/*! 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} Q.99 Sixty-eight percent of online co... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Q.99 (page 553)

Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California鈥檚 percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data.

Short Answer

Expert verified

Confidence Interval: (0.65, 0.90)

Step by step solution

01

Find H0 and Ha: We want to test if California community colleges take 68% of the online classes thought by full-time faculties, or not.

H0:p=0.68;Ha:p鈮�0.68

02

Determine the distribution needed: In words, CLEARLY state what your random variable P′ represents.Let P′ = the proportion of online courses at LBCC that are taught by full-time faculty.

Normal:N0.68,(0.68)(1-0.68)44

Test Statistic:z=0.1873

03

Calculate the p-value using the normal distribution for proportions:

p-value=0.1873

In one to two complete sentences, explain what the p-value means for this problem.

If the null hypothesis is true (the proportion is 0.68), then there is a 0.1873 probability that the sample (estimated) proportion is 0.773 or more.

04

Compare α and the p-value:Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences.

alphadecisionreason for decision
0.05Do not reject the null hypothesis.
p-value>0.05

Conclusion: At the 5% level, the data do not provide statistically significant evidence that the true proportion of online courses taught by full-time faculty is not 68%.

05

Confidence Interval

Confidence Interval: (0.65, 0.90): The 鈥減lus-4s鈥� confidence interval is (0.6275, 0.8725).

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

A teacher believes that 85%of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50students and 39reply that they would want to go to the zoo. For the hypothesis test, use a 1%level of significance.

First, determine what type of test this is, set up the hypothesis test, find the p-value, sketch the graph, and state your conclusion.

The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the 伪=0.05 level in Kentucky? Are the results applicable across the country? Why?

A poll done for Newsweek found that 13%of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent's survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results.

State the Type I and Type II errors in complete sentences given the following statements.

a. The mean number of years Americans work before retiring is 34.

b. At most 60% of Americans vote in presidential elections.

c. The mean starting salary for San Jose State University graduates is at least \(100,000 per year.

d. Twenty-nine percent of high school seniors get drunk each month.

e. Fewer than 5% of adults ride the bus to work in Los Angeles.

f. The mean number of cars a person owns in his or her lifetime is not more than ten.

g. About half of Americans prefer to live away from cities, given the choice.

h. Europeans have a mean paid vacation each year of six weeks.

i. The chance of developing breast cancer is under 11% for women.

j. Private universities mean tuition cost is more than \)20,000 per year.

A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is0.83You want to test to see if the claim is correct. State the null and that alternative hypothesis.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Probability and 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.