertainty or variability in our estimates. When calculating a confidence interval for a population proportion, the goal is to create an interval that captures the true population proportion a specific percentage of the time—typically 95% or 99%. The width of this interval depends on several factors, including sample size, variability in the population, and the desired confidence level. One key point when using a confidence interval for proportions is to ensure that the underlying distribution conditions are suitable for approximation by a normal distribution. If these conditions are met, it allows us to confidently infer population parameters from sample data.
Population Proportion
Population proportion refers to the fraction or percentage of a population that possesses a certain characteristic of interest. For example, if you are examining how many people in a city own a pet, the population proportion would be the percentage of pet owners relative to the total population.Calculating this proportion in real-world studies involves sampling, as surveying an entire population is often impractical. The sample proportion, denoted as \( \hat{p} \), serves as an estimate of the true population proportion \( p \). The accuracy of this estimate can be assessed and adjusted using confidence intervals based on the sample data.In order to apply the normal approximation for computing confidence intervals of \( p \), the sample size \( n \) should be large enough so that both \( n p \) and \( n (1-p) \) are 5 or greater. This ensures that the distribution of \( \hat{p} \) becomes approximately normal.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental concept in statistics that explains why many distributions tend to approximate a normal distribution when the sample size becomes large. According to the CLT, regardless of the original distribution of the data, the sampling distribution of the sample mean will approach a normal distribution as the sample size \( n \) becomes large enough. This theorem is crucial for estimating probabilities and making inferences about a population based on sample data.For population proportions, the CLT supports that the sampling distribution of sample proportions is approximately normal when the size of the sample is sufficiently large, specifically when both \( n p \) and \( n (1-p) \) are at least 5. This enables statisticians to apply normal distribution methods to construct confidence intervals and perform hypothesis testing on proportions.
Sampling Distribution
A sampling distribution is the probability distribution of a given statistic based on a random sample. When we repeatedly take samples from a larger population and calculate a statistic, such as the mean or proportion, the values of this statistic form a distribution known as the sampling distribution.The sampling distribution of the sample proportion \( \hat{p} \) allows for inference about the population proportion \( p \). It becomes approximately normal under certain conditions, especially for large samples due to the Central Limit Theorem. The normal approximation can be safely used when both \( n p \) and \( n (1-p) \) are greater than 5. This ensures the shape of the sampling distribution is symmetric and bell-shaped, reflecting typical characteristics of a normal distribution. This approximation is pivotal for constructing confidence intervals and performing other statistical analyses.

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Most popular questions from this chapter

Marketing: Customer Loyalty In a marketing survey, a random sample of 730 women shoppers revealed that 628 remained loyal to their favorite supermarket during the past year (i.e., did not switch stores) (Source: Trends in the United States: Consumer Attitudes and the Supermarket, The Research Department, Food Marketing Institute). (a) Let \(p\) represent the proportion of all women shoppers who remain loyal to their favorite supermarket. Find a point estimate for \(p\). (b) Find a \(95 \%\) confidence interval for \(p .\) Give a brief explanation of the meaning of the interval. (c) Interpretation As a news writer, how would you report the survey results regarding the percentage of women supermarket shoppers who remained loyal to their favorite supermarket during the past year? What is the margin of error based on a \(95 \%\) confidence interval?

Pro Football and Basketball: Heights of Players Independent random samples of professional football and basketball players gave the following information (References: Sports Encyclopedia of Pro Football and Official NBA Basketball Encyclopedia). Note: These data are also available for download at the Online Study Center. $$ \begin{aligned} &\text { Heights (in } \mathrm{ft} \text { ) of pro football players: } x_{1} ; n_{1}=45\\\ &\begin{array}{llllllllll} 6.33 & 6.50 & 6.50 & 6.25 & 6.50 & 6.33 & 6.25 & 6.17 & 6.42 & 6.33 \\ 6.42 & 6.58 & 6.08 & 6.58 & 6.50 & 6.42 & 6.25 & 6.67 & 5.91 & 6.00 \\ 5.83 & 6.00 & 5.83 & 5.08 & 6.75 & 5.83 & 6.17 & 5.75 & 6.00 & 5.75 \\ 6.50 & 5.83 & 5.91 & 5.67 & 6.00 & 6.08 & 6.17 & 6.58 & 6.50 & 6.25 \\ 6.33 & 5.25 & 6.67 & 6.50 & 5.83 & & & & & \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Heights (in } \mathrm{ft} \text { ) of pro basketball players: } x_{2} ; n_{2}=40\\\ &\begin{array}{llllllllll} 6.08 & 6.58 & 6.25 & 6.58 & 6.25 & 5.92 & 7.00 & 6.41 & 6.75 & 6.25 \\ 6.00 & 6.92 & 6.83 & 6.58 & 6.41 & 6.67 & 6.67 & 5.75 & 6.25 & 6.25 \\ 6.50 & 6.00 & 6.92 & 6.25 & 6.42 & 6.58 & 6.58 & 6.08 & 6.75 & 6.50 \\ 6.83 & 6.08 & 6.92 & 6.00 & 6.33 & 6.50 & 6.58 & 6.83 & 6.50 & 6.58 \end{array} \end{aligned} $$ (a) Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1} \approx 6.179, s_{1} \approx 0.366, \bar{x}_{2} \approx 6.453\), and \(s_{2} \approx 0.314 .\) (b) Let \(\mu_{1}\) be the population mean for \(x_{1}\) and let \(\mu_{2}\) be the population mean for \(x_{2}\). Find a \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (c) Interpretation Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the \(90 \%\) level of confidence, do professional football players tend to have a higher population mean height than professional basketball players? (d) Check Requirements Which distribution (standard normal or Student's \(t)\) did you use? Why? Do you need information about the height distributions? Explain.

Medical: Plasma Compress At Community Hospital, the burn center is experimenting with a new plasma compress treatment. A random sample of \(n_{1}=316\) patients with minor burns received the plasma compress treatment. Of these patients, it was found that 259 had no visible scars after treatment. Another random sample of \(n_{2}=419\) patients with minor burns received no plasma compress treatment. For this group, it was found that 94 had no visible scars after treatment. Let \(p_{1}\) be the population proportion of all patients with minor burns receiving the plasma compress treatment who have no visible scars. Let \(p_{2}\) be the population proportion of all patients with minor burns not receiving the plasma compress treatment who have no visible scars. (a) Check Requirements Can a normal distribution be used to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(95 \%\) confidence interval for \(p_{1}-p_{2}\). (c) Interpretation Explain the meaning of the confidence interval found in part (b) in the context of the problem. Does the interval contain numbers that are all positive? all negative? both positive and negative? At the \(95 \%\) level of confidence, does treatment with plasma compresses seem to make a difference in the proportion of patients with visible scars from minor burns?

Profits: Banks Jobs and productivity! How do banks rate? One way to answer this question is to examine annual profits per employee. Forbes Top Companies, edited by J. T. Davis (John Wiley \& Sons), gave the following data about annual profits per employee (in units of one thousand dollars per employee) for representative companies in financial services. Companies such as Wells Fargo, First Bank System, and Key Banks were included. Assume \(\sigma \approx 10.2\) thousand dollars.$$ \begin{array}{lllllllllll} 42.9 & 43.8 & 48.2 & 60.6 & 54.9 & 55.1 & 52.9 & 54.9 & 42.5 & 33.0 & 33.6 \\ 36.9 & 27.0 & 47.1 & 33.8 & 28.1 & 28.5 & 29.1 & 36.5 & 36.1 & 26.9 & 27.8 \\ 28.8 & 29.3 & 31.5 & 31.7 & 31.1 & 38.0 & 32.0 & 31.7 & 32.9 & 23.1 & 54.9 \\ 43.8 & 36.9 & 31.9 & 25.5 & 23.2 & 29.8 & 22.3 & 26.5 & 26.7 & & \end{array} $$ (a) Use a calculator or appropriate computer software to verify that, for the preceding data, \(\bar{x} \approx 36.0\). (b) Let us say that the preceding data are representative of the entire sector of (successful) financial services corporations. Find a \(75 \%\) confidence interval for \(\mu\), the average annual profit per employee for all successful banks. (c) Interpretation Let us say that you are the manager of a local bank with a large number of employees. Suppose the annual profits per employee are less than 30 thousand dollars per employee. Do you think this might be somewhat low compared with other successful financial institutions? Explain by referring to the confidence interval you computed in part (b). (d) Interpretation Suppose the annual profits are more than 40 thousand dollars per employee. As manager of the bank, would you feel somewhat better? Explain by referring to the confidence interval you computed in part (b). (e) Repeat parts (b), (c), and (d) for a \(90 \%\) confidence level.

Focus Problem: Wood Duck Nests In the Focus Problem at the beginning of this chapter, a study was described comparing the hatch ratios of wood duck nesting boxes. Group I nesting boxes were well separated from each other and well hidden by available brush. There were a total of 474 eggs in group I boxes, of which a field count showed about 270 had hatched. Group II nesting boxes were placed in highly visible locations and grouped closely together. There were a total of 805 eggs in group II boxes, of which a field count showed about 270 had hatched. (a) Find a point estimate \(\hat{p}_{1}\) for \(p_{1}\), the proportion of eggs that hatched in group I nest box placements. Find a \(95 \%\) confidence interval for \(p_{1}\). (b) Find a point estimate \(\hat{p}_{2}\) for \(p_{2}\), the proportion of eggs that hatched in group II nest box placements. Find a \(95 \%\) confidence interval for \(p_{2}\). (c) Find a \(95 \%\) confidence interval for \(p_{1}-p_{2}\). Does the interval indicate that the proportion of eggs hatched from group I nest boxes is higher than, lower than, or equal to the proportion of eggs hatched from group II nest boxes? (d) Interpretation What conclusions about placement of nest boxes can be drawn? In the article discussed in the Focus Problem, additional concerns are raised about the higher cost of placing and maintaining group I nest box placements. Also at issue is the cost efficiency per successful wood duck hatch.

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