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How hard is it to reach a businessperson by phone? Let \(p\) be the proportion of calls to businesspeople for which the caller reaches the person being called on the first try. (a) If you have no preliminary estimate for \(p\), how many business phone calls should you include in a random sample to be \(80 \%\) sure that the point estimate \(\hat{p}\) will be within a distance of \(0.03\) from \(p ?\) (b) The Book of Odds, by Shook and Shook (Signet), reports that businesspeople can be reached by a single phone call approximately \(17 \%\) of the time. Using this (national) estimate for \(p\), answer part (a).

Short Answer

Expert verified
(a) At least 456 calls; (b) at least 258 calls.

Step by step solution

01

Understanding the Problem

To solve the problem, we need to determine the sample size required for a certain level of confidence (80%) so that the point estimate, \( \hat{p} \), is within 0.03 of the true proportion \( p \). Part (a) assumes no preliminary estimate of \( p \), while part (b) uses a known estimate of \( p = 0.17 \).
02

Clarify the Necessary Formula

The formula for determining sample size \( n \) when estimating a proportion \( p \) is:\[ n = \left(\frac{Z_{\alpha/2} \cdot \sigma}{E}\right)^2 \]where \( Z_{\alpha/2} \) is the z-score for a given confidence level, \( E \) is the margin of error, and \( \sigma \) is the standard deviation of the sample proportion, which is \( \sqrt{p(1-p)} \). For 80% confidence level, \( Z_{\alpha/2} = 1.28 \).
03

Set Values Without Estimate for p (Part a)

When no preliminary value for \( p \) is available, the conservative approach is to use \( p = 0.5 \) as it maximizes \( \sigma \). Thus, we have:\[ E = 0.03, \quad Z_{\alpha/2} = 1.28, \quad p = 0.5 \]Then the formula for \( n \) becomes:\[ n = \left(\frac{1.28 \times \sqrt{0.5 \cdot 0.5}}{0.03}\right)^2 \]
04

Calculate n Without Preliminary Estimate

Evaluate the calculation step by step:1. Calculate the standard deviation: \( \sqrt{0.5 \times 0.5} = 0.5 \)2. Compute the numerator: \( 1.28 \times 0.5 = 0.64 \)3. Compute \( n \):\[ n = \left(\frac{0.64}{0.03}\right)^2 \approx 455.11 \] Thus, at least 456 calls are needed without a preliminary estimate of \( p \).
05

Set Values with Estimate for p (Part b)

Using a preliminary estimate \( p = 0.17 \), the parameters are:\[ E = 0.03, \quad Z_{\alpha/2} = 1.28, \quad p = 0.17 \]The adjusted \( \sigma \) is \( \sqrt{0.17 \cdot (1 - 0.17)} \).
06

Calculate n With Preliminary Estimate

Calculate the sample size using the estimate:1. Standard deviation: \( \sqrt{0.17 \times 0.83} = \sqrt{0.1411} \approx 0.3756 \)2. Compute the numerator: \( 1.28 \times 0.3756 = 0.480768 \)3. Compute \( n \):\[ n = \left(\frac{0.480768}{0.03}\right)^2 \approx 257.28 \] Thus, if the estimate \( p = 0.17 \) is used, at least 258 calls are needed.

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Key Concepts

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

Confidence Level
In statistical analysis, the confidence level is a key term that measures how certain we are that our sample accurately represents the entire population. Think of it as a "safety net" for our estimates. A confidence level of 80% means that if we were to take 100 different samples, about 80 of them would contain the true proportion, p. Therefore, it's an expression of how confident we are in the results of our study.

When dealing with sample size calculations, this concept helps in determining the z-score, which is derived from the standard normal distribution. For an 80% confidence level, the corresponding z-score is approximately 1.28. The z-score is critical in calculating the margin of error and, ultimately, the needed sample size.

To improve your understanding, remember:
  • The higher the confidence level, the larger the sample size needed, as we need more data to be sure our sample reflects the population.
  • A good balance is essential because overly high confidence levels will demand impractically large sample sizes, whereas too low may result in unreliable results.
Margin of Error
The margin of error provides a range within which we expect the true population parameter to fall. In simpler words, it's an allowance for uncertainty.

The margin of error (E) is directly linked to the confidence level and the variability in the data. In the problem you're studying, the margin of error is set to 0.03, which means that we want the difference between our estimate and the true proportion to be no more than 3%.

Key points about the Margin of Error include:
  • The greater the desired precision (smaller margin of error), the larger the necessary sample size.
  • It balances precision and practicality, as tighter margins mean gathering more data, which might not always be feasible.
  • The formula for calculating margin of error in population proportion scenarios is:
    \[ E = Z_{\alpha/2} \cdot \sigma \]

Remember, choosing the margin of error is about mixing the significance of precision with the reality of resources, time, and other such constraints.
Proportion Estimation
Estimating a population proportion accurately can be understood as calculating what fraction of a group possesses a certain trait. For example, in this exercise, we are estimating the proportion (\( p \)) of businesspeople who can be reached in a phone call on the first attempt.

In the absence of prior knowledge, a conservative guess is often made to ensure the most significant variability, maximizing the needed sample size to get a reliable estimate. This is why the default assumption for \( p \) is set to 0.5 when there's no preliminary estimate. This approach guarantees that whatever the true proportion is, the sample size is sufficient to meet the accuracy requirements.

Substituting known values of \( p \), like the 17% estimate from the problem, allows us to refine our calculations for more precision without overestimating sample size. This mathematical strategy ensures that our resources are used wisely.

Useful facts about proportion estimation:
  • A preliminary estimate helps refine the required sample size, making the calculation more efficient.
  • Using an uncertain estimate like \( p = 0.5 \) is a safeguard in statistical practices to avoid under-sampling.
  • The precision in estimating this proportion is crucial for making valid assumptions and drawing meaningful conclusions about the entire population.

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Most popular questions from this chapter

In order to use a normal distribution to compute confidence intervals for \(p\), what conditions on \(n p\) and \(n q\) need to be satisfied?

Case studies showed that out of 10,351 convicts who escaped from U.S. prisons, only 7867 were recaptured (The Book of Odds, by Shook and Shook, Signet). (a) Let \(p\) represent the proportion of all escaped convicts who will eventually be recaptured. Find a point estimate for \(p\). (b) Find a \(99 \%\) confidence interval for \(p .\) Give a brief statement of the meaning of the confidence interval. (c) Is use of the normal approximation to the binomial justified in this problem? Explain.

Lorraine was in a hurry when she computed a confidence interval for \(\mu .\) Because \(\sigma\) was not known, she used a Student's \(t\) distribution. However, she accidentally used degrees of freedom \(n\) instead of \(n-1 .\) Will her confidence interval be longer or shorter than one found using the correct degrees of freedom \(n-1 ?\) Explain.

(a) Suppose a \(95 \%\) confidence interval for the difference of means contains both positive and negative numbers. Will a \(99 \%\) confidence interval based on the same data necessarily contain both positive and negative numbers? Explain. What about a \(90 \%\) confidence interval? Explain. (b) Suppose a \(95 \%\) confidence interval for the difference of proportions contains all positive numbers. Will a \(99 \%\) confidence interval based on the same data necessarily contain all positive numbers as well? Explain. What about a \(90 \%\) confidence interval? Explain.

In the Focus Problem at the beginning of this chapter, a study was described comparing the hatch ratios of wood duck nesting boxes. Group I nesting boxes were well separated from each other and well hidden by available brush. There were a total of 474 eggs in group I boxes, of which a field count showed about 270 hatched. Group II nesting boxes were placed in highly visible locations and grouped closely together. There were a total of 805 eggs in group II boxes, of which a field count showed about 270 hatched. (a) Find a point estimate \(\hat{p}_{1}\) for \(p_{1}\), the proportion of eggs that hatch in group I nest box placements. Find a \(95 \%\) confidence interval for \(p_{1}\). (b) Find a point estimate \(\hat{p}_{2}\) for \(p_{2}\), the proportion of eggs that hatch in group II nest box placements. Find a \(95 \%\) confidence interval for \(p_{2}\). (c) Find a \(95 \%\) confidence interval for \(p_{1}-p_{2} .\) Does the interval indicate that the proportion of eggs hatched from group I nest boxes is higher than, lower than, or equal to the proportion of eggs hatched from group II nest boxes? (d) What conclusions about placement of nest boxes can be drawn? In the article discussed in the Focus Problem, additional concerns are raised about the higher cost of placing and maintaining group I nest box placements. Also at issue is the cost efficiency per successful wood duck hatch.

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