/*! 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 19 For a population data set, \(\si... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 19

For a population data set, \(\sigma=12.5\). a. How large a sample should be selected so that the margin of error of estimate for a \(99 \%\) confidence interval for \(\mu\) is \(2.50\) ? b. How large a sample should be selected so that the margin of error of estimate for a \(96 \%\) confidence interval for \(\mu\) is \(3.20 ?\)

Short Answer

Expert verified
For a margin of error of 2.50 at a 99% confidence interval, a sample size of 107 is necessary. For a margin of error of 3.20 at a 96% confidence interval, a sample size of 42 is needed.

Step by step solution

01

Solve for sample size at 99% confidence level

First, determine the Z-score for a 99% confidence level. This Z-score is approximately 2.576. Then, substitute these values into the formula to solve for \( n \): \( E = 2.50 = 2.576 \cdot \frac{12.5}{\sqrt{n}} \). This can be rearranged to \( n = \left(\frac{2.576 \cdot 12.5}{2.50}\right)^2 \)
02

Calculate the sample size

Perform the mathematical operations: \( n = \left(\frac{2.576 \cdot 12.5}{2.50}\right)^2 = 106.57 \). However, since the sample size should be a whole number, we round it up to the nearest whole, getting \( n = 107 \). The rounding up is done to ensure the margin of error does not exceed 2.50. This means with a sample size of 107, one can be 99% confident that the estimate for \( \mu \) will have a margin of error of at most 2.50.
03

Solve for the sample size at 96% confidence level

For 96% confidence level, the Z-score is approximately 2.054. As such, substitute these into the formula, \( E = 3.20 = 2.054 \cdot \frac{12.5}{\sqrt{n}} \). Solving for \( n \) gives, \( n = \left(\frac{2.054 \cdot 12.5}{3.20}\right)^2 \)
04

Calculate the sample size

Calculate \( n \) by simplifying the equation: \( n = \left(\frac{2.054 \cdot 12.5}{3.20}\right)^2 = 41.23 \). After rounding up to the nearest whole number, we get \( n = 42 \). So with a sample size of 42, one can be 96% confident that the estimate for \( \mu \) will have a margin of error of at most 3.20.

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

A Centers for Disease Control and Prevention survey about cell phone use noted that \(14.7 \%\) of U.S. households are wireless-only, which means that the household members use only cell phones and do not have a landline. Suppose that this percentage is based on a random sample of 855 U.S. households (Source: http://www.cdc.gov/nchs/data/nhsr/nhsr014.pdf). a. Construct a \(95 \%\) confidence interval for the proportion of all U.S. households that are wireless-only. b. Explain why we need to construct a confidence interval. Why can we not simply say that \(14.7 \%\) of all U.S. households are wireless-only?

A jumbo mortgage is a mortgage with a loan amount above the industry-standard definition of conyentional conforming loan limits. As of January 2009, approximately \(2.57 \%\) of people who took out a jumbo mortgage during the previous 12 months were at least 60 days late on their payments. Suppose that this percentage is based on a random sample of 1430 people who took out a jumbo mortgage during the previous 12 months. a. Construct a \(95 \%\) confidence interval for the proportion of all people who took out a jumbo mortgage during the previous 12 months and were at least 60 days late on their payments. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

KidPix Entertainment is in the planning stages of producing a new computer- animated movie for national release, so they need to determine the production time (labor-hours necessary) to produce the movie. The mean production time for a random sample of 14 big-screen computer-animated movies is found to be 53,550 labor-hours. Suppose that the population standard deviation is known to be 7462 labor-hours and the distribution of production times is normal. a. Construct a \(98 \%\) confidence interval for the mean production time to produce a big-screen computer-animated movie. b. Explain why we need to make the confidence interval. Why is it not correct to say that the average production time needed to produce all big-screen computer-animated movies is 53,550 labor-hours?

A sample of 1500 homes sold recently in a state gave the mean price of homes equal to \(\$ 299,720\). The population standard deviation of the prices of homes in this state is \(\$ 68,650\). Construct a \(99 \%\) confidence interval for the mean price of all homes in this state.

Briefly explain the similarities and the differences between the standard normal distribution and the \(t\) distribution.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

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.