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Chapter 9: Problem 18

Home Ownership An urban economist wishes to estimate the proportion of Americans who own their house. What size sample should be obtained if he wishes the estimate to be within 0.02 with 90\% confidence if (a) he uses an estimate of 0.675 from the fourth quarter of 2000 obtained from the U.S. Census Bureau? (b) he does not use any prior estimates?

Short Answer

Expert verified
Sample size: (a) 457, (b) 1692.

Step by step solution

01

Understand the Problem

We need to determine the sample size required to estimate the proportion of Americans who own their house with a margin of error of 0.02 at a 90% confidence level for two scenarios: (a) using a prior estimate of 0.675, and (b) without using any prior estimates.
02

Identify the Formula

The formula for sample size (\(n\)) in proportion estimation is: \[ n = \left( \frac{Z_{\alpha/2}^2 \cdot p \cdot (1-p)}{E^2} \right) \]Where: \(Z_{\alpha/2}\) is the z-value corresponding to the desired confidence level, p is the estimated proportion, and E is the margin of error.
03

Identify Given Values and Z-value

Given: p = 0.675 (for part a) E = 0.02 For a 90% confidence level, \(Z_{\alpha/2}\) is approximately 1.645.
04

Calculate Sample Size for Part (a)

Substitute the values into the formula: \[ n = \left( \frac{1.645^2 \cdot 0.675 \cdot (1-0.675)}{0.02^2} \right) \]Calculating gives: \[ n \approx 1.645^2 \times 0.675 \times 0.325 / 0.0004 \approx 456.56 \]Thus, the sample size needed is approximately 457.
05

Calculate Sample Size for Part (b)

When no prior estimate p\text{ is available, use }p = 0.5. Using the same formula with p = 0.5: \[ n = \left( \frac{1.645^2 \cdot 0.5 \cdot 0.5}{0.02^2} \right) \]Calculating gives: \[ n \approx 1.645^2 \times 0.25 / 0.0004 \approx 1691.39 \]Thus, the sample size needed is approximately 1692.

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values that is likely to contain the population parameter with a specified level of confidence. For instance, in our exercise, we use a 90% confidence level to estimate the proportion of Americans who own their house.
The confidence interval provides an upper and lower bound within which we expect the true proportion to lie. This helps researchers account for the variability and uncertainty inherent in sample data.
In simpler terms, if you were to repeat the same sampling process 100 times, for 90 of those samples, the true population proportion would fall within the calculated interval.
Margin of Error
The margin of error is a measure of the precision of an estimate. It is the maximum amount by which the sample estimate may differ from the true population parameter.
In our problem, the margin of error is specified as 0.02, meaning the economist wants the estimate to be within 2% of the true proportion.
The margin of error is influenced by the sample size and the confidence level. Generally, a smaller margin of error requires a larger sample size. By controlling the margin of error, researchers ensure that their estimates are reasonably close to the true population value.
Proportion Estimation
Proportion estimation involves calculating the proportion of a certain characteristic within a population. In our case, it is the proportion of American homeowners.
The estimate is usually derived from a sample taken from the population. Proportions can provide insights into population characteristics and are often used in surveys and polls.
When using prior estimates, such as 0.675 in part (a) of our exercise, the formula for sample size can yield more accurate results. However, if no prior estimate is available, assuming a proportion of 0.5 (as in part (b)) is a common practice because it maximizes the required sample size, providing a more conservative estimate.
Sample Size Formula
Determining the right sample size is crucial for obtaining reliable estimates. The formula we used in the exercise is:
\[ n = \frac{{Z_{\frac{\text{伪}}{2}}^2 \times p \times (1-p)}}{E^2} \]
Here鈥檚 a breakdown:
  • \( n \): The required sample size
  • \( Z_{\frac{\text{伪}}{2}} \): The z-value associated with your confidence level (for 90%, \( Z_{\frac{\text{伪}}{2}} \) is 1.645)
  • \( p \): The estimated proportion (0.675 for part (a), 0.5 for part (b))
  • \( 1-p \): The complement of the estimated proportion
  • \( E \): The margin of error (0.02 in our problem)
By substituting the known values into the formula, we calculated the sample sizes for both scenarios, ensuring that our estimates will be within the specified margin of error with the desired level of confidence.

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Most popular questions from this chapter

In a random sample of 100 estate tax returns that was audited by the Internal Revenue Service, it was determined that the mean amount of additional tax owed was \(\$ 3137 .\) Assuming the population standard deviation of the additional amount owed is \(\$ 2694,\) construct and interpret a \(90 \%\) confidence interval for the mean additional amount of tax owed for estate tax returns.

Partial Birth Abortions In an October 2003 poll conducted by the Gallup Organization, 684 of 1006 randomly selected adults aged 18 years old or older stated they think the government should make partial birth abortions illegal, except in cases necessary to save the life of the mother. (a) Obtain a point estimate for the proportion of adults aged 18 or older who think the government should make partial birth abortions illegal, except in cases necessary to save the life of the mother. (b) Verify that the requirements for constructing a confidence interval about \(\hat{p}\) are satisfied. (c) Construct a \(98 \%\) confidence interval for the proportion of adults aged 18 or older who think the government should make partial birth abortions illegal, except in cases necessary to save the life of the mother. Interpret the interval.

A simple random sample of size \(n\) is drawn. The sample mean, \(\bar{x},\) is found to be \(18.4,\) and the sample standard deviation, \(s\), is found to be \(4.5 .\) (a) Construct a \(95 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is 35 (b) Construct a \(95 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(50 .\) How does increasing the sample size affect the margin of error, \(E ?\) (c) Construct a \(99 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(35 .\) Compare the results to those obtained in part (a). How does increasing the level of confidence affect the margin of error, \(E ?\) (d) If the sample size is \(n=15,\) what conditions must be satisfied to compute the confidence interval?

IQ scores as measured by the Stanford-Binet IQ test are normally distributed with \(\mu=100\) and \(\boldsymbol{\sigma}=16\) (a) Simulate obtaining 20 samples of size \(n=15\) from this population. (b) Construct \(95 \%\) confidence intervals for each of the 20 samples. (c) How many of the intervals do you expect to include the population mean? How many actually contain the population mean?

Fifty rounds of a new type of ammunition were fired from a test weapon, and the muzzle velocity of the projectile was measured. The sample had a mean muzzle velocity of 863 meters per second, with a standard deviation of 2.7 meters per second. Construct and interpret a \(99 \%\) confidence interval for the mean muzzle velocity.
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