/*! 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 69 A study claims that \(65 \%\) of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 69

A study claims that \(65 \%\) of students at all colleges and universities hold off-campus (part-time or full-time) jobs. You want to check if the percentage of students at your school who hold off-campus jobs is different from \(65 \%\). Briefly explain how you would conduct such a test. Collect data from 40 students at your school 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, test the hypothesis. Select your own significance level.

Short Answer

Expert verified
The answer depends on the results of the hypothesis test, particularly the calculated Z value and the associated probability. Based on these, you will either reject or fail to reject the null hypothesis that the proportion of students at your school holding off-campus jobs is 65%.

Step by step solution

01

State the Hypotheses

For a hypothesis test, the first step is to state the null hypothesis (\(H_0\)) and alternative hypothesis (\(H_1\)). In this case, \(H_0\) is that the proportion of students at your school who hold off-campus jobs is 65%, or \(P=0.65\). The alternative hypothesis \(H_1\) is that the proportion is different from 65%, or \(P\neq0.65\).
02

Conduct a Survey

In order to test these hypotheses, data needs to be collected. For this, conduct a survey among 40 students at your school to find out if they hold off-campus jobs.
03

Calculate the Sample Proportion

Once the survey data is collected, calculate the sample proportion. This is done by dividing the number of students who hold off-campus jobs by the total number of students surveyed.
04

Perform a Hypothesis Test

Now, compare the sample proportion with the claimed proportion of 0.65 under the null hypothesis. Since no significance level is given, you are free to choose a standard one, like 0.05. Use a Z-test to determine the test statistic. The Z value will be calculated using the formula: \(Z = (\text{{sample proportion}} - \text{{population proportion}}) / \sqrt{((\text{{population proportion}})(1 - \text{{population proportion}}))/\text{{sample size}}}\)
05

Determine the Decision

Finally, look up the Z value in the Z-table to find the associated probability. If the probability is less than the chosen significance level, reject the null hypothesis. If it is greater, fail to reject the null hypothesis. This will be the conclusion of the hypothesis test.

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

According to the analysis of Federal Reserve statistics and other government data, American households with credit card debts owed an average of \(\$ 15,706\) on their credit cards in August 2015 (www.nerdwallet.com). A recent random sample of 500 American households with credit card debts produced a mean credit card debt of \(\$ 16,377\) with a standard deviation of \(\$ 3800 .\) Do these data provide significant evidence at a \(1 \%\) significance level to conclude that the current mean credit card debt of American households with credit card debts is higher than \(\$ 15,706 ?\) Use both the \(p\) -value approach and the critical-value approach.

A telephone company claims that the mean duration of all long-distance phone calls made by its residential customers is 10 minutes. A random sample of 100 long-distance calls made by its residential customers taken from the records of this company showed that the mean duration of calls for this sample is \(9.20\) minutes. The population standard deviation is known to be \(3.80\) minutes. a. Find the \(p\) -value for the test that the mean duration of all longdistance calls made by residential customers of this company is different from 10 minutes. If \(\alpha=.02\), based on this \(p\) -value, would you reject the null hypothesis? Explain. What if \(\alpha=.05 ?\) b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.02 .\) Does your conclusion change if \(\alpha=.05\) ?

Write the null and alternative hypotheses for each of the following examples. Determine if each is a case of a two-tailed, a left-tailed, or a right-tailed test. a. To test if the mean amount of time spent per week watching sports on television by all adult men is different from \(9.5\) hours b. To test if the mean amount of money spent by all customers at a supermarket is less than \(\$ 105\) c. To test whether the mean starting salary of college graduates is higher than \(\$ 47,000\) per year d. To test if the mean waiting time at the drive-through window at a fast food restaurant during rush hour differs from 10 minutes e. To test if the mean time spent per week on house chores by all housewives is less than 30 hours

A soft-drink manufacturer claims that its 12 -ounce cans do not contain, on average, more than 30 calories. A random sample of 64 cans of this soft drink, which were checked for calories, contained a mean of 32 calories with a standard deviation of 3 calories. Does the sample information support the alternative hypothesis that the manufacturer's claim is false? Use a significance level of \(5 \%\). Find the range for the \(p\) -value for this test. What will your conclusion be using this \(p\) -value and \(\alpha=.05\) ?

Consider the null hypothesis \(H_{0}: \mu=625 .\) Suppose that a random sample of 29 observations is taken from a normally distributed population with \(\sigma=32 .\) Using a significance level of \(.01\), show the rejection and nonrejection regions on the sampling distribution curve of the sample mean and find the critical value(s) of \(z\) when the alternative hypothesis is as follows. a. \(H_{1}: \mu \neq 625\) b. \(H_{1}: \mu>625\) c. \(H_{1}: \mu<625\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

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.