Chapter 8: Problem 38
Briefly explain the meaning of the degrees of freedom for a \(t\) distribution. Give one example.
/*! 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}
Learning Materials
Features
Discover
Chapter 8: Problem 38
Briefly explain the meaning of the degrees of freedom for a \(t\) distribution. Give one example.
All the tools & learning materials you need for study success - in one app.
Get started for free
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.
You want to estimate the percentage of students at your college or university who are satisfied with the campus food services. Briefly explain how you will make such an estimate. Select a sample of 30 students and ask them whether or not they are satisfied with the campus food services. Then calculate the percentage of students in the sample who are satisfied. Using this information, find the confidence interval for the corresponding population percentage. Select your own confidence level.
a. A sample of 100 observations taken from a population produced a sample mean equal to \(55.32\) and a standard deviation equal to \(8.4 .\) Make a \(90 \%\) confidence interval for \(\mu .\) b. Another sample of 100 observations taken from the same population produced a sample mean equal to \(57.40\) and a standard deviation equal to \(7.5 .\) Make a \(90 \%\) confidence interval for \(\mu .\) c. A third sample of 100 observations taken from the same population produced a sample mean equal to \(56.25\) and a standard deviation equal to \(7.9 .\) Make a \(90 \%\) confidence interval for \(\mu .\) d. The true population mean for this population is \(55.80 .\) Which of the confidence intervals constructed in parts a through c cover this population mean and which do not?
A consumer agency that proposes that lawyers' rates are too high wanted to estimate the mean hourly rate for all lawyers in New York City. A sample of 70 lawyers taken from New York City showed that the mean hourly rate charged by them is \(\$ 570\). The population standard deviation of hourly charges for all lawyers in New York City is \(\$ 110\). a. Construct a \(99 \%\) confidence interval for the mean hourly charges for all lawyers in New York City. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?
A sample selected from a population gave a sample proportion equal to \(.73\) a. Make a \(99 \%\) confidence interval for \(p\) assuming \(n=100\). b. Construct a \(99 \%\) confidence interval for \(p\) assuming \(n=600\). c. Make a \(99 \%\) confidence interval for \(p\) assuming \(n=1500\). d. Does the width of the confidence intervals constructed in parts a through \(\mathrm{c}\) decrease as the sample size increases? If yes, explain why.
What do you think about this solution?
We value your feedback to improve our textbook solutions.