/*! 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 50 A certain pen has been designed ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 50

A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least \(10 \mathrm{hr}\). A random sample of 18 pens is selected, the writing lifetime of each is determined, and a normal probability plot of the resulting data support the use of a one-sample \(t\) test. The relevant hypotheses are \(H_{0}: \mu=10\) versus \(H_{a}: \mu<10\). a. If \(t=-2.3\) and \(\alpha=.05\) is selected, what conclusion is appropriate? b. If \(t=-1.83\) and \(\alpha=.01\) is selected, what conclusion is appropriate? c. If \(t=0.47\), what conclusion is appropriate?

Short Answer

Expert verified
a. Reject the null hypothesis, the average is less than 10 hours. b. Fail to reject the null hypothesis, not enough evidence that the average is less than 10 hours. c. Fail to reject the null hypothesis, the average is not less than 10 hours.

Step by step solution

01

Understand the hypotheses and the nature of the t-test

We are using a one-sample t-test, which compares a sample mean to a known population mean. In this case, it is used to test whether the average writing lifetime of pens is at least 10 hours. The null hypothesis, \(H_{0} : \mu = 10\), assumes the average is 10 hours. The alternative hypothesis, \(H_{a} : \mu < 10\), assumes the average is less than 10 hours. The test is left-tailed as we are checking for a value lesser than the hypothesized mean.
02

Conclusion for t = -2.3, \(\alpha = .05\)

First, look up the critical t-value for \(\alpha = .05\) on 17 degrees of freedom (one less than the sample size of 18). A t-table shows the critical value is about -1.74. Since the calculated t-value of -2.3 is less than the critical t-value, we would reject the null hypothesis. This means that the average writing lifetime is less than 10 hours.
03

Conclusion for t = -1.83, \(\alpha = .01\)

Now, look up the critical t-value for \(\alpha = .01\) on 17 degrees of freedom. The critical value is roughly -2.567. Since the calculated t-value of -1.83 is greater than the critical t-value, we fail to reject null hypothesis. This means that there is not enough evidence to conclude that the average writing lifetime is less than 10 hours.
04

Conclusion for t = 0.47

Lastly, for any given \(\alpha\), the calculated t-value (0.47) falls in the acceptance region, since it is greater than any negative critical t-value. This implies that the null hypothesis cannot be rejected.

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

Medical personnel are required to report suspected cases of child abuse. Because some diseases have symptoms that mimic those of child abuse, doctors who see a child with these symptoms must decide between two competing hypotheses: \(H_{0}\) " symptoms are due to child abuse \(H_{a}:\) symptoms are due to disease (Although these are not hypotheses about a population characteristic, this exercise illustrates the definitions of Type I and Type II errors.) The article "Blurred Line Between Illness, Abuse Creates Problem for Authorities" (Macon Telegraph, February 28,2000 ) included the following quote from a doctor in Atlanta regarding the consequences of making an incorrect decision: "If it's disease, the worst you have is an angry family. If it is abuse, the other kids (in the family) are in deadly danger." a. For the given hypotheses, describe Type I and Type II errors. b. Based on the quote regarding consequences of the two kinds of error, which type of error does the doctor quoted consider more serious? Explain.

Past experience has indicated that the true response rate is \(40 \%\) when individuals are approached with a request to fill out and return a particular questionnaire in a stamped and addressed envelope. An investigator believes that if the person distributing the questionnaire is stigmatized in some obvious way, potential respondents would feel sorry for the distributor and thus tend to respond at a rate higher than \(40 \%\). To investigate this theory, a distributor is fitted with an eye patch. Of the 200 questionnaires distributed by this individual, 109 were returned. Does this strongly suggest that the response rate in this situation exceeds the rate in the past? State and test the appropriate hypotheses at significance level . 05 .

A county commissioner must vote on a resolution that would commit substantial resources to the construction of a sewer in an outlying residential area. Her fiscal decisions have been criticized in the past, so she decides to take a survey of constituents to find out whether they favor spending money for a sewer system. She will vote to appropriate funds only if she can be fairly certain that a majority of the people in her district favor the measure. What hypotheses should she test?

A survey of teenagers and parents in Canada conducted by the polling organization Ipsos ("Untangling the Web: The Facts About Kids and the Internet," January \(25 .\) 2006) included questions about Internet use. It was reported that for a sample of 534 randomly selected teens, the mean number of hours per week spent online was \(14.6\) and the standard deviation was \(11.6\). a. What does the large standard deviation, \(11.6\) hours, tell you about the distribution of online times for this sample of teens? b. Do the sample data provide convincing evidence that the mean number of hours that teens spend online is greater than 10 hours per week?

Students at the Akademia Podlaka conducted an experiment to determine whether the Belgium-minted Euro coin was equally likely to land heads up or tails up. Coins were spun on a smooth surface, and in 250 spins, 140 landed with the heads side up (New Scientist, January 4 , 2002). Should the students interpret this result as convincing evidence that the proportion of the time the coin would land heads up is not .5? Test the relevant hypotheses using \(\alpha=.01\). Would your conclusion be different if a significance level of \(.05\) had been used? Explain.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Mechanics 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.