/*! 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] | 91Ó°ÊÓ

91Ó°ÊÓ

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 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

What are the five steps of a test of hypothesis using the critical value approach? Explain briefly,

A real estate agent claims that the mean living area of all singlefamily homes in his county is at most 2400 square feet. A random sample of 50 such homes selected from this county produced the mean living area of 2540 square feet and a standard deviation of 472 square feet. a. Using \(\alpha=.05\), can you conclude that the real estate agent's claim is true? b. What will your conclusion be if \(\alpha=.01 ?\) Comment on the results of parts a and \(\mathrm{b}\).

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. \(H_{1}: \mu<625\)

Thirty percent of all people who are inoculated with the current vaccine that is used to prevent a disease contract the disease within a year. The developer of a new vaccine that is intended to prevent this disease wishes to test for significant evidence that the new vaccine is more effective. a. Determine the appropriate null and altemative hypotheses. b. The developer decides to study 100 randomly selected people by inoculating them with the new vaccine. If 84 or more of them do not contract the disease within a year, the developer will conclude that the new vaccine is superior to the old one. What significance level is the developer using for the test? c. Suppose 20 people inoculated with the new vaccine are studied and the new vaccine is concluded to be better than the old one if fewer than 3 people contract the disease within a year. What is the significance level of the test?

According to the U.S. Bureau of Labor Statistics, all workers in America who had a bachelor's degree and were employed earned an average of \(\$ 1224\) a week in 2014 . A recent sample of 400 American workers who have a bachelor's degree showed that they earn an average of \(\$ 1260\) per week. Suppose that the population standard deviation of such earnings is \(\$ 160\). a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the current mean weekly earning of American workers who have a bachelor's degree is higher than \(\$ 1224\). Will you reject the null hypothesis at \(\alpha=.025 ?\) b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.025\).

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.