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Chapter 8: Problem 7

Describe the sampling distribution of \(\hat{p}\). Assume that the size of the population is 25,000 for each problem. $$ n=500, p=0.4 $$

Short Answer

Expert verified
The sampling distribution of \(\hat{p}\) is approximately normal with \( \mu_{\hat{p}} = 0.4\) and \( \sigma_{\hat{p}} \approx 0.0219\).

Step by step solution

01

- Understand the problem

The problem is asking to describe the sampling distribution of the sample proportion \(\hat{p}\). We are given the sample size \(n = 500\) and the population proportion \(p = 0.4\).
02

- Determine the mean of the sampling distribution

The mean of the sampling distribution of \(\hat{p}\) is equal to the population proportion \(p\). Therefore, the mean \( \mu_{\hat{p}} \) is: \[ \mu_{\hat{p}} = p = 0.4 \]
03

- Calculate the standard deviation of the sampling distribution

The standard deviation of the sampling distribution of \(\hat{p}\) can be found using the formula: \[ \sigma_{\hat{p}} = \sqrt{ \frac{p(1-p)}{n} } \] Plugging in the values, we get: \[ \sigma_{\hat{p}} = \sqrt{ \frac{0.4(1-0.4)}{500} } = \sqrt{ \frac{0.24}{500} } = \sqrt{ 0.00048 } \approx 0.0219 \]
04

- Check normality

To determine if the sampling distribution of \(\hat{p}\) is approximately normal, check if \(np \geq 10\) and \(n(1-p) \geq 10\). Here: \[ np = 500 \cdot 0.4 = 200 \geq 10 \] \[ n(1-p) = 500 \cdot (1-0.4) = 300 \geq 10 \] Since both conditions are satisfied, the sampling distribution of \(\hat{p}\) is approximately normal.
05

- Summarize the sampling distribution

The sampling distribution of \(\hat{p}\) is approximately normal with a mean \(\mu_{\hat{p}} = 0.4\) and a standard deviation \( \sigma_{\hat{p}} \approx 0.0219 \).

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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 statistic that estimates the true proportion within a population. It is calculated by taking the ratio of the number of successes to the total sample size. For instance, if we draw a sample of 500 individuals from a population and find that 200 out of those 500 are left-handed, our sample proportion \(\hat{p}\) is \(\frac{200}{500} = 0.4\). This value serves as an estimate of the true population proportion.
Population proportion
The population proportion, denoted as \(p\), is the true proportion of individuals within the entire population that have a specific characteristic. Unlike the sample proportion, population proportion is a fixed value but often unknown. In practice, we use the sample proportion to make inferences about the population proportion. For example, if we know that 40% of all individuals in a population of 25,000 are left-handed, our population proportion \(p = 0.4\). This is used as a benchmark to compare against our sample proportion.
Standard deviation
The standard deviation of the sampling distribution of \(\hat{p}\) measures how much the sample proportion \(\hat{p}\) is expected to vary from sample to sample. It is calculated using the formula \[ \sigma_{\hat{p}} = \sqrt{ \frac{p(1-p)}{n} } \] where \(\hat{p}\) is the sample proportion, \(p\) is the population proportion, and \(n\) is the sample size. For instance, with a population proportion \(p = 0.4\) and sample size \(n = 500\), the standard deviation \[ \sigma_{\hat{p}} = \sqrt{ \frac{0.4(1-0.4)}{500} } = \sqrt{ \frac{0.24}{500} } = 0.0219 \] This value indicates the likely amount of variation we will see in different sample proportions.
Normality condition
To determine if the sampling distribution of \(\hat{p}\) is approximately normal, we use the normality condition. This involves checking that \(np \geq 10\) and \(n(1-p) \geq 10\). These conditions ensure that the sample size is large enough for the Central Limit Theorem to apply, making the distribution of the sample proportion approximately normal. In our example, with \(n = 500\) and \(p = 0.4\), the conditions are: \[ np = 500 \cdot 0.4 = 200 \geq 10 \] \[ n(1-p) = 500 \cdot 0.6 = 300 \geq 10 \] Since both conditions are satisfied, the sampling distribution of \(\hat{p}\) can be considered approximately normal.

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Most popular questions from this chapter

The amount of time Americans spend watching television is closely monitored by firms such as \(\mathrm{AC}\) Nielsen because this helps determine advertising pricing for commercials. (a) Do you think the variable "weekly time spent watching television" would be normally distributed? If not, what shape would you expect the variable to have? (b) According to the American Time Use Survey, adult Americans spend 2.35 hours per day watching television on a weekday. Assume that the standard deviation for "time spent watching television on a weekday" is 1.93 hours. If a random sample of 40 adult Americans is obtained, describe the sampling distribution of \(\bar{x},\) the mean amount of time spent watching television on a weekday. (c) Determine the probability that a random sample of 40 adult Americans results in a mean time watching television on a weekday of between 2 and 3 hours. (d) One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 35 individuals who consider themselves to be avid Internet users results in a mean time of 1.89 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 1.89 hours or less from a population whose mean is presumed to be 2.35 hours. Based on the result obtained, do you think avid Internet users watch less television?

The quality-control manager of a Long John Silver's restaurant wants to analyze the length of time that a car spends at the drive-through window waiting for an order. It is determined that the mean time spent at the window is 59.3 seconds with a standard deviation of 13.1 seconds. The distribution of time at the window is skewed right (data based on information provided by Danica Williams, student at Joliet Junior College). (a) To obtain probabilities regarding a sample mean using the normal model, what size sample is required? (b) The quality-control manager wishes to use a new delivery system designed to get cars through the drive-through system faster. A random sample of 40 cars results in a sample mean time spent at the window of 56.8 seconds. What is the probability of obtaining a sample mean of 56.8 seconds or less, assuming that the population mean is 59.3 seconds? Do you think that the new system is effective? (c) Treat the next 50 cars that arrive as a simple random sample. There is a \(15 \%\) chance the mean time will be at or below ____ seconds

Suppose a simple random sample of size \(n\) is drawn from a large population with mean \(\mu\) and standard deviation \(\sigma .\) The sampling distribution of \(\bar{x}\) has mean \(\mu_{\bar{x}}=\) ______ and standard deviation \(\sigma_{\bar{x}}=\) ______.

The following data represent the ages of the winners of the Academy Award for Best Actor for the years \(2004-2009\) $$ \begin{array}{lc} \hline \text { 2004: Jamie Foxx } & 37 \\ \hline \text { 2005: Philip Seymour Hoffman } & 38 \\ \hline \text { 2006: Forest Whitaker } & 45 \\ \hline \text { 2007: Daniel Day-Lewis } & 50 \\ \hline \text { 2008: Sean Penn } & 48 \\ \hline \text { 2009: Jeff Bridges } & 60 \\ \hline \end{array} $$ (a) Compute the population mean, \(\mu\). (b) List all possible samples with size \(n=2 .\) There should be \({ }_{6} C_{2}=15\) samples. (c) Construct a sampling distribution for the mean by listing the sample means and their corresponding probabilities. (d) Compute the mean of the sampling distribution. (e) Compute the probability that the sample mean is within 3 years of the population mean age. (f) Repeat parts (b)-(e) using samples of size \(n=3\). Comment on the effect of increasing the sample size.

Describe the sampling distribution of \(\hat{p}\). Assume that the size of the population is 25,000 for each problem. $$ n=1000, p=0.103 $$
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