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Basic Computation: Sample Size What is the minimal sample size needed for a \(99 \%\) confidence interval to have a maximal margin of error of \(0.06\) (a) if a preliminary estimate for \(p\) is \(0.8\) ? (b) if there is no preliminary estimate for \(p ?\)

Short Answer

Expert verified
(a) 295 samples are needed; (b) 738 samples are needed.

Step by step solution

01

Understanding Margin of Error Formula

The formula for the margin of error (E) in a confidence interval for a proportion is given by \[ E = z \sqrt{\frac{p(1-p)}{n}} \]where \(z\) is the z-score corresponding to the desired confidence level, \(p\) is the sample proportion, and \(n\) is the sample size. For a 99% confidence level, the z-score is approximately 2.576.
02

Solving for Sample Size with Given Estimate

Type (a) uses the given estimate for \(p\). Plug \(E = 0.06\), \(z = 2.576\), and \(p = 0.8\) into the formula:\[ 0.06 = 2.576 \sqrt{\frac{0.8 \cdot (1 - 0.8)}{n}} \]Squaring both sides, we get:\[ 0.0036 = \frac{6.635776 \times 0.16}{n} \]\[ n = \frac{6.635776 \times 0.16}{0.0036} \]\[ n \approx 294.51 \]Since \(n\) must be a whole number, round up to \(n = 295\).
03

Solving for Sample Size without Estimate

Type (b) assumes no preliminary estimate for \(p\), meaning the maximum product of \(p(1-p)\) occurs when \(p = 0.5\). Plug \(E = 0.06\), \(z = 2.576\), and \(p = 0.5\) into the formula:\[ 0.06 = 2.576 \sqrt{\frac{0.5 \cdot (1 - 0.5)}{n}} \]Squaring both sides, we have:\[ 0.0036 = \frac{6.635776 \cdot 0.25}{n} \]\[ n = \frac{6.635776 \cdot 0.25}{0.0036} \]\[ n \approx 737.29 \]Round up to \(n = 738\).

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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 estimates a parameter with a certain level of confidence. Imagine you are surveying a city to understand what proportion of the population favors building a new park. Since surveying the entire population is impractical, you survey a smaller sample and calculate a confidence interval to estimate the true population proportion.
The confidence interval tells us how likely it is that the interval contains the true population parameter. For example, a 99% confidence interval means that we expect 99% of such intervals to contain the real proportion if the survey is repeated numerous times. Higher confidence levels, like 99%, result in wider intervals, implying a trade-off between confidence and precision.
In the problem, we are using a 99% confidence interval, which influences the z-score in our computations. A higher confidence level increases the z-score, affecting both the width of the interval and the required sample size.
Margin of Error
The margin of error quantifies the uncertainty around the point estimate in our sample statistics. It describes how close the sample proportion is to the true population proportion. A smaller margin of error gives more precise estimates. However, achieving a smaller margin of error often requires a larger sample size.
In mathematical terms, the margin of error (E) for a proportion is given by the formula: \[ E = z \sqrt{\frac{p(1-p)}{n}} \]where:
  • \(E\) is the margin of error,
  • \(z\) is the z-score corresponding to the confidence level,
  • \(p\) is the estimated proportion, and
  • \(n\) is the sample size.
In the exercise, we see a margin of error of 0.06. When the sample size is determined to achieve this margin of error, the values of \(p\) (either from a preliminary estimate or assumed as 0.5) and \(z\) (2.576 for a 99% confidence level) are crucial to get accurate results.
Preliminary Estimate of Proportion
A preliminary estimate of proportion, denoted by \(p\), is an initial guess or previous study result used in the margin of error formula. This estimate helps in calculating the sample size needed for a given confidence level and margin of error. Having a preliminary estimate can make a significant difference.
For example, in the exercise, we have two scenarios: with a preliminary estimate \(p = 0.8\) and without one.
  • When a preliminary estimate of \(p = 0.8\) is available, you use it directly in calculations to determine the sample size more precisely.
  • Without a preliminary estimate, the most conservative approach is to use \(p = 0.5\), as it maximizes the product \(p(1-p)\), ensuring enough sample size for any possible proportion value.
Using a preliminary estimate of \(p\) can help optimize the sample size, making the data collection process more efficient and cost-effective.

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Most popular questions from this chapter

Finance: \(\mathrm{P} / \mathrm{E}\) Ratio The price of a share of stock divided by a company's estimated future earnings per share is called the P/E ratio. High P/E ratios usually indicate "growth" stocks, or maybe stocks that are simply overpriced. Low \(\mathrm{P} / \mathrm{E}\) ratios indicate "value" stocks or bargain stocks. A random sample of 51 of the largest companies in the United States gave the following P/E ratios (Reference: Forbes). $$ \begin{array}{rrrrrrrrrrrr} 11 & 35 & 19 & 13 & 15 & 21 & 40 & 18 & 60 & 72 & 9 & 20 \\ 29 & 53 & 16 & 26 & 21 & 14 & 21 & 27 & 10 & 12 & 47 & 14 \\ 33 & 14 & 18 & 17 & 20 & 19 & 13 & 25 & 23 & 27 & 5 & 16 \\ 8 & 49 & 44 & 20 & 27 & 8 & 19 & 12 & 31 & 67 & 51 & 26 \\ 19 & 18 & 32 & & & & & & & & & \end{array} $$ (a) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x} \approx 25.2\) and \(s \approx 15.5\). (b) Find a \(90 \%\) confidence interval for the \(\mathrm{P} / \mathrm{E}\) population mean \(\mu\) of all large U.S. companies. (c) Find a \(99 \%\) confidence interval for the \(\mathrm{P} / \mathrm{E}\) population mean \(\mu\) of all large U.S. companies. (d) Interpretation Bank One (now merged with J.P. Morgan) had a P/E of 12 , AT\&T Wireless had a \(\mathrm{P} / \mathrm{E}\) of 72 , and Disney had a \(\mathrm{P} / \mathrm{E}\) of 24 . Examine the confidence intervals in parts (b) and (c). How would you describe these stocks at the time the sample was taken? (e) Check Requirements In previous problems, we assumed the \(x\) distribution was normal or approximately normal. Do we need to make such an assumption in this problem? Why or why not? Hint: See the central limit theorem in Section \(6.5\).

Jerry tested 30 laptop computers owned by classmates enrolled in a large computer science class and discovered that 22 were infected with keystroke- tracking spyware. Is it appropriate for Jerry to use his data to estimate the proportion of all laptops infected with such spyware? Explain.

Answer true or false. Explain your answer. If the sample mean \(\bar{x}\) of a random sample from an \(x\) distribution is relatively small, then the confidence interval for \(\mu\) will be relatively short.

Suppose \(x\) has a mound-shaped distribution with \(\sigma=9\). A random sample of size 36 has sample mean 20 . (a) Check Requirements Is it appropriate to use a normal distribution to compute a confidence interval for the population mean \(\mu\) ? Explain. (b) Find a \(95 \%\) confidence interval for \(\mu .\) (c) Interpretation Explain the meaning of the confidence interval you computed

Health Care: Colorado Physicians A random sample of 5792 physicians in Colorado showed that 3139 provide at least some charity care (i.e., treat poor people at no cost). These data are based on information from State Health Care Data: Utilization, Spending, and Characteristics (American Medical Association). (a) Let \(p\) represent the proportion of all Colorado physicians who provide some charity care. Find a point estimate for \(p\). (b) Find a \(99 \%\) confidence interval for \(p .\) Give a brief explanation of the meaning of your answer in the context of this problem. (c) Check Requirements Is the normal approximation to the binomial justified in this problem? Explain.

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