/*! 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 Medical: Blood Type A random sam... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Medical: Blood Type 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 B (Reference: Manual of Laboratory and Diagnostic Tests by F. Fischbach).

Short Answer

Expert verified
Without an estimate, 208 files are needed; with a preliminary estimate, 52 files are sufficient.

Step by step solution

01

Identify Requirements

We aim to find the sample size \( n \) needed to estimate the proportion \( p \) of people with blood type B, ensuring the point estimate \( \hat{p} \) is within 0.05 of \( p \) with 85% confidence. Without an initial estimate, use \( p = 0.5 \) for maximum variability.
02

Determine Critical Value

The confidence level is 85%, which corresponds to a confidence interval of [0.075, 0.925]. Using the standard normal distribution, find \( z \) such that 85% of the area lies between \( -z \) and \( z \), resulting in \( z \approx 1.44 \).
03

Calculate Sample Size Without Preliminary Estimate

Use the formula for sample size: \[ n = \left( \frac{z^2 \cdot p \cdot (1-p)}{E^2} \right) \] With \( z = 1.44 \), \( p = 0.5 \), and error \( E = 0.05 \), calculate: \[ n = \left( \frac{1.44^2 \times 0.5 \times 0.5}{0.05^2} \right) \approx 207.36 \] Round up to get \( n \approx 208 \).
04

Compute Preliminary Estimate

Given 8 out of 90 people have blood type B, calculate the preliminary estimate: \[ \hat{p} = \frac{8}{90} = 0.0889 \]
05

Calculate Sample Size With Preliminary Estimate

Use the sample size formula with preliminary estimate: \[ n = \left( \frac{z^2 \cdot \hat{p} \cdot (1-\hat{p})}{E^2} \right) \] Substitute \( \hat{p} = 0.0889 \), \( z = 1.44 \), and \( E = 0.05 \): \[ n = \left( \frac{1.44^2 \times 0.0889 \times 0.9111}{0.05^2} \right) \approx 51.8 \] Round up to get \( n \approx 52 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Sample Size Calculation
Calculating the sample size is an essential step in statistics when you want to estimate a population parameter with a specified level of confidence and precision. When trying to determine how large your sample needs to be, especially when you have no preliminary estimate, you can use a conservative approach by assuming the worst-case scenario. This typically involves assuming that the proportion parameter, denoted by \(p\), is 0.5 because it maximizes the variability, thus requiring the largest sample size to achieve a certain margin of error and confidence level.

This ensures that your sample size is sufficient regardless of the true proportion. To determine this, you can use the formula: \[ n = \left( \frac{z^2 \cdot p \cdot (1-p)}{E^2} \right) \] where \(z\) is the z-value corresponding to your desired confidence level, \(p\) is the estimated proportion, and \(E\) is the margin of error you are comfortable with. For example, if you want your point estimate to be within 0.05 of the actual proportion with 85% confidence, you substitute \(z\approx 1.44\) for an 85% confidence interval, \(p = 0.5\), and \(E = 0.05\) into the formula. The result informs you about how many sample observations are necessary to adequately estimate the population proportion.
Confidence Interval
A confidence interval represents a range of values within which you expect the true population parameter to fall, based on your sample data. This interval gives an estimated range constructed using a point estimate and margin of error.

Datewise, if you repeatedly draw samples and calculate the confidence interval for each sample, a significant percentage of those intervals will contain the true population parameter. The percentage of these intervals covering the true parameter is known as the confidence level.

For example, with an 85% confidence level, you want to ensure that if you sampled multiple times, 85 out of 100 confidence intervals would contain the actual population mean or proportion. The z-value corresponding to 85% confidence level usually approximates to 1.44. Using the z-value alongside your sample's data helps you determine how wide your confidence interval should be, giving you a sense of how sure you can be about your estimation. The equation used to calculate the confidence interval is:
\[ \text{Point Estimate} \pm z \times \text{Standard Error} \] where the standard error is based on the variability of your data.
Point Estimate
The point estimate is a single value that serves as the best approximation of an unknown population parameter based on sample data. In the context of proportion estimation, a point estimate is commonly represented as \( \hat{p} \), which is calculated from your sample data.

For instance, if you are estimating the proportion of individuals with a certain characteristic, such as having blood type B, you would count the number of individuals with blood type B in your sample and divide by the total number of people in the sample.

If the sample provides 8 individuals with blood type B out of a sample size of 90, your point estimate \( \hat{p} \) becomes \( \frac{8}{90} = 0.0889 \). This number is your best guess at what the true proportion would be in the entire population.

Point estimates are instrumental because they allow you to apply additional statistical techniques, like creating confidence intervals or conducting hypothesis testing, to further understand the parameter in question.
Proportion Estimation
Estimating the proportion of a population that possesses a certain characteristic is a fundamental task in statistics. Proportion estimation involves taking a sample from a population and using the observed characteristics to infer about the broader group.

This type of estimation is critical, especially in cases where data about the entire population is not feasible or easy to collect. For example, when estimating how many people have a specific blood type, you collect a sample and calculate the proportion of that sample exhibiting the characteristic.

The formal calculation involves dividing the number of occurrences of the characteristic in your sample by the total number of items in your sample, forming the previously mentioned point estimate \( \hat{p} \). Ensuring your sample is random and large enough based on the desired margin of error and confidence level makes your proportion estimate more robust and reflective of the true population proportion. This approach enables data-driven decision-making, as you can better understand the true parameter by factoring in natural variability and sample size calculation. The insights gained from such estimations support predictions, quality control processes, and strategic planning across various disciplines.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

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 \(6.5\).

Myers-Briggs: Marriage Counseling Isabel Myers was a pioneer in the study of personality types. She identified four basic personality preferences, which are described at length in the book A Guide to the Development and Use of the Myers-Briggs Type Indicator by Myers and McCaulley (Consulting Psychologists Press). Marriage counselors know that couples who have none of the four preferences in common may have a stormy marriage. Myers took a random sample of 375 married couples and found that 289 had two or more personality preferences in common. In another random sample of 571 married couples, it was found that only 23 had no preferences in common. Let \(p_{1}\) be the population proportion of all married couples who have two or more personality preferences in common. Let \(p_{2}\) be the population proportion of all married couples who have no personality perferences in common. (a) Check Requirements Can a normal distribution be used to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(99 \%\) confidence interval for \(p_{1}-p_{2}\). (c) Interpretation Explain the meaning of the confidence interval in part (a) in the context of this problem. Does the confidence interval contain all positive, all negative, or both positive and negative numbers? What does this tell you (at the \(99 \%\) confidence level) about the proportion of married couples with two or more personality preferences in common compared with the proportion of married couples sharing no personality preferences in common?

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).

Assume that the population of \(x\) values has an approximately normal distribution. Diagnostic Tests: Total Calcium Over the past several months, an adult patient has been treated for tetany (severe muscle spasms). This condition is associated with an average total calcium level below \(6 \mathrm{mg} / \mathrm{dl}\) (Reference: Manual of Laboratory and Diagnostic Tests by F. Fischbach). Recently, the patient's total calcium tests gave the following readings (in \(\mathrm{mg} / \mathrm{d}\) l). $$ \begin{array}{rrrrrrr} 9.3 & 8.8 & 10.1 & 8.9 & 9.4 & 9.8 & 10.0 \\ 9.9 & 11.2 & 12.1 & & & & \end{array} $$ (a) Use a calculator to verify that \(\bar{x}=9.95\) and \(s \approx 1.02\). (b) Find a \(99.9 \%\) confidence interval for the population mean of total calcium in this patient's blood. (c) Interpretation Based on your results in part (b), does it seem that this patient still has a calcium deficiency? Explain.

For all these problems, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding. In order to use a normal distribution to compute confidence intervals for \(p\), what conditions on \(n p\) and \(n q\) need to be satisfied?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.