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A population proportion is \(.40 .\) A simple random sample of size 200 will be taken and the sample proportion \(\bar{p}\) will be used to estimate the population proportion. a. What is the probability that the sample proportion will be within ±.03 of the population proportion? b. What is the probability that the sample proportion will be within ±.05 of the population proportion?

Short Answer

Expert verified
a. Probability is 0.8078; b. Probability is 0.8531.

Step by step solution

01

Understand the Scenario

We're given a population proportion \( p = 0.40 \). We're sampling 200 individuals from this population. The sample proportion \( \bar{p} \) will likely differ from the population proportion due to sampling variability. We're tasked with finding out how likely \( \bar{p} \) will be close to \( p \).
02

Define the Sample Proportion Distribution

The sample proportion \( \bar{p} \) is approximately normally distributed because of the Central Limit Theorem (assuming the sample size is large enough). The mean of this distribution is the population proportion \( p = 0.40 \). The standard deviation (standard error) of \( \bar{p} \) is given by \( \sigma_{\bar{p}} = \sqrt{ \frac{p(1-p)}{n} } \), where \( n \) is the sample size. Here, \( \sigma_{\bar{p}} = \sqrt{ \frac{0.40(0.60)}{200} } = 0.0346 \).
03

Calculate Z-Score for \.03 Range

For part a, we want to find the probability that \( \bar{p} \) is within \( \pm 0.03 \) of \( 0.40 \). Calculate the z-scores for \( 0.40 \pm 0.03 \):\[ z = \frac{0.43 - 0.40}{0.0346} = 0.87 \] \[ z = \frac{0.37 - 0.40}{0.0346} = -0.87 \].
04

Find Probability for \.03 Range

Using a standard normal distribution table or calculator, find the probability between z-scores of \( -0.87 \) and \( 0.87 \). This corresponds to \( P(-0.87 \leq Z \leq 0.87) \). From standard normal distribution tables, this probability is approximately \( 0.8078 \).
05

Calculate Z-Score for \.05 Range

For part b, we want to find the probability that \( \bar{p} \) is within \( \pm 0.05 \) of \( 0.40 \). Calculate the z-scores for the range: \[ z = \frac{0.45 - 0.40}{0.0346} = 1.45 \] \[ z = \frac{0.35 - 0.40}{0.0346} = -1.45 \].
06

Find Probability for \.05 Range

Using a standard normal distribution table or calculator, find the probability between z-scores of \( -1.45 \) and \( 1.45 \). This corresponds to \( P(-1.45 \leq Z \leq 1.45) \). From standard normal distribution tables, this probability is approximately \( 0.8531 \).

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem is a crucial concept in statistics. It states that the sampling distribution of the sample mean (or sample proportion) approaches a normal distribution as the sample size increases, regardless of the population's original distribution. Imagine you have a large number of samples, each drawn randomly from a population with a known proportion of a certain characteristic. If you calculate the mean proportion for each sample, those means will form their own distribution. As the sample size grows, the shape of this distribution will increasingly resemble a normal one.

This theorem is essential because it allows us to use normal probability methods even when the population distribution is not normal. Therefore, when we are given sample proportions like in the exercise, we assume it follows a normal distribution. This simplifies the calculation of probabilities related to sample means or proportions.

The Central Limit Theorem holds true under certain conditions, notably:
  • The samples should be randomly selected.
  • The sample size should be large enough, often considered to be 30 or more.
  • The samples should be independent from one another.
These conditions ensure the validity of applying the theorem and make the results reliable.
Sample Proportion
Sample proportion is a way of estimating a population proportion by randomly selecting a sample from the population and calculating the proportion within that sample. It's represented by \( \bar{p} \). In our exercise, if 40% of a population has a certain characteristic, and you draw a sample of 200 individuals - the sample proportion helps us estimate this 40%. Although the sample proportion will differ from sample to sample, it offers a way to approximate the real population proportion.

The accuracy of the sample proportion depends on the sample size. Bigger samples tend to give more reliable estimates. However, due to natural sampling variability, sample proportions won't always match the true population proportion exactly. Instead, statisticians calculate the likely range within which the true proportion will lie. This is where the Central Limit Theorem and standard error come into play by providing the tools to quantify this variability.

To understand sample proportions better:
  • Recognize that they're just estimates of the population proportion.
  • They depend heavily on sampling methods and sample size.
  • The larger the sample, the more precise the estimate.
Standard Error
Standard error is a statistic that describes the variability of a sampling distribution. When calculating the probability of a sample statistic (like a mean or proportion) falling within a certain range of the population statistic, the standard error is key. It tells us how much we expect the sample proportion to vary from the true population proportion.

Specifically, the standard error of the sample proportion is calculated as:\[\sigma_{\bar{p}} = \sqrt{\frac{p(1-p)}{n}}\]where \( p \) is the population proportion and \( n \) is the sample size. In the given exercise, the standard error was calculated to be approximately 0.0346. This value helps determine the z-scores needed to find probabilities of the sample proportion being close to the population proportion.

By understanding standard error, you can:
  • Gauge how "spread out" your sample proportion estimates are likely to be.
  • Predict the confidence level for your estimates.
  • Understand the level of precision afforded by your sample size.
With more awareness of the standard error, you'll appreciate its role in forming inferences about population parameters.

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Most popular questions from this chapter

To estimate the mean age for a population of 4000 employees, a simple random sample of 40 employees is selected. a. Would you use the finite population correction factor in calculating the standard error of the mean? Explain. b. If the population standard deviation is \(\sigma=8.2\) years, compute the standard error both with and without the finite population correction factor. What is the rationale for ignoring the finite population correction factor whenever \(n / N \leq .05 ?\) c. What is the probability that the sample mean age of the employees will be within ±2 years of the population mean age?

People end up tossing \(12 \%\) of what they buy at the grocery store (Reader 's Digest, March, 2009 ). Assume this is the true population proportion and that you plan to take a sample survey of 540 grocery shoppers to further investigate their behavior. a. Show the sampling distribution of \(\bar{p},\) the proportion of groceries thrown out by your sample respondents. b. What is the probability that your survey will provide a sample proportion within ±.03 of the population proportion? c. What is the probability that your survey will provide a sample proportion within ±.015 of the population proportion?

After deducting grants based on need, the average cost to attend the University of Southern California (USC) is \(\$ 27,175\) (U.S. News \& World Report, America 's Best Colleges, \(2009 \text { ed. }) .\) Assume the population standard deviation is \(\$ 7400 .\) Suppose that a random sample of 60 USC students will be taken from this population. a. What is the value of the standard error of the mean? b. What is the probability that the sample mean will be more than \(\$ 27,175 ?\) c. What is the probability that the sample mean will be within \(\$ 1000\) of the population mean? d. How would the probability in part (c) change if the sample size were increased to \(100 ?\)

The Food Marketing Institute shows that \(17 \%\) of households spend more than \(\$ 100\) per week on groceries. Assume the population proportion is \(p=.17\) and a simple random sample of 800 households will be selected from the population. a. Show the sampling distribution of \(\bar{p}\), the sample proportion of households spending more than \(\$ 100\) per week on groceries. b. What is the probability that the sample proportion will be within ±.02 of the population proportion? c. Answer part (b) for a sample of 1600 households.

A survey question for a sample of 150 individuals yielded 75 Yes responses, 55 No responses, and 20 No Opinions. a. What is the point estimate of the proportion in the population who respond Yes? b. What is the point estimate of the proportion in the population who respond No?

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