/*! 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 43 For each of the following, find ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For each of the following, find the area in the appropriate tail of the \(t\) distribution. a. \(t=-1.302\) and \(d f=42\) b. \(t=2.797\) and \(n=25\) c. \(t=1.397\) and \(n=9\) d. \(t=-2.383\) and \(d f=67\)

Short Answer

Expert verified
The solution to this exercise involves using a \(t\)-distribution resource to find the areas for the given \(t\) scores and degrees of freedom. The exact p-values (areas) would depend on the specific calculator or \(t\)-distribution table used.

Step by step solution

01

Interpret the given \(t\) and degrees of freedom

Firstly, observe the given values for \(t\) and either the degrees of freedom (df or n). The \(t\) score represents the calculated difference represented in units of standard error. Positive \(t\) indicates the sample mean is greater than the hypothesized population mean, whereas negative \(t\) signals the opposite. The degrees of freedom (often shown as n or df in \(t\) tests) are related to the sample size and affect the shape of the \(t\)-distribution.
02

Use a \(t\)-distribution chart or calculator

In the next step, a \(t\)-distribution chart or calculator is used to find the area in the tail. Note that if \(t\) is negative, look into the left tail of the distribution, if \(t\) is positive, look into the right tail. Enter the given \(t\) score and degrees of freedom (df or n) into your resource to find the p-value (area).
03

Repeat for each given scenario

Repeat the first two steps for each given set of \(t\) and df/n. Record the calculated area for each.

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 91Ó°ÊÓ!

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

According to the 2010 Time Use Survey conducted by the U.S. Bureau of Labor Statistics, Americans of age 15 years and older spent an average of 164 minutes per day watching TV in 2010 (USA TODAY, June 23,2011 ). Suppose a recent sample of 25 people of age 15 years and older selected from a city showed that they spend an average of 172 minutes per day watching TV with a standard deviation of 28 minutes. Make a \(90 \%\) confidence interval for the average time that all people of age 15 years and older in this city spend per day watching TV. Assume that the times spent by all people of age 15 years and older in this city watching TV have a normal distribution.

Suppose, for a sample selected from a population, \(\bar{x}=25.5\) and \(s=4.9\). a. Construct a \(95 \%\) confidence interval for \(\mu\) assuming \(n=47\). b. Construct a \(99 \%\) confidence interval for \(\mu\) assuming \(n=47\). Is the width of the \(99 \%\) confidence interval larger than the width of the \(95 \%\) confidence interval calculated in part a? If yes, explain why. c. Find a \(95 \%\) confidence interval for \(\mu\) assuming \(n=32 .\) Is the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=32\) larger than the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=47\) calculated in part a? If so, why? Explain.

An insurance company selected a sample of 50 auto claims filed with it and investigated those claims carefully. The company found that \(12 \%\) of those claims were fraudulent. a. What is the point estimate of the percentage of all auto claims filed with this company that are fraudulent? b. Make a \(99 \%\) confidence interval for the percentage of all auto claims filed with this company that are fraudulent.

You want to estimate the proportion of students at your college who hold off- campus (part-time or fulltime) jobs. Briefly explain how you will make such an estimate. Collect data from 40 students at your college on whether or not they hold off-campus jobs. Then calculate the proportion of students in this sample who hold off-campus jobs. Using this information, estimate the population proportion. Select your own confidence level.

Almost all employees working for financial companies in New York City receive large bonuses at the end of the year. A sample of 65 employees selected from financial companies in New York City showed that they received an average bonus of \(\$ 55,000\) last year with a standard deviation of \(\$ 18,000\). Construct a \(95 \%\) confidence interval for the average bonus that all employees working for financial companies in New York City received last year.

See all solutions

Recommended explanations on Math Textbooks

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.