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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

Introduction to the Problem

We are asked if \( \hat{p} \) is an unbiased estimator for \( p \) given that \( n p > 5 \) and \( n q > 5 \). An unbiased estimator means the expected value or mean of the estimator equals the actual population parameter.
02

Defining Variables

Let \( \hat{p} \) be the sample proportion, \( p \) be the population proportion, \( n \) the sample size, and \( q = 1 - p \). We start by defining these variables, ensuring they align with the conditions \( n p > 5 \) and \( n q > 5 \).
03

Understanding Expected Value

We need to find \( E(\hat{p}) \). The expected value of the sample proportion \( \hat{p} \) is calculated as:\[ E(\hat{p}) = E\left(\frac{X}{n}\right) = \frac{E(X)}{n}, \]where \( X \) is a random variable representing the number of successes in \( n \) trials.
04

Expected Value of X

Since \( X \) is a binomial random variable with parameters \( n \) and \( p \), we know:\[ E(X) = n \cdot p. \]
05

Calculate Expected Value of Sample Proportion

Substitute \( E(X) = n \cdot p \) into the expression for \( E(\hat{p}) \):\[ E(\hat{p}) = \frac{n \cdot p}{n} = p. \]
06

Conclusion

Since \( E(\hat{p}) = p \), the sample proportion \( \hat{p} \) is indeed an unbiased estimator for the population proportion \( p \) when \( n p > 5 \) and \( n q > 5 \).

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Key Concepts

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

Sample Proportion
A sample proportion, often denoted as \( \hat{p} \), is a statistical measure that represents the fraction or percentage of successes in a sample relative to all possible outcomes in the sample. For example, if you survey 100 people and 40 of them like ice cream, the sample proportion of ice cream enthusiasts would be \( \hat{p} = 40/100 = 0.40 \). The sample proportion is widely used to make inferences about the population proportion \( p \), which is the true fraction of successes in the entire population. It helps statisticians estimate characteristics of a large population using a smaller, manageable sample. These estimates are only reliable if certain conditions are met, such as having a sufficiently large sample size, typically indicated by the rules \( n p > 5 \) and \( n q > 5 \), where \( q = 1 - p \). These conditions ensure that the distribution of the sample proportion is approximately normal, allowing for more accurate inferences about the population.One critical attribute of \( \hat{p} \) is that it can serve as an unbiased estimator for \( p \), meaning its expected value matches the population proportion exactly, provided the aforementioned conditions are satisfied.
Expected Value
The expected value is a fundamental concept in probability and statistics, representing the average or mean value that a random variable is expected to assume over numerous trials. For a sample proportion \( \hat{p} \), calculating the expected value helps determine if \( \hat{p} \) is an unbiased estimator of the population proportion \( p \). The expected value of \( \hat{p} \) is calculated using the formula:\[E(\hat{p}) = E\left(\frac{X}{n}\right) = \frac{E(X)}{n},\] where \( X \) is the number of successes in \( n \) independent trials.Since \( X \) is considered a binomial random variable with parameters \( n \) and \( p \), the expected value of \( X \) can be easily computed using the formula:\[E(X) = n \cdot p.\]By substituting \( E(X) \) back into the equation for \( E(\hat{p}) \), we find:\[E(\hat{p}) = \frac{n \cdot p}{n} = p.\]Hence, the expected value of the sample proportion equals the population proportion, confirming that \( \hat{p} \) is an unbiased estimator. This characteristic is crucial as it ensures that, on average, the sample proportion provides an accurate estimate of the population proportion across repeated sampling.
Binomial Distribution
The binomial distribution is a discrete probability distribution describing the number of successes in a fixed number of independent trials, each with the same probability of success. For the example of flipping a coin 10 times, each flip represents a trial, and if the coin is fair, the probability of landing heads (a success) is 0.5. Mathematically, if \( X \) denotes the number of successes out of \( n \) trials, the probability mass function of \( X \), when it follows a binomial distribution with parameters \( n \) and \( p \), is given by:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k},\]where \( k \) represents the number of successes, \( p \) the probability of success in each trial, and \( \binom{n}{k} \) is the binomial coefficient.The binomial distribution is essential for understanding sample proportions because \( \hat{p} = \frac{X}{n} \) involves \( X \), which follows a binomial distribution. When \( n \) is large and \( p \) is moderate, the binomial distribution approximates a normal distribution, which is vital for making the sample proportion \( \hat{p} \) a reliable estimator of the population proportion \( p \). This approximation allows us to perform statistical analyses that rely on normality assumptions to make decisions or construct confidence intervals about \( p \).

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Most popular questions from this chapter

The amount of money spent weekly on cleaning, maintenance, and repairs at a large restaurant was observed over a long period of time to be approximately normally distributed, with mean \(\mu=\$ 615\) and standard deviation \(\sigma=\$ 42\). (a) If \(\$ 646\) is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount? (b) How much should be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a given week is only 0.10?

Suppose you want to eat lunch at a popular restaurant. The restaurant does not take reservations, so there is usually a waiting time before you can be seated. Let \(x\) represent the length of time waiting to be seated. From past experience, you know that the mean waiting time is \(\mu=18\) minutes with \(\sigma=4\) minutes. You assume that the \(x\) distribution is approximately normal. (a) What is the probability that the waiting time will exceed 20 minutes, given that it has exceeded 15 minutes? Hint: Compute \(P(x>20 \mid x>15)\). (b) What is the probability that the waiting time will exceed 25 minutes, given that it has exceeded 18 minutes? Hint: Compute \(P(x>25 \mid x>18)\). (c) Hint for solution: Review item 6 , conditional probability, in the summary of basic probability rules at the end of Section 4.2. Note that $$ P(A \mid B)=\frac{P(A \text { and } B)}{P(B)} $$ and show that in part (a), $$ P(x>20 \mid x>15)=\frac{P((x>20) \text { and }(x>15))}{P(x>15)}=\frac{P(x>20)}{P(x>15)} $$

Distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us. (a) Suppose \(n=33\) and \(p=0.21\). Can we approximate the \(\hat{p}\) distribution by a normal distribution? Why? What are the values of \(\mu_{\hat{p}}\) and \(\sigma_{\hat{p}} ?\) (b) Suppose \(n=25\) and \(p=0.15 .\) Can we safely approximate the \(\hat{p}\) distribution by a normal distribution? Why or why not? (c) Suppose \(n=48\) and \(p=0.15 .\) Can we approximate the \(\hat{p}\) distribution by a normal distribution? Why? What are the values of \(\mu_{\hat{p}}\) and \(\sigma_{\hat{p}} ?\)

Police response time to an emergency call is the difference between the time the call is first received by the dispatcher and the time a patrol car radios that it has arrived at the scene (based on information from The Denver Post). Over a long period of time, it has been determined that the police response time has a normal distribution with a mean of \(8.4\) minutes and a standard deviation of \(1.7\) minutes. For a randomly received emergency call, what is the probability that the response time will be (a) between 5 and 10 minutes? (b) less than 5 minutes? (c) more than 10 minutes?

Let \(x\) be a random variable that represents checkout time (time spent in the actual checkout process) in minutes in the express lane of a large grocery. Based on a consumer survey, the mean of the \(x\) distribution is about \(\mu=2.7\) minutes, with standard deviation \(\sigma=0.6\) minute. Assume that the express lane always has customers waiting to be checked out and that the distribution of \(x\) values is more or less symmetrical and mound-shaped. What is the probability that the total checkout time for the next 30 customers is less than 90 minutes? Let us solve this problem in steps. (a) Let \(x_{i}\) (for \(\left.i=1,2,3, \ldots, 30\right)\) represent the checkout time for each customer. For example, \(x_{1}\) is the checkout time for the first customer, \(x_{2}\) is the checkout time for the second customer, and so forth. Each \(x_{i}\) has mean \(\mu=2.7\) minutes and standard deviation \(\sigma=0.6\) minute. Let \(w=x_{1}+x_{2}+\cdots+x_{30}\) Explain why the problem is asking us to compute the probability that \(w\) is less than 90 . (b) Use a little algebra and explain why \(w<90\) is mathematically equivalent to \(w / 30<3 .\) Since \(w\) is the total of the \(30 x\) values, then \(w / 30=\bar{x}\). Therefore, the statement \(\bar{x}<3\) is equivalent to the statement \(w<90\). From this we conclude that the probabilities \(P(\bar{x}<3)\) and \(P(w<90)\) are equal. (c) What does the central limit theorem say about the probability distribution of \(\bar{x} ?\) Is it approximately normal? What are the mean and standard deviation of the \(\bar{x}\) distribution? (d) Use the result of part (c) to compute \(P(\bar{x}<3)\). What does this result tell you about \(P(w<90)\) ?

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