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What percentage of your campus student body is female? Let \(p\) be the proportion of women students on your campus. (a) If no preliminary study is made to estimate \(p,\) how large a sample is needed to be \(99 \%\) sure that a point estimate \(\hat{p}\) will be within a distance of 0.05 from \(p ?\) (b) The Statistical Abstract of the United States, 1 12th edition, indicates that approximately \(54 \%\) of college students are female. Answer part (a) using this estimate for \(p\).

Short Answer

Expert verified
(a) 665; (b) 607.

Step by step solution

01

Understanding the Problem

To solve the problem, we need to determine the sample size required to estimate the proportion of female students on campus with 99% confidence and a margin of error of 0.05. We have two scenarios: one without an initial guess for proportion (part a) and one with an initial guess (part b).
02

Formula for Sample Size

The general formula for determining sample size for estimating a proportion is: \[ n = \left(\frac{Z^2 \cdot p \cdot (1-p)}{E^2}\right) \]where \(Z\) is the Z-score corresponding to the desired confidence level, \(p\) is the estimated proportion, and \(E\) is the margin of error.
03

Calculate Z-score for 99% Confidence

For a 99% confidence level, the Z-score \(Z\) is approximately 2.576, as retrieved from Z-tables.
04

Solve Part (a) Without a Preliminary Estimate

Without a prior estimate of \(p\), we assume \(p = 0.5\) to maximize the product \(p(1-p)\). Plugging into the formula:\[ n = \left(\frac{2.576^2 \times 0.5 \times (1-0.5)}{0.05^2}\right) \approx 664.64 \]We round up to ensure the sample size is adequate, so \(n = 665\).
05

Solve Part (b) Using Preliminary Estimate

With a preliminary estimate \(p = 0.54\), the formula becomes:\[ n = \left(\frac{2.576^2 \times 0.54 \times (1-0.54)}{0.05^2}\right) \approx 606.89 \]Rounding up gives \(n = 607\).
06

Conclusion

For part (a), without an initial estimate, a sample size of 665 is needed. For part (b), using the estimate that 54% of students are female, the sample size needed is 607.

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Key Concepts

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

Sample size calculation
Calculating the right sample size is crucial for conducting accurate statistical analyses, especially when estimating proportions. The sample size is basically the number of observations or respondents needed to achieve reliable results with a certain level of confidence. Let's imagine you're trying to determine the proportion of female students on a college campus. To ensure your findings are trustworthy, you must start with a proper sample size calculation.
This process often utilizes a specific formula that takes into account:
  • The desired confidence level (how sure you want to be about your results).
  • The margin of error (how precise you want the estimate to be).
  • An estimate of the proportion (if available).
For instance, to calculate the sample size for a 99% confidence level and a margin of error of 0.05, you use the formula: \[ n = \left(\frac{Z^2 \cdot p(1-p)}{E^2}\right) \] Where:
  • Z = Z-score (for 99% confidence, Z is 2.576).
  • p = estimated proportion, often assumed as 0.5 if no prior estimate is available, to maximize variability.
  • E = margin of error (in this scenario, 0.05).
If no preliminary estimate is available, assuming a proportion of 0.5, you'll need around 665 samples as calculated in the exercise. However, if you have a preliminary estimate, like 0.54, the required sample reduces to 607. This is because knowing more about the proportion can slightly change the variability expectations, reducing uncertainty.
Confidence interval
Confidence intervals provide a range of values within which the real proportion is expected to fall. When you calculate a confidence interval, you're essentially expressing how sure you are about the estimate of the population parameter. Imagine you want to estimate the proportion of female students at a college; a confidence interval will help determine a range where the actual proportion probably lies.
To calculate a confidence interval for proportion estimation, you'll use: \[ \hat{p} \pm Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]
Where:
  • \(\hat{p}\) is the estimated proportion of female students.
  • Z is the Z-score that corresponds to the desired level of confidence (such as 2.576 for 99% confidence).
  • n is the sample size.
For instance, if your sample's estimated proportion is 0.54 and your sample size is 607 with 99% confidence, the interval will provide a range that likely contains the true proportion of female students on campus. The broader your confidence interval, the less precise your estimate, while a narrower interval suggests more precision. The width of this interval reflects the level of certainty you have about the population parameter your sample estimate is predicting.
Proportion estimation
Proportion estimation plays a critical role in determining how common a particular attribute is within a population. When you wish to find out what percentage of students in a college is female, you're essentially trying to estimate a proportion. Here, proportion is represented by \(p\).
To calculate this, you follow a simple ratio of the number of successes (e.g., female students) to the total number in the sample. This gives rise to what statisticians call \(\hat{p}\), the sample proportion. It not only estimates the population proportion but also provides a groundwork for further statistical analysis like building confidence intervals or conducting hypothesis tests. Closing in on a reliable estimate involves several steps:
  • First, determine the size of your sample. This is crucial because a larger sample size generally offers a more accurate reflection of the true population proportion.
  • Then, collect data carefully to avoid biases.
  • Calculate the sample proportion \(\hat{p}\) using the formula: \[\hat{p} = \frac{x}{n}\] Where \(x\) is the number of favorable outcomes (e.g., female students) and \(n\) is the total sample size.
Remember, while \(\hat{p}\) serves as the best estimate of \(p\), it remains susceptible to sampling variability, which can be mitigated by using an appropriate sample size and calculating confidence intervals.

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Most popular questions from this chapter

Consider \(n=100\) binomial trials with \(r=30\) successes. (a) Is it appropriate to use a normal distribution to approximate the \(\hat{p}\) distribution? (b) Find a \(90 \%\) confidence interval for the population proportion of successes \(p\). (c) Explain the meaning of the confidence interval you computed.

In a survey of 1000 large corporations, 250 said that, given a choice between a job candidate who smokes and an equally qualified nonsmoker, the nonsmoker would get the job (USA Today). (a) Let \(p\) represent the proportion of all corporations preferring a nonsmoking candidate. Find a point estimate for \(p\). (b) Find a 0.95 confidence interval for \(p\). (c) As a news writer, how would you report the survey results regarding the proportion of corporations that hire the equally qualified nonsmoker? What is the margin of error based on a \(95 \%\) confidence interval?

If a \(90 \%\) confidence interval for the difference of proportions contains some positive and some negative values, what can we conclude about the relationship between \(p_{1}\) and \(p_{2}\) at the \(90 \%\) confidence level?

Jobs and productivity! How do banks rate? One way to answer this question is to examine annual profits per employee. Forbes Top Companies, edited by J. T. Davis (John Wiley \& Sons), gave the following data about annual profits per employee (in units of 1 thousand dollars per employee) for representative companies in financial services. Companies such as Wells Fargo, First Bank System, and Key Banks were included. Assume \(\sigma \approx 10.2\) thousand dollars. $$\begin{array}{lllllllllll} 42.9 & 43.8 & 48.2 & 60.6 & 54.9 & 55.1 & 52.9 & 54.9 & 42.5 & 33.0 & 33.6 \\ 36.9 & 27.0 & 47.1 & 33.8 & 28.1 & 28.5 & 29.1 & 36.5 & 36.1 & 26.9 & 27.8 \\ 28.8 & 29.3 & 31.5 & 31.7 & 31.1 & 38.0 & 32.0 & 31.7 & 32.9 & 23.1 & 54.9 \\ 43.8 & 36.9 & 31.9 & 25.5 & 23.2 & 29.8 & 22.3 & 26.5 & 26.7 \end{array}$$ (a) Use a calculator or appropriate computer software to verify that, for the preceding data, \(\bar{x} \approx 36.0\) (b) Let us say that the preceding data are representative of the entire sector of (successful) financial services corporations. Find a \(75 \%\) confidence interval for \(\mu,\) the average annual profit per employec for all successful banks. (c) Interpretation Let us say that you are the manager of a local bank with a large number of employees. Suppose the annual profits per employee are less than 30 thousand dollars per employee. Do you think this might be somewhat low compared with other successful financial institutions? Explain by referring to the confidence interval you computed in part (b). (d) Interpretation Suppose the annual profits are more than 40 thousand dollars per employee. As manager of the bank, would you feel somewhat better? Explain by referring to the confidence interval you computed in part (b). (e) Repeat parts (b), (c), and (d) for a \(90 \%\) confidence level.

A random sample is drawn from a population with \(\sigma=12 .\) The sample mean is 30 (a) Compute a \(95 \%\) confidence interval for \(\mu\) based on a sample of size 49 What is the value of the margin of error? (b) Compute a \(95 \%\) confidence interval for \(\mu\) based on a sample of size \(100 .\) What is the value of the margin of error? (c) Compute a \(95 \%\) confidence interval for \(\mu\) based on a sample of size \(225 .\) What is the value of the margin of error? (d) Compare the margins of error for parts (a) through (c). As the sample size increases, does the margin of error decrease? (e) Critical Thinking Compare the lengths of the confidence intervals for parts (a) through (c). As the sample size increases, does the length of a \(90 \%\) confidence interval decrease?

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