/*! 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 18 In a survey of 1000 randomly sel... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 18

In a survey of 1000 randomly selected adults in the United States, participants were asked what their most favorite and what their least favorite subject was when they were in school (Associated Press, August 17,2005\() .\) In what might seem like a contradiction, math was chosen more often than any other subject in both categories! Math was chosen by 230 of the 1000 as the favorite subject, and it was also chosen by 370 of the 1000 as the least favorite subject. a. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults for whom math was the favorite subject in school. b. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults for whom math was the least favorite subject.

Short Answer

Expert verified
a. The 95% confidence interval for the proportion of U.S. adults for whom math was the favorite subject in school is approximately (0.203, 0.257). b. The 95% confidence interval for the proportion of U.S. adults for whom math was the least favorite subject is approximately (0.343, 0.397).

Step by step solution

01

Calculate proportions

First, need to calculate the proportions of U.S adults whose favorite and least favorite subject was math. For favorites, the proportion ( \( p_1 \) ) is 230 out of 1000, or \( p_1 = 230/1000 = 0.23 \). For least favorites, the proportion (\( p_2 \)) is 370 out of 1000, or \( p_2 = 370/1000 = 0.37 \).
02

Construct the confidence interval for favorite subject

Plug \( p_1 \), the z-value (1.96), and the sample size (1000) into the formula for the confidence interval: \( p_1 \pm 1.96 \sqrt{p_1(1-p_1)/1000} \). After solving, this gives \( 0.23 \pm 1.96 \sqrt{0.23(1-0.23)/1000} \). Calculating the square root and carrying out the operation, find the confidence interval to be approximately (0.203, 0.257).
03

Construct the confidence interval for least favorite subject

Using \( p_2 \), the z-value (1.96), and the sample size (1000) into the formula for the confidence interval: \( p_2 \pm 1.96 \sqrt{p_2(1-p_2)/1000} \). After solving, this gives \( 0.37 \pm 1.96 \sqrt{0.37(1-0.37)/1000} \). Calculating the square root and carrying out the operation, find the confidence interval to be approximately (0.343, 0.397).

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

Fat contents (in percentage) for 10 randomly selected hot dogs were given in the article "Sensory and Mechanical Assessment of the Quality of Frankfurters" (Journal of Texture Studies \([1990]: 395-409\) ). Use the following data to construct a \(90 \%\) confidence interval for the true mean fat percentage of hot dogs: \(\begin{array}{lllllllllllll}25.2 & 21.3 & 22.8 & 17.0 & 29.8 & 21.0 & 25.5 & 16.0 & 20.9 & 19.5\end{array}\)

The formula used to compute a large-sample confidence interval for \(\pi\) is $$ p \pm(z \text { critical value }) \sqrt{\frac{p(1-p)}{n}} $$ What is the appropriate \(z\) critical value for each of the following confidence levels? a. \(95 \%\) d. \(80 \%\) b. \(90 \%\) e. \(85 \%\) c. \(99 \%\)

For each of the following choices, explain which would result in a wider large-sample confidence interval for \(\pi\) : a. \(90 \%\) confidence level or \(95 \%\) confidence level b. \(n=100\) or \(n=400\)

Acrylic bone cement is sometimes used in hip and knee replacements to fix an artificial joint in place. The force required to break an acrylic bone cement bond was measured for six specimens under specified conditions, and the resulting mean and standard deviation were \(306.09\) Newtons and \(41.97\) Newtons, respectively. Assuming that it is reasonable to assume that breaking force under these conditions has a distribution that is approximately normal, estimate the true average breaking force for acrylic bone cement under the specified conditions.

The article "Consumers Show Increased Liking for Diesel Autos" (USA Today, January 29,2003 ) reported that \(27 \%\) of U.S. consumers would opt for a diesel car if it ran as cleanly and performed as well as a car with a gas engine. Suppose that you suspect that the proportion might be different in your area and that you want to conduct a survey to estimate this proportion for the adult residents of your city. What is the required sample size if you want to estimate this proportion to within \(.05\) with \(95 \%\) confidence? Compute the required sample size first using 27 as a preliminary estimate of \(\pi\) and then using the conservative value of \(.5 .\) How do the two sample sizes compare? What sample size would you recommend for this study?
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

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.