/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 A random sample of medical files... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A random sample of medical files is used to estimate the proportion \(p\) of all people who have blood type B. (a) If you have no preliminary estimate for \(p,\) how many medical files should you include in a random sample in order to be \(85 \%\) sure that the point estimate \(\hat{p}\) will be within a distance of 0.05 from \(p ?\) (b) Answer part (a) if you use the preliminary estimate that about 8 out of 90 people have blood type \(\mathrm{B}\) (Reference: Manual of Laboratory and Diagnostic Tests by F. Fischbach).

Short Answer

Expert verified
(a) 83 medical files; (b) 44 medical files.

Step by step solution

01

Understanding the Question

We are asked to determine the sample size needed to estimate a proportion \( p \) with a certain level of confidence and margin of error. This involves using the formula for sample size determination in statistics.
02

Formula for Sample Size Without Preliminary Estimate

When no preliminary estimate is available, the sample size \( n \) can be calculated using the formula: \[ n = \left( \frac{Z^2_{\alpha/2} \cdot 0.25}{E^2} \right) \] where \( Z_{\alpha/2} \) is the Z-value for the desired confidence level (in this case, 85%), and \( E \) is the margin of error (0.05).
03

Find the Critical Value

For an 85% confidence level, the critical value \( Z_{\alpha/2} \) is approximately 1.44. This value can be found in statistical Z-tables or using statistical software.
04

Calculate the Sample Size Without Preliminary Estimate

Using the values in the formula: \[ n = \left( \frac{1.44^2 \cdot 0.25}{0.05^2} \right) \approx 82.944 \]. Hence, you should sample at least 83 medical files.
05

Incorporate the Preliminary Estimate

With a preliminary estimate that 8 out of 90 people have blood type B, \( \hat{p} = \frac{8}{90} \approx 0.0889 \). The sample size formula is then: \[ n = \left( \frac{Z^2_{\alpha/2} \cdot \hat{p}(1-\hat{p})}{E^2} \right) \]
06

Calculate the Sample Size With Preliminary Estimate

Plugging in the known values: \[ n = \left( \frac{1.44^2 \cdot 0.0889 \cdot (1 - 0.0889)}{0.05^2} \right) \approx 43.974 \]. Therefore, you should sample at least 44 medical files.

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Key Concepts

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

Confidence Level
Understanding the confidence level is essential when determining how accurately you want to estimate a parameter, like a proportion for blood type B in a population. Essentially, the confidence level specifies the degree of certainty that the true parameter value will fall within the calculated interval.
- For example, in statistics, an 85% confidence level means you can be 85% confident that your calculated range (or interval) includes the true proportion.
The confidence level you choose affects the sample size required for your study. - Higher confidence levels lead to wider confidence intervals, - Consequently, require larger samples to maintain the same margin of error.
Common confidence levels include 90%, 95%, and 99%. For this exercise, an 85% confidence level was used, reflecting a moderate desire for accuracy and efficiency.
Margin of Error
The margin of error reflects how much you expect your sample estimate might differ from the true population parameter. In simple terms, it tells us the maximum expected difference between the sample proportion and the true population proportion.
- For example, a margin of error of 0.05 means that the estimate is expected to vary by plus or minus 5% from the true value.
Margin of error is critical because it affects the reliability of your results. A smaller margin of error gives more precise results, but:
  • It typically requires a larger sample size.
  • Thus, it might lead to increased time and cost.
The margin of error is chosen based on how much uncertainty you are willing to accept. In this scenario, a margin of error of 0.05 was used, signifying a balance between precision and practicality.
Preliminary Estimate
A preliminary estimate is an initial guess or calculated estimate of the parameter being studied before proceeding with detailed analysis. This serves as a useful baseline.
- Having a preliminary estimate can guide more efficient sample size determination. With an estimated proportion
For example, if previous studies or expert opinions provide you with initial data, you can use these to refine calculations.
In this case, the preliminary estimate suggests that 8 out of 90 people have blood type B, giving a baseline proportion estimate of 0.0889. By incorporating this into the sample size formula, a more accurate and often smaller sample size can be chosen, reducing unnecessary work and costs.
Proportion Estimation
Proportion estimation involves predicting the proportion of a population that possesses a certain characteristic, like having blood type B. This often requires analyzing a sample to infer the population-wide percentage.
- This exercise focuses on determining the proportion of individuals with a specific blood type.
Using statistical formulas, you can calculate the required sample size to ensure that your estimated proportion is reliable and close to the actual value. The formula requires inputs like the margin of error, confidence level, and either a preliminary estimate or a worst-case scenario estimate.
Getting accurate estimates is not just academic; it can impact resource allocation, policy decisions, and strategic planning. Proportion estimation helps simplify and focus complex datasets into actionable insights.

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Most popular questions from this chapter

A requirement for using the normal distribution to approximate the \(\hat{p}\) distribution is that both \(n p > 5\) and \(n q > 5\) since we usually do not know \(p,\) we estimate \(p\) by \(\hat{p}\) and \(q\) by \(\hat{q}=1-\hat{p}\) Then we require that \(n \hat{p}>5\) and \(n \hat{q}>5 .\) Show that the conditions \(n \hat{p}>5\) and \(n \hat{q}>5\) are equivalent to the condition that out of \(n\) binomial trials, both the number of successes \(r\) and the number of failures \(n-r\) must exceed 5 Hint: In the inequality \(n \hat{p}>5,\) replace \(\hat{p}\) by \(r / n\) and solve for \(r .\) In the inequality \(n \hat{q}>5,\) replace \(\hat{q}\) by \((n-r) / n\) and solve for \(n-r\).

Lorraine computed a confidence interval for \(\mu\) based on a sample of size \(41 .\) Since she did not know \(\sigma,\) she used \(s\) in her calculations. Lorraine used the normal distribution for the confidence interval instead of a Student's \(t\) distribution. Was her interval longer or shorter than one obtained by using an appropriate Student's \(t\) distribution? Explain.

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At Community Hospital, the burn center is experimenting with a new plasma compress treatment. A random sample of \(n_{1}=316\) patients with minor burns received the plasma compress treatment. Of these patients, it was found that 259 had no visible scars after treatment. Another random sample of \(n_{2}=419\) patients with minor burns received no plasma compress treatment. For this group, it was found that 94 had no visible scars after treatment. Let \(p_{1}\) be the population proportion of all patients with minor burns receiving the plasma compress treatment who have no visible scars. Let \(p_{2}\) be the population proportion of all patients with minor burns not receiving the plasma compress treatment who have no visible scars. (a) Can a normal distribution be used to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(95 \%\) confidence interval for \(p_{1}-p_{2}\) (c) Explain the meaning of the confidence interval found in part (b) in the context of the problem. Does the interval contain numbers that are all positive? all negative? both positive and negative? At the \(95 \%\) level of confidence, does treatment with plasma compresses seem to make a difference in the proportion of patients with visible scars from minor burns?

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