/*! 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 82 A researcher wants to determine ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 82

A researcher wants to determine a \(99 \%\) confidence interval for the mean number of hours that adults spend per week doing community service. How large a sample should the researcher select so that the estimate is within \(1.2\) hours of the population mean? Assume that the standard deviation for time spent per week doing community service by all adults is 3 hours.

Short Answer

Expert verified
The required sample size, rounded up to the nearest whole number, is calculated to be 35.

Step by step solution

01

Identify the Z-Score for the Desired Confidence Level

The 99% confidence level corresponds to a standard normal distribution (z-distribution) percentile. For a two-sided 99% confidence interval, you would look up the z-value that leaves 0.5% of the distribution's area on each tail. This z-value is also called the critical value. From the standard normal distribution table, or using a z-score calculator, you find that \(Z = 2.575\).
02

Identify the Desired Margin of Error (E)

The margin of error, E, also called tolerance, is the maximum amount you're willing to be wrong. It's the amount that you're willing to deviate from the population mean. It's given as 1.2 hours in this problem.
03

Identify the population standard deviation (蟽)

The standard deviation for time spent per week doing community service by all adults is given as 3 hours.
04

Apply the formula for sample size

The formula for sample size when the population standard deviation is known and you're estimating a population mean is \(n = \left(\frac{Z蟽}{E}\right)^2\). Subsituting the values you have (\(Z = 2.575\), \(蟽 = 3\), \(E = 1.2\)), you calculate: \(n = \left(\frac{2.575*3}{1.2}\right)^2\).
05

Round up to the nearest whole number

Always round up your sample size to the nearest whole number. This is because you can't have a fraction of a person (or item) in your sample.

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 91影视!

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

It is said that happy and healthy workers are efficient and productive. A company that manufactures exercising machines wanted to know the percentage of large companies that provide on-site health club facilities. A random sample of 240 such companies showed that 96 of them provide such facilities on site. a. What is the point estimate of the percentage of all such companies that provide such facilities on site? b. Construct a \(97 \%\) confidence interval for the percentage of all such companies that provide such facilities on site. What is the margin of error for this estimate?

For a data set obtained from a random sample, \(n=81\) and \(\bar{x}=48.25\). It is known that \(\sigma=4.8\). a. What is the point estimate of \(\mu\) ? b. Make a \(95 \%\) confidence interval for \(\mu\). c. What is the margin of error of estimate for part b?

What assumption(s) must hold true to use the normal distribution to make a confidence interval for the population proportion, \(p\) ?

A department store manager wants to estimate the mean amount spent by all customers at this store at a \(98 \%\) confidence level. The manager knows that the standard deviation of amounts spent by all customers at this store is \(\$ 31\). What minimum sample size should he choose so that the estimate is within \(\$ 3\) of the population mean?

An insurance company selected a random sample of 50 auto claims filed with it and investigated those claims carefully. The company found that \(12 \%\) of those claims were fraudulent. a. What is the point estimate of the percentage of all auto claims filed with this company that are fraudulent? b. Make a \(99 \%\) confidence interval for the percentage of all auto claims filed with this company that are fraudulent.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation
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.