Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Q. 2.1 (page 755)

The previous Check Your Understanding (page 750) described some results from a study of body weights and backpack weights. Here, again, is the Minitab output from the least-squares regression analysis for these data.

1. Do these data provide convincing evidence of a linear relationship between pack weight and body weight in the population of ninth-grade students at the school? Carry out a test at the a=0.05significance level. Assume that the conditions for regression inference are met.

Short Answer

Expert verified

There is sufficient evidence to support the claim of a linear relationship between pack weight and body weight in the population of the ninth-grade students at the school.

Step by step solution

01

Given Information

It is given in the output as:

n=8

b=0.09080

SEb=0.02831

02

Explanation

Let's define the hypothesis by:

H0:β=0

Ha:β≠0

The value of the test statistics is as:

t=b-βSEb

=0.09080-00.02831

=3.2073

And the degrees of freedom are:df=n-2

8-2=6

Thus, the P-value is as:

0.006<P<0.018

The null hypothesis is rejected if the P-value is less than or equal to the significance level, as follows:

P<0.05⇒RejectH0

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

Prey attracts predators Here is one way in which nature regulates the size of animal populations: high population density attracts predators, which remove a higher proportion of the population than when the density of the prey is low. One study looked at kelp perch and their common predator, the kelp bass. The researcher set up four large circular pens on sandy ocean bottoms off the coast of southern California. He chose young perch at random from a large group and placed 10,20,40 and60 perch in the four pens. Then he dropped the nets protecting the pens, allowing the bass to swarm in, and counted the perch left after two hours. Here are data on the proportions of perch eaten in four repetitions of this setup .

The explanatory variable is the number of perch (the prey) in a confined area. The response variable is the proportion of perch killed by bass (the predator) in two hours when the bass are allowed access to the perch. A scatterplot of the data shows a linear relationship.

We used Minitab software to carry out a least-squares regression analysis for these data. A residual plot and a histogram of the residuals are shown below. Check whether the conditions for performing inference about the regression model are met.

The P-value for the test in Question 5 is 0.0087. A correct interpretation of this result is that

(a) the probability that there is no linear relationship between an average number of putts per hole and total winnings for these 69players is 0.0087.

(b) the probability that there is no linear relationship between the average number of putts per hole and total winnings for all players on the PGA Tour’s world money list is 0.0087.

(c) if there is no linear relationship between an average number of putts per hole and total winnings for the players in the sample, the probability of getting a random sample of 69players yields a least-squares regression line with a slope of -4139198or less is 0.0087.

(d) if there is no linear relationship between an average number of putts per hole and total winnings for the players on the PGA Tour’s world money list, the probability of getting a random sample of 69 players yields a least-squares regression line with a slope of -4139198or less is 0.0087.

(e) the probability of making a Type II error is0.0087.

Which of the following sampling plans for estimating the proportion of all adults in a medium-sized town who favor a tax increase to support the local school system does not suffer from undercoverage bias?

(a) A random sample of 250names from the local phone book

(b) A random sample of 200parents whose children attend one of the local schools

(c) A sample consisting of 500people from the city who take an online survey about the issue

(d) A random sample of 300homeowners in the town

(e) A random sample of 100people from an alphabetical list of all adults who live in the town

Some high school physics students dropped a ball and measured the distance fallen (in centimeters) at various times (in seconds) after its release. If you have studied physics, then you probably know that the theoretical relationship between the variables is distance =490(time)2. A scatterplot of the students’ data showed a clear curved pattern. Which of the following single transformations should linearize the relationship? I. time2II. distance2III. distance

(a) I only

(c) III only

(e) I and III only

(b) II only

(d) I and II only

We record data on the population of a particular country from 1960to 2010. A scatterplot reveals a clear curved relationship between population and year. However, a different scatterplot reveals a strong linear relationship between the logarithm (base 10) of the population and the year. The least-squares regression line for the transformed data is,

log ( population) =-13.5+0.01(years).

Based on this equation, the population of the country in the year 2020 should be about

(a)6.7

(b) 812

(c) 5,000,000

(d) 6,700,000

(e) 8,120,000.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Decision Maths

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