/*! 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} Q54E Lobster trap placement. Refer to... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Q54E (page 417)

Lobster trap placement. Refer to the Bulletin of Marine Science (April 2010) observational study of lobster trap placement by teams fishing for the red spiny lobster in Baja California Sur, Mexico, Exercise 6.29 (p. 348). Trap-spacing measurements (in meters) for a sample of seven teams of red spiny lobster fishermen are reproduced in the accompanying table. Let \(\mu \) represent the average of the trap-spacing measurements for the population of red spiny lobster fishermen fishing in Baja California Sur, Mexico. In Exercise 6.29 you computed the mean and standard deviation of the sample measurements to be \(\bar x = 89.9\) meters and s = 11.6 meters, respectively. Suppose you want to determine if the true value of \(\mu \) differs from 95 meters

a. Specify the null and alternative hypotheses for this test.

Short Answer

Expert verified

a)

\(\begin{array}{}{H_0}:\mu = 95\\{H_a}:\mu \ne 95\end{array}\)

Step by step solution

01

Given Information

The sample size is 7.

The mean and standard deviation is given. i.e.

\(\bar x = 89.9\,and\,s = 11.6\)

02

Standard deviation

The standard deviation is a metric that calculates as the sum of squares of the variance as well as quantifies the dispersion of a data compared to its mean. The standard deviation is determined as the sum of squares of the variance by calculating the departure of each piece of data from the mean.In descriptive statistics, the standard deviation is the amount of dispersal as well as scattering of sample points compared to their mean.

03

Step 3:

The hypothesis is given by

\(\begin{array}{}{H_0}:\mu = 95\\{H_a}:\mu \ne 95\end{array}\)

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

Which hypothesis, the null or the alternative, is the status-quo hypothesis? Which is the research hypothesis?

EPA limits on vinyl chloride. The EPA sets an airborne limit of 5 parts per million (ppm) on vinyl chloride, a colorless gas used to make plastics, adhesives, and other chemicals. It is both a carcinogen and a mutagen (New Jersey Department of Health, Hazardous Substance Fact Sheet, 2010). A major plastics manufacturer, attempting to control the amount of vinyl chloride its workers are exposed to, has given instructions to halt production if the mean amount of vinyl chloride in the air exceeds 3.0 ppm. A random sample of 50 air specimens produced the following statistics: \(\overline x = 3.1\)ppm,\(s = 0.5\)ppm.

a. Do these statistics provide sufficient evidence to halt the production process? Use\(\alpha = 0.01\).

b. If you were the plant manager, would you want to use a large or a small value for\(\alpha \)the test in part a? Explain.

c. Find the p-value for the test and interpret its value

The National Association of Realtors (NAR) reported the results of an April 2015 survey of home buyers. In a random sample of 1,971 residential properties purchased during the year, 414 were purchased as a vacation home. Five years ago, 10% of residential properties were vacation homes.

a. Do the survey results allow the NAR to conclude (at 伪=.01) that the percentage of all residential properties purchased for vacation homes is greater than 10%?

b. In a previous year, the NAR sent the survey questionnaire to a nationwide sample of 45,000 new home owners, of which 1,982 responded to the survey. How might this bias the results? [Note: In the most recent survey, the NAR used a more valid sampling method.

In a test of H0:渭=100againstHa:渭>100, the sample data yielded the test statistic z = 2.17. Find and interpret the p-value for the test.

Customers who participate in a store鈥檚 free loyalty card program save money on their purchases but allow the store to keep track of the customer鈥檚 shopping habits and potentially sell these data to third parties. A Pew Internet & American Life Project Survey (January 2016) revealed that 225 of a random sample of 250 U.S. adults would agree to participate in a store loyalty card program, despite the potential for information sharing. Letp represent the true proportion of all customers who would participate in a store loyalty card program.

a. Compute a point estimate ofp

b. Consider a store owner who claims that more than 80% of all customers would participate in a loyalty card program. Set up the null and alternative hypotheses for testing whether the true proportion of all customers who would participate in a store loyalty card program exceeds .8

c. Compute the test statistic for part b.

d. Find the rejection region for the test if 伪=.01.

e. Find the p-value for the test.

f. Make the appropriate conclusion using the rejection region.

g. Make the appropriate conclusion using the p-value.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Applied Mathematics

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.