Chapter 8: Problem 1
Briefly explain the meaning of an estimator and an estimate.
/*! 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}
Learning Materials
Features
Discover
Chapter 8: Problem 1
Briefly explain the meaning of an estimator and an estimate.
All the tools & learning materials you need for study success - in one app.
Get started for free
You want to estimate the percentage of students at your college or university who are satisfied with the campus food services. Briefly explain how you will make such an estimate. Select a sample of 30 students and ask them whether or not they are satisfied with the campus food services. Then calculate the percentage of students in the sample who are satisfied. Using this information, find the confidence interval for the corresponding population percentage. Select your own confidence level.
For each of the following, find the area in the appropriate tail of the \(t\) distribution. a. \(t=-1.302\) and \(d f=42\) b. \(t=2.797\) and \(n=25\) c. \(t=1.397\) and \(n=9\) d. \(t=-2.383\) and \(d f=67\)
The management of a health insurance company wants to know the percentage of its policyholders who have tried alternative treatments (such as acupuncture, herbal therapy, etc.). A random sample of 24 of the company's policyholders were asked whether or not they have ever tried such treatments. The following are their responses. \(\begin{array}{llllllll}\text { Yes } & \text { No } & \text { No } & \text { Yes } & \text { No } & \text { Yes } & \text { No } & \text { No } \\ \text { No } & \text { Yes } & \text { No } & \text { No } & \text { Yes } & \text { No } & \text { Yes } & \text { No } \\ \text { No } & \text { No } & \text { Yes } & \text { No } & \text { No } & \text { No } & \text { Yes } & \text { No }\end{array}\) a. What is the point estimate of the corresponding population proportion? b. Construct a \(99 \%\) confidence interval for the percentage of this company's policyholders who have tried alternative treatments.
A consumer agency wants to estimate the proportion of all drivers who wear seat belts while driving. Assume that a preliminary study has shown that \(76 \%\) of drivers wear seat belts while driving. How large should the sample size be so that the \(99 \%\) confidence interval for the population proportion has a margin of error of \(.03\) ?
Determine the most conservative sample size for the estimation of the population proportion for the following. a. \(E=.025\), confidence level \(=95 \%\) b. \(E=.05, \quad\) confidence level \(=90 \%\) c. \(E=.015\), confidence level \(=99 \%\)
What do you think about this solution?
We value your feedback to improve our textbook solutions.