/*! 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 42 The president of a university cl... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 42

The president of a university claims that the mean time spent partying by all students at this university is not more than 7 hours per week. A random sample of 40 students taken from this university showed that they spent an average of \(9.50\) hours partying the previous week with a standard deviation of \(2.3\) hours. Test at a \(2.5 \%\) significance level whether the president's claim is true. Explain your conclusion in words.

Short Answer

Expert verified
The conclusion depends on the calculated value of t-score. If the calculated t-score is greater than about \(1.685\), we reject the null hypothesis stating that students party on average no more than \(7\) hours and conclude that the mean time students spend partying per week is significantly more than \(7\) hours. Otherwise, there is not enough statistical evidence to reject the president's claim.

Step by step solution

01

Set up the null and alternative hypotheses

The null hypothesis (\(H_0\)) is what the president claims - that the students spend on average \(7\) hours or less partying each week. Thus, \(H_0: \mu \leq 7\). The alternative hypothesis (\(H_1\)) is that the students party more than \(7\) hours a week, so \(H_1: \mu > 7\).
02

Calculate the test statistic

The t-score is calculated as \(t = \frac{\bar{x} - \mu_0}{S/\sqrt{n}}\), where \(\bar{x}\) is the sample mean (9.5 hours), \(\mu_0\) is the hypothesized population mean (7 hours), \(S\) is the sample standard deviation (2.3 hours), and \(n\) is the sample size (40 students). Substituting these values, we find \(t = \frac{9.5 - 7}{2.3 / \sqrt{40}}\).
03

Determine the critical t-value

We look up the critical t-value that corresponds to a \(2.5\%\) significance level (in our case we need one-sided critical value because our alternative hypothesis is that the mean is greater than \(7\) hours) and \(40 - 1= 39\) degrees of freedom. From t-distribution table, the critical value for one-tailed test at \(2.5\%\), with \(39\) degrees of freedom is approximately \(1.685\).
04

Compare the t-score with the critical t-value

If the calculated t-score is greater than the critical t-value, we reject the null hypothesis. If it's less, we fail to reject the null hypothesis.
05

Conclusion

Based on our t-score and critical t-value, we make a conclusion about whether the data provide enough evidence to reject the null hypothesis and accept the alternative hypothesis, or if we do not have enough evidence to reject the null hypothesis. We explain the conclusion in terms of the time students spend partying each week.

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

Consider \(H_{0}: \mu=72\) versus \(H_{1}: \mu>72 . \Lambda\) random sample of 16 observations taken from this population produced a sample mean of \(75.2\). The population is normally distributed with \(\sigma=6\). a. Calculate the \(p\) -value. b. Considering the \(p\) -value of part a, would you reject the null hypothesis if the test were made at a significance level of \(.01\) ? c. Considering the \(p\) -value of part a, would you reject the null hypothesis if the test were made at a significance level of \(.025 ?\)

A tool manufacturing company claims that its top-of-the-line machine that is used to manufacture bolts produces an average of 88 or more bolts per hour. A company that is interested in buying this machine wants to check this claim. Suppose you are asked to conduct this test. Briefly explain how you would do so when \(\sigma\) is not known.

For each of the following significance levels, what is the probability of making a Type I error? a. \(\alpha=.025\) b. \(\alpha=.05\) c. \(\alpha=.01\)

\(\quad\) Consider \(H_{0}: p=.70\) versus \(H_{1}: p \neq .70\). a. A random sample of 600 observations produced a sample proportion equal to .68. Using \(\alpha=.01\), would you reject the null hypothesis? b. Another random sample of 600 observations taken from the same population produced a sample proportion equal to \(.76\). Using \(a=.01\), would you reject the null hypothesis? Comment on the results of parts a and b.

According to a 2014 CIRP Your First College Year Survey, \(88 \%\) of the first- year college students said that their college experience exposed them to diverse opinions, cultures, and values (www. heri.ucla.edu). Suppose in a recent poll of 1800 first-year college students, \(91 \%\) said that their college experience exposed them to diverse opinions, cultures, and values. Perform a hypothesis test to determine if it is reasonable to conclude that the current percentage of all firstyear college students who will say that their college experience exposed them to diverse opinions, cultures, and values is higher than \(88 \% .\) Use a \(2 \%\) significance level, and use both the \(p\) -value and the critical-value approaches.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

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.