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Chapter 8: Problem 49

How many businesses fail? A study is planned to estimate the proportion of businesses started in the year 2006 that had failed within five years of their start-up. How large a sample size is needed to guarantee estimating this proportion correct to within a. 0.10 with probability \(0.95 ?\) b. 0.05 with probability \(0.95 ?\) c. 0.05 with probability \(0.99 ?\) d. Compare sample sizes for parts a and \(\mathrm{b},\) and \(\mathrm{b}\) and \(\mathrm{c}\), and summarize the effects of decreasing the margin of error and increasing the confidence level.

Short Answer

Expert verified
a. 97; b. 385; c. 664. Larger sample size is needed for lower errors and higher confidence.

Step by step solution

01

Understanding the Formula

The formula for the sample size needed to estimate a proportion with a specified confidence level and margin of error is given by:\[ n = \left( \frac{Z^2 \cdot p \cdot (1-p)}{E^2} \right) \]where \( n \) is the sample size, \( Z \) is the Z-value from the standard normal distribution for the given confidence level, \( p \) is the estimated proportion (if unknown, use 0.5 for maximum variability), and \( E \) is the margin of error.
02

Calculate Sample Size for Part a

For part a, the margin of error \( E \) is 0.10, and the confidence level is 95%, so we use the Z-value for 95%, which is 1.96. Assume \( p = 0.5 \) for maximum variability.Plug into the formula:\[ n = \left( \frac{1.96^2 \times 0.5 \times 0.5}{0.10^2} \right) \approx 96.04 \]Thus, the sample size needed is approximately 97.
03

Calculate Sample Size for Part b

For part b, the margin of error \( E \) is 0.05, and the confidence level remains at 95% with a Z-value of 1.96. Assume \( p = 0.5 \) for maximum variability.Using the formula:\[ n = \left( \frac{1.96^2 \times 0.5 \times 0.5}{0.05^2} \right) \approx 384.16 \]Thus, the sample size required is approximately 385.
04

Calculate Sample Size for Part c

For part c, the margin of error \( E \) is 0.05, and the confidence level is 99% with a Z-value of 2.576. Assume \( p = 0.5 \) for maximum variability.Using the formula:\[ n = \left( \frac{2.576^2 \times 0.5 \times 0.5}{0.05^2} \right) \approx 663.57 \]Therefore, the sample size required is approximately 664.
05

Comparing Sample Sizes and Effects

Comparing the calculated sample sizes: - Part a requires 97 samples, Part b requires 385. - Part b requires 385 samples, Part c requires 664. Decreasing the margin of error from 0.10 to 0.05 increases the sample size considerably, showcasing that a closer estimate requires more data. Increasing the confidence level from 95% to 99% also increases the sample size, illustrating the trade-off between certainty and required data.

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Key Concepts

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

Margin of Error
The margin of error is a critical aspect of sample size calculations. It measures the amount of random sampling error in a survey's results. In simpler terms, it's an expression of the amount by which the estimated proportion might differ from the true population proportion.
It is represented by the symbol \( E \) in the formula. A smaller margin of error means the estimate is more precise, but it also requires a larger sample size.
  • For example, in the exercise, a margin of error of 0.10 indicates that the estimate could be off by 10% in either direction.
  • Reducing the margin of error to 0.05 narrows this gap, demanding a larger sample to maintain accuracy.
Understanding the margin of error helps to determine how closely an estimate will reflect the actual population parameter. Smaller margins suggest greater precision but come at a cost of needing more data.
Confidence Level
The confidence level reflects how confident one is that the true population parameter lies within the confidence interval calculated from the sample data. It is usually expressed as a percentage, such as 95% or 99%. A higher confidence level indicates more certainty about the estimate's accuracy but typically requires larger sample sizes.
  • For instance, a 95% confidence level means that if we were to draw 100 different samples, we expect that 95 out of those 100 samples would produce a confidence interval that includes the true proportion.
  • In the exercise scenarios, the confidence level of 95% corresponds to a Z-value of 1.96, while at 99%, it corresponds to a Z-value of 2.576.
Choosing the right confidence level is crucial as it informs how much trust you place in the results, with higher confidence levels requiring more data to support that increased certainty.
Z-value
The Z-value, stemming from the standard normal distribution, is pivotal in determining both the width of the confidence interval and the necessary sample size. It corresponds directly to the desired confidence level and quantifies the number of standard deviations a data point is from the mean.
  • For example, a Z-value of 1.96 is used for a 95% confidence level, meaning that the interval spans approximately 1.96 standard deviations from the mean on both sides.
  • Similarly, a confidence level of 99% involves a Z-value of 2.576, reflecting a wider interval and demanding more data for accuracy.
The Z-value is a key component in statistical calculations as it allows researchers to set the confidence level and, in tandem, affects how much data is needed to achieve the intended precision with confidence.
Proportion Estimation
Proportion estimation involves predicting the probability or fraction of a population possessing a particular characteristic. When the true proportion is unknown, as often the case in initial studies, an estimated proportion \( p \) of 0.5 is generally used to provide the most conservative sample size estimate.
  • The choice of 0.5 results in the largest possible sample size because it represents the highest variability scenario.
  • In the exercise, assuming \( p = 0.5 \) is strategic, ensuring the sample size is sufficient even if the actual proportion varies.
This approach helps guarantee that the sample size will be adequate under most circumstances. Understanding how to estimate proportions accurately is essential for producing reliable and valid statistical analyses.

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Most popular questions from this chapter

Alleviate PMS? A pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome). In the first stages of a clinical trial, it was successful for 7 out of 10 women. a. Construct an appropriate \(95 \%\) confidence interval for the population proportion. b. Is it plausible that it's successful for only half the population? Explain.

Multiple choice: CI property 2 Other things being equal, increasing \(n\) causes the margin of error of a confidence interval to (a) increase, (b) decrease, (c) stay the same.

Median as point estimate When the population distribution is normal, the population mean equals the population median. How good is the sample median as a point estimate of this common value? For a random sample, the estimated standard error of the sample median equals \(1.25(s / \sqrt{n}) .\) If the population is normal, explain why the sample mean tends to be a better estimate than the sample median.

Working mother In response to the statement on a recent General Social Survey, "A preschool child is likely to suffer if his or her mother works," suppose the response categories (strongly agree, agree, disagree, strongly disagree) had counts \((104,370,665,169) .\) Scores (2,1,-1,-2) were assigned to the four categories, to treat the variable as quantitative. Software reported a. Explain what this choice of scoring assumes about relative distances between categories of the scale. b. Based on this scoring, how would you interpret the sample mean of \(-0.1261 ?\) c. Explain how you could also make an inference about proportions for these data.

Grandmas using e-mail For the question about e-mail in the previous exercise, suppose seven females in the GSS sample of age at least 80 had the responses $$ 0,0,1,2,5,7,14 $$ a. Using software or a calculator, find the sample mean and standard deviation and the standard error of the sample mean. b. Find and interpret a \(90 \%\) confidence interval for the population mean. c. Explain why the population distribution may be skewed right. If this is the case, is the interval you obtained in part b useless, or is it still valid? Explain.
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