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Chapter 7: Problem 24

Business: Phone Contact 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) 457 calls; (b) 207 calls.

Step by step solution

01

Understanding the Margin of Error and Confidence Level

To determine the sample size, we need a formula that relates the margin of error (ME), the proportion \( p \), and the confidence level. The formula for sample size \( n \) when estimating a population proportion is given by: \( n = \left( \frac{Z^2 \, p \, (1-p)}{ME^2} \right) \). Here, \( Z \) is the Z-score corresponding to the desired confidence level, \( p \) is the estimated proportion, and \( ME \) is the margin of error which is 0.03 in this problem.
02

Calculate Z-score for 80% Confidence Level

Check a Z-table or use a calculator to find the Z-score that corresponds to an 80% confidence level. For an 80% confidence level, the Z-score is approximately 1.28.
03

Solving Part (a) - No Preliminary Estimate

If there is no preliminary estimate for \( p \), use the most conservative estimate for \( p \), which is 0.5, as it maximizes the product \( p(1-p) \). Substitute \( p = 0.5 \), \( ME = 0.03 \), and \( Z = 1.28 \) into the formula: \[ n = \left( \frac{1.28^2 \times 0.5 \times 0.5}{0.03^2} \right) \approx 456.53\]Round up to the nearest whole number, resulting in \( n = 457 \).
04

Solving Part (b) - Using Estimate of 17%

Use the proportion \( p = 0.17 \) as the national estimate. Substitute \( p = 0.17 \), \( ME = 0.03 \), and \( Z = 1.28 \) into the formula:\[n = \left( \frac{1.28^2 \times 0.17 \times 0.83}{0.03^2} \right) \approx 206.90 \]Round up to the nearest whole number, resulting in \( n = 207 \).

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Key Concepts

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

Confidence Level
In statistics, the confidence level is a crucial concept when making estimates about a population parameter based on a sample data set. It signifies how sure you can be that the true parameter lies within your calculated confidence interval.
For example, in the context of our problem, using an 80% confidence level implies that if you were to take 100 different samples and construct a confidence interval from each one, approximately 80 of those intervals would contain the true proportion of calls successfully reaching a businessperson on the first try.
  • Why 80%? Selecting an 80% confidence level balances the certainty of the interval and the width of the interval. Higher confidence levels like 95% require larger sample sizes but give more assurance. For easier, quicker estimates, 80% works well, as seen in this exercise.
  • Z-Score: The Z-score associated with an 80% confidence level is about 1.28, derived from the standard normal distribution table.
Understanding this concept allows statisticians and researchers to convey statistical certainty and develop meaningful insights from data.
Margin of Error
The margin of error is a measure reflecting the range of values above and below the sample statistic in a confidence interval. A smaller margin of error suggests greater precision in your estimation of the population parameter.
Within the problem exercise, the margin of error is specified as 0.03. This means that the estimated proportion \( \hat{p} \) from the sampled business calls is only allowed to deviate by 0.03 from the true population proportion \( p \).
  • Impact of Margin of Error: A smaller margin of error would need a larger sample size to maintain the same confidence level, whereas a larger margin allows for smaller samples.
  • Application in the Exercise: With a margin of error of 0.03, the sample size calculated must ensure the prediction remains within these limits for reliable decisions.
The margin of error is a vital component, especially in decision-making, as it gives a quantitative measure of the confidence in your sample estimate.
Sample Size
A sample size determines the number of observations or data points gathered from a population to make statistical inferences. The accuracy and reliability of an estimate often rely heavily on choosing an appropriate sample size.
In the given exercise, once we know the confidence level and margin of error, the sample size is derived using the formula \[ n = \left( \frac{Z^2 \, p \, (1-p)}{ME^2} \right) \].
  • Importance of Sample Size: Sample size directly affects the precision and stability of the results. Too small a sample can lead to inaccurate results, while a overly large sample may be unnecessarily costly and time-consuming.
  • No Preliminary Estimate: When no prior estimate of \( p \) exists, a conservative estimate of \( p = 0.5 \) ensures the largest possible \( n \) is calculated, accounting for maximum variability.
  • Using an Estimate: With a prior estimate like 17%, the sample size can be adjusted to be smaller while still maintaining statistical reliability.
Being aware of how sample size affects statistical results leads to better data-driven decision-making, crucial in fields like business analysis.

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Most popular questions from this chapter

Basic Computation: Confidence Interval for \(p_{1}-p_{2}\) Consider two independent binomial experiments. In the first one, 40 trials had 10 successes. In the second one, 50 trials had 15 successes. (a) Check Requirements Is it appropriate to use a normal distribution to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(90 \%\) confidence interval for \(p_{1}-p_{2}\). (c) Interpretation Based on the confidence interval you computed, can you be \(90 \%\) confident that \(p_{1}\) is less than \(p_{2}\) ? Explain.

Basic Computation: Confidence Interval for \(\mu_{1}-\mu_{2}\) Consider two independent distributions that are mound-shaped. A random sample of size \(n_{1}=36\) from the first distribution showed \(\bar{x}_{1}=15\), and a random sample of size \(n_{2}=40\) from the second distribution showed \(\bar{x}_{2}=14\) (a) Check Requirements If \(\sigma_{1}\) and \(\sigma_{2}\) are known, what distribution does \(\bar{x}_{1}-\bar{x}_{2}\) follow? Explain. (b) Given \(\sigma_{1}=3\) and \(\sigma_{2}=4\), find a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (c) Check Requirements Suppose \(\sigma_{1}\) and \(\sigma_{2}\) are both unknown, but from the random samples, you know \(s_{1}=3\) and \(s_{2}=4\). What distribution approximates the \(\bar{x}_{1}-\bar{x}_{2}\) distribution? What are the degrees of freedom? Explain. (d) With \(s_{1}=3\) and \(s_{2}=4\), find a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (e) If you have an appropriate calculator or computer software, find a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using degrees of freedom based on Satterthwaite's approximation. (f) Interpretation Based on the confidence intervals you computed, can you be \(95 \%\) confident that \(\mu_{1}\) is larger than \(\mu_{2} ?\) Explain.

Marketing: Customer Loyalty In a marketing survey, a random sample of 730 women shoppers revealed that 628 remained loyal to their favorite supermarket during the past year (i.e., did not switch stores) (Source: Trends in the United States: Consumer Attitudes and the Supermarket, The Research Department, Food Marketing Institute). (a) Let \(p\) represent the proportion of all women shoppers who remain loyal to their favorite supermarket. Find a point estimate for \(p\). (b) Find a \(95 \%\) confidence interval for \(p .\) Give a brief explanation of the meaning of the interval. (c) Interpretation As a news writer, how would you report the survey results regarding the percentage of women supermarket shoppers who remained loyal to their favorite supermarket during the past year? What is the margin of error based on a \(95 \%\) confidence interval?

Basic Computation: Confidence Interval Suppose \(x\) has a mound-shaped distribution. A random sample of size 16 has sample mean 10 and sample standard deviation \(2 .\) (a) Check Requirements Is it appropriate to use a Student's \(t\) distribution to compute a confidence interval for the population mean \(\mu ?\) Explain. (b) Find a \(90 \%\) confidence interval for \(\mu .\) (c) Interpretation Explain the meaning of the confidence interval you computed.

Answer true or false. Explain your answer. The value \(z_{c}\) is a value from the standard normal distribution such that \(P\left(-z_{c}
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