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91Ó°ÊÓ

Is \(\hat{p}\) an unbiased estimator for \(p\) when \(n p>5\) and \(n q>5 ?\) Recall that a statistic is an unbiased estimator of the corresponding parameter if the mean of the sampling distribution equals the parameter in question.

Short Answer

Expert verified
Yes, \( \hat{p} \) is an unbiased estimator for \( p \).

Step by step solution

01

Understanding the Problem

We are asked to determine if \( \hat{p} \) is an unbiased estimator for \( p \). We know that a statistic is an unbiased estimator if the expected value of the statistic equals the parameter it estimates. Here, we must show that the expected value of \( \hat{p} \) equals \( p \).
02

Define the Estimator

\( \hat{p} \) represents the sample proportion, calculated as \( \hat{p} = \frac{X}{n} \), where \( X \) is the number of successes in the sample, and \( n \) is the total number of trials. \( X \) follows a Binomial distribution with parameters \( n \) and \( p \).
03

Calculate the Expected Value of \( \hat{p} \)

The expected value of \( \hat{p} \) can be calculated as \( E(\hat{p}) = E\left(\frac{X}{n}\right) = \frac{1}{n} E(X) \). Since \( X \) is Binomially distributed, \( E(X) = np \). Thus, \( E(\hat{p}) = \frac{np}{n} = p \).
04

Conclusion

Since \( E(\hat{p}) = p \), \( \hat{p} \) is an unbiased estimator of \( p \), given the conditions \( nq > 5 \) and \( np > 5 \). These conditions ensure that the Binomial distribution approximates a Normal distribution well, providing further justification in practice, though they do not affect the unbiased nature inherently.

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Key Concepts

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

Sample Proportion
The sample proportion, denoted as \( \hat{p} \), is a statistical measure that tells us the fraction of the sample that has a particular attribute. It's calculated by taking the number of successes \( X \) in the sample and dividing it by the total number of trials \( n \): \( \hat{p} = \frac{X}{n} \).
This simple formula gives us a proportion that represents how common a specific event is within the sample. Think about a scenario where you're flipping a coin 100 times to see how often heads appear. If you get heads 60 times, the sample proportion is \( 0.6 \) or 60%.
Using the sample proportion allows statisticians to estimate how a population behaves based on a sample. It's a key component in many statistical methods, particularly when trying to infer population parameters from sample data.
Expected Value
The expected value is a fundamental concept in probability and statistics. It represents the average outcome of a random variable if you were to repeat an experiment infinitely many times. In simpler terms, it's what you'd "expect" to see, on average.
Mathematically, for discrete random variables, the expected value is calculated by multiplying each possible outcome by its probability and summing these products. When dealing with a sample proportion, like \( \hat{p} = \frac{X}{n} \), the expected value helps in understanding whether \( \hat{p} \) is a fair estimate for the population proportion \( p \).
In this case, you might calculate the expected value of \( \, \hat{p} \) \, as \, \( \, E(\hat{p}) = p \, \), which shows that on average, \( \hat{p} \) will give us the true population proportion. This characteristic when \( E(\hat{p}) = p \) confirms that our estimator is unbiased.
Binomial Distribution
The binomial distribution is a probability distribution that models the number of successes in a fixed number of independent and identically distributed Bernoulli trials. It’s defined by two parameters: \( n \), the number of trials, and \( p \), the probability of success in each trial.
This distribution is often used to model scenarios where there are two possible outcomes, such as success or failure. A common example is flipping a coin, where success could mean "heads".
The expected value of a binomial distribution, \( X \), is calculated as \( E(X) = np \). This is a crucial connection because it leads us to conclude that the sample proportion \( \hat{p} = \frac{X}{n} \) has an expected value of \( p \), supporting its unbiased nature.
For practical purposes, the conditions \( np > 5 \) and \( nq > 5 \), where \( q = 1-p \), ensure the distribution approximates a normal distribution, making it easier to work with in practice, particularly in hypothesis testing and confidence interval estimation.

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Most popular questions from this chapter

A person's blood glucose level and diabetes are closely related. Let \(x\) be a random variable measured in milligrams of glucose per deciliter \((1 / 10 \text { of a liter })\) of blood. After a 12 -hour fast, the random variable \(x\) will have a distribution that is approximately normal with mean \(\mu=85\) and standard deviation \(\sigma=25\) (Source: Diagnostic Tests with Nursing Implications, edited by S. Loeb, Springhouse Press). Note: After 50 years of age, both the mean and standard deviation tend to increase. What is the probability that, for an adult (under 50 years old) after a 12 -hour fast, (a) \(x\) is more than \(60 ?\) (b) \(x\) is less than \(110 ?\) (c) \(x\) is between 60 and \(110 ?\) (d) \(x\) is greater than 125 (borderline diabetes starts at 125 )?

What is a population parameter? Give three examples.

Find the \(z\) value described and sketch the area described.Find the \(z\) value such that \(98 \%\) of the standard normal curve lies between \(-z\) and \(z\).

Insurance: Claims Do you try to pad an insurance claim to cover your deductible? About \(40 \%\) of all U.S. adults will try to pad their insurance claims! (Source: Are You Normal?, by Bernice Kanner, St. Martin's Press.) Suppose that you are the director of an insurance adjustment office. Your office has just received 128 insurance claims to be processed in the next few days. What is the probability that (a) half or more of the claims have been padded? (b) fewer than 45 of the claims have been padded? (c) from 40 to 64 of the claims have been padded? (d) more than 80 of the claims have not been padded?

Coal is carried from a mine in West Virginia to a power plant in New York in hopper cars on a long train. The automatic hopper car loader is set to put 75 tons of coal into each car. The actual weights of coal loaded into each car are normally distributed, with mean \(\mu=75\) tons and standard deviation \(\sigma=0.8\) ton. (a) What is the probability that one car chosen at random will have less than 74.5 tons of coal? (b) What is the probability that 20 cars chosen at random will have a mean load weight \(\bar{x}\) of less than 74.5 tons of coal? (c) Interpretation Suppose the weight of coal in one car was less than 74.5 tons. Would that fact make you suspect that the loader had slipped out of adjustment? Suppose the weight of coal in 20 cars sclected at random had an average \(\bar{x}\) of less than 74.5 tons. Would that fact make you suspect that the loader had slipped out of adjustment? Why?

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