/*! 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 16 Use the given sample data and co... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use the given sample data and confidence level. In each case, (a) find the best point estimate of the population proportion \(p ;(b)\) identify the value of the margin of error \(E ;(c)\) construct the confidence interval; (d) write a statement that correctly interprets the confidence interval. In a study of 1228 randomly selected medical malpractice lawsuits, it was found that 856 of them were dropped or dismissed (based on data from the Physicians Insurers Association of America). Construct a \(95 \%\) confidence interval for the proportion of medical malpractice lawsuits that are dropped or dismissed.

Short Answer

Expert verified
The point estimate is \(0.696\). The margin of error is \(0.026\). The \(95\%\) confidence interval is \[0.67, 0.722\].

Step by step solution

01

Find the Point Estimate of the Population Proportion

The point estimate for the population proportion, denoted as \(\hat{p}\), is found using the formula \(\hat{p} = \frac{x}{n}\), where \(x\) is the number of successes (lawsuits dropped or dismissed) and \(n\) is the total number of trials (total lawsuits). For our data, \(x = 856\) and \(n = 1228\). Therefore, \(\hat{p} = \frac{856}{1228} = 0.696\).
02

Calculate the Margin of Error

To calculate the margin of error \(E\), use the formula \[E = z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\], where \(z_{\alpha/2}\) is the critical value for the given confidence level. For a \(95\%\) confidence level, \(z_{\alpha/2} \) is approximately \(1.96\). Therefore, \[E = 1.96 \sqrt{\frac{0.696(1-0.696)}{1228}} = 0.026\].
03

Construct the Confidence Interval

The confidence interval is given by \[\hat{p} - E < p < \hat{p} + E\]. Substituting the values found, \[0.696 - 0.026 < p < 0.696 + 0.026\]. Therefore, the confidence interval is \[0.67 < p < 0.722\].
04

Interpret the Confidence Interval

The \(95\%\) confidence interval for the proportion of medical malpractice lawsuits that are dropped or dismissed is \[0.67, 0.722\]. This means that we are \(95\%\) confident that the true proportion of dropped or dismissed lawsuits lies within this interval.

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.

point estimate
The point estimate is a single value that serves as an estimate of an unknown population parameter.
In this case, we are estimating the population proportion (denoted as \(p\)), which is the proportion of medical malpractice lawsuits that are dropped or dismissed.
The point estimate for the population proportion is calculated as \( \hat{p} = \frac{x}{n} \), where \(x\) is the number of lawsuits dropped or dismissed, and \(n\) is the total number of lawsuits.

For our dataset, \(x = 856\) and \(n = 1228\). Thus, the point estimate is \(\hat{p} = \frac{856}{1228} = 0.696\). This point estimate suggests that approximately 69.6% of medical malpractice lawsuits are dropped or dismissed.
margin of error
The margin of error (denoted as \(E\)) provides a measure of the accuracy of the point estimate.
It indicates the range in which the true population parameter is expected to lie, given a certain level of confidence. To calculate the margin of error for a proportion, use the formula \[E = z_{\frac{\alpha}{2}} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\], where \(z_{\frac{\alpha}{2}}\) is the critical value for the desired confidence level.

For a 95% confidence level, \(z_{\frac{\alpha}{2}}\) is approximately 1.96. Plugging in the values, we get \[E = 1.96 \sqrt{\frac{0.696(1-0.696)}{1228}} = 0.026\].
This means the margin of error is 0.026, indicating a 2.6% deviation on either side of the point estimate.
confidence interval interpretation
Interpreting the confidence interval is crucial to understanding the range of possible values for the population proportion \(p\).
The confidence interval is constructed as \[\hat{p} - E < p < \hat{p} + E\]. Using our calculated values: \[0.696 - 0.026 < p < 0.696 + 0.026\].
Therefore, the confidence interval is \[0.67 < p < 0.722\].

This interval tells us that we are 95% confident that the true proportion of medical malpractice lawsuits that are dropped or dismissed lies between 67% and 72.2%. In other words, if we were to take many samples and construct a confidence interval from each, 95% of these intervals would contain the true population proportion.
sample data analysis
Sample data analysis helps us derive meaningful insights about a population based on a subset of that population.
In our scenario, we analyzed a sample of 1228 medical malpractice lawsuits. We used this sample to estimate the true proportion of lawsuits that are dropped or dismissed in the entire population.

Analyzing sample data involves several steps:
  • Calculate a point estimate, which, in our case, is the proportion of dropped or dismissed lawsuits.
  • Determine the margin of error, which gives a sense of the estimate's accuracy.
  • Construct a confidence interval to provide a range of plausible values for the true population parameter.
By understanding these elements, we can make informed decisions and interpretations about the population based on our sample data.

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

Construct the confidence interval estimate of the mean. Listed below are measured amounts of caffeine (mg per 12 oz of drink) obtained in one can from each of 20 brands (7UP, A\&W Root Beer, Cherry Coke, \(\dots,\) TaB). Use a confidence level of \(99 \% .\) Does the confidence interval give us good information about the population of all cans of the same 20 brands that are consumed? Does the sample appear to be from a normally distributed population? If not, how are the results affected? $$\begin{array}{rrrrrrrrrrrrrrrrrrr} 0 & 0 & 34 & 34 & 34 & 45 & 41 & 51 & 55 & 36 & 47 & 41 & 0 & 0 & 53 & 54 & 38 & 0 & 41 & 47 \end{array}$$

Find the sample size required to estimate the population mean. Data Set 1 "Body Data" in Appendix B includes pulse rates of 153 randomly selected adult males, and those pulse rates vary from a low of 40 bpm to a high of 104 bpm. Find the minimum sample size required to estimate the mean pulse rate of adult males. Assume that we want \(99 \%\) confidence that the sample mean is within 2 bpm of the population mean. a. Find the sample size using the range rule of thumb to estimate \(\sigma .\) b. Assume that \(\sigma=11.3\) bpm, based on the value of \(s=11.3\) bpm for the sample of 153 male pulse rates. c. Compare the results from parts (a) and (b). Which result is likely to be better?

Finding Critical Values and Confidence Intervals. In Exercises \(5-8,\) use the given information to find the number of degrees of freedom, the critical values \(\mathcal{X}_{L}^{2}\) and \(\mathcal{X}_{R}^{2},\) and the confidence interval estimate of \(\boldsymbol{\sigma} .\) The samples are from Appendix \(\boldsymbol{B}\) and it is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Platelet Counts of Women \(99 \%\) confidence; \(n=147, s=65.4\)

Use the given sample data and confidence level. In each case, (a) find the best point estimate of the population proportion \(p ;(b)\) identify the value of the margin of error \(E ;(c)\) construct the confidence interval; (d) write a statement that correctly interprets the confidence interval. In a study of cell phone use and brain hemispheric dominance, an Intemet survey was e-mailed to 5000 subjects randomly selected from an online group involved with ears. 717 surveys were returned. Construct a \(90 \%\) confidence interval for the proportion of retumed surveys.

Use the data and confidence level to construct a confidence interval estimate of \(p,\) then address the given question. Before its clinical trials were discontinued, the Genetics \& IVF Institute conducted a clinical trial of the XSORT method designed to increase the probability of conceiving a girl and, among the 945 babies born to parents using the XSORT method, there were 879 girls. The YSORT method was designed to increase the probability of conceiving a boy and, among the 291 babies born to parents using the YSORT method, there were 239 boys. Construct the two \(95 \%\) confidence interval estimates of the percentages of success. Compare the results. What do you conclude?

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.