Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Q. 69 (page 659)

A study of road rage asked samples of 596men and 523women about their behaviour while driving. Based on their answers, each person was assigned a road rage score on a scale of 0to20. The participants were chosen by random digit dialling of telephone numbers. We suspect that men are more prone to road rage than women. To see if this is true, test these hypotheses for the mean road rage scores of all male and female drivers

(a) H0:μM=μFversusH0:μM>μF

(b) H0:μM=μFversusH0:μM≠μF

(c) role="math" localid="1650363760526" H0:μM=μFversusH0:μM<μF

(d) H0:x¯M=x¯FversusH0:x¯M>x¯F

(e)H0:x¯M=x¯FversusH0:x¯M<xF

Short Answer

Expert verified

The answer is option (a)H0:μM=μF,H0:μM>μF

Step by step solution

01

Given information

A study of road rage samples

Men is596

Women is 523

It is suspected that men are more prone to road rage than women and see if this is true.

02

Explanation 

The hypothesis are claims that are made concerning a population parameter (s).

The population meanμan is the population parameter.

According to the null hypothesis, the two population parameters are equal:

H0:μM=μF

According to the claim, the alternative hypothesis expresses the polar opposite of the null hypothesis. Men are more prone to road rage than women, according to the claim:

H0:μM>μF, so it is option (a).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

National Park rangers keep data on the bears that inhabit their park. Below is a histogram of the weights of 143bears measured in a recent year.

Which statement below is correct?

(a) The median will lie in the interval (140,180), and the mean will lie in the interval (180,220).

(b) The median will lie in the interval (140,180), and the mean will lie in the interval (260,300).

(c) The median will lie in the interval (100,140), and the mean will lie in the interval (180,220).

(d) The mean will lie in the interval (140,180), and the median will lie in the interval (260,300).

(e) The mean will lie in the interval (100,140), and the median will lie in the interval (180,200).

A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state’s driver’s license exam. An incoming class of 100students is randomly assigned to two groups, each of size 50. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, 30of Instructor A’s students and 22of Instructor B’s students pass the state exam. Do these results give convincing evidence that Instructor A is more effective?

Min Jae carried out the significance test shown below to answer this question. Unfortunately, he made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one.

State: I want to perform a test of

H0:p1-p2=0

Ha:p1-p2>0

where p1=the proportion of Instructor A's students that passed the state exam and p2=the proportion of Instructor B's students that passed the state exam. Since no significance level was stated, I'll use σ=0.05

Plan: If conditions are met, I’ll do a two-sample ztest for comparing two proportions.

Random The data came from two random samples of 50students.

- Normal The counts of successes and failures in the two groups -30,20,22, and 28-are all at least 10.

- Independent There are at least 1000 students who take this driving school's class.

Do: From the data, p^1=2050=0.40and p^2=3050=0.60. So the pooled proportion of successes is

p^C=22+3050+50=0.52

- Test statistic

localid="1650450621864" z=(0.40-0.60)-00.52(0.48)100+0.52(0.48)100=-2.83

- p-value From Table A, localid="1650450641188" P(z≥-2.83)=1-0.0023=0.9977.

Conclude: The p-value, 0.9977, is greater than α=0.05, so we fail to reject the null hypothesis. There is no convincing evidence that Instructor A's pass rate is higher than Instructor B's.

Which of the following is not a property of a binomial setting?

(a) Outcomes of different trials are independent.

(b) The chance process consists of a fixed number of trials,n

(c) The probability of success is the same for each trial.

(d) If we use a sample size of 30, the binomial distribution will be approximately Normal.

(e) Each trial can result in either a success or a failure.

A large toy company introduces a lot of new toys to its product line each year. The company wants to predict the demand as measured by y, first-year sales (in millions of dollars) using x, awareness of the product (as measured by the proportion of customers who had heard of the product by the end of the second month after its introduction). A random sample of 65new products was taken, and a correlation of 0.96was computed. Which of the following is a correct interpretation of this value?

(a) Ninety-six percent of the time, the least-squares regression line accurately predicts first-year sales.

(b) About 92%of the time, the proportion of people who have heard of the product by the end of the second month will correctly predict first-year sales.

(c) About 92%of first-year sales can be explained by the proportion of people who have heard of the product by the end of the second month.

(d) The least-squares regression line relating the proportion of people who have heard of the product by the end of the second month and first-year sales will have a slope of 0.96.

(e) Ninety-two percent of the variation in first-year sales can be explained by the least-squares regression line with proportion of people who have heard of the product by the end of the second month as the explanatory variable.

Thirty-five people from a random sample of 125workers from Company A admitted to using sick leave when they weren’t really ill. Seventeen employees from a random sample of 68workers from Company B admitted that they had used sick leave when they weren’t ill. A 95% confidence interval for the difference in the proportions of workers at the two companies who would admit to using sick leave when they weren’t ill is

(a) 0.03±(0.28)(0.72)125+(0.25)(0.75)68

(b) localid="1650367573248" 0.03±1.96(0.28)(0.72)125+(0.25)(0.75)68

(c) 0.03±1.645(0.28)(0.72)125+(0.25)(0.75)68

(d)

0.03±1.96(0.269)(0.731)125+(0.269)(0.731)68

(e) 0.03±1.645(0.269)(0.731)125+(0.269)(0.731)68
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Applied Mathematics

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.

Sign-up for free