Effectiveness of online courses. The Survey of Online Learning, “Grade Level: Tracking Online Education in the United States, 2014,” reported that 74% of college leaders believe that their online education courses are as good as or superior to courses that use traditional, face-to-face instruction.

a. Give the null hypothesis for testing the claim made by the survey.

Short Answer

Expert verified

a. The null hypothesis is, \({H_0}:p = 0.74.\)

Step by step solution

01

Given Information

74% of college leaders believe that their online education courses are as good as or superior to courses that use traditional, face-to-face instruction.

02

State the large sample test of the hypothesis about p.

The test statistic used to obtaining the large sample test of the hypothesis about p is,

\({Z_c} = \frac{{\left( {\hat p - {p_0}} \right)}}{{{\sigma _{\hat p}}}}\)

The condition required for a valid large sample hypothesis test for p are:

  • The sample size n is large.
  • A random sample is selected from a binomial population.
03

State the null hypothesis of the test.

a.

Let p be the true proportion of college leaders who believe that their online education courses are as good as or superior to courses that use traditional, face-to-face instruction.

The value of the p is computed as:

\(\begin{aligned}p = \frac{{74}}{{100}}\\ = 0.74\end{aligned}\)

The required null hypothesis is,\({H_0}:p = 0.74.\)

Hence, the null hypothesis is that the true proportion of college leaders who believe that their online education courses are as good as or superior to courses that use traditional, face-to-face instruction is 0.74.

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 Vaia!

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

In a test of the hypothesis \({H_0}:\mu = 10\) versus \({H_a}:\mu \ne 10\), a sample of n = 50 observations possessed mean \(\bar x = 10.7\) and standard deviation s = 3.1. Find and interpret the p-value for this test.

Complete the following statement: The smaller the p-value associated with a test of hypothesis, the stronger the support for the _____ hypothesis. Explain your answer.

Improving the productivity of chickens. Refer to the Applied Animal Behaviour Science (October 2000) study of the color of string preferred by pecking domestic chickens, Exercise 6.124 (p. 376). Recall that n = 72 chickens were exposed to blue string and the number of pecks each chicken took at the string over a specified time interval had a mean of\(\overline x = 1.13\,\)pecks and a standard deviation of s = 2.21 pecks. Also recall that previous research had shown that m = 7.5 pecks if chickens are exposed to white string.

a. Conduct a test (at\(\alpha = 0.01\)) to determine if the true mean number of pecks at blue string is less than\(\mu = 7.5\)pecks.
b. In Exercise 6.122, you used a 99% confidence interval as evidence that chickens are more apt to peck at white string than blue string. Do the test results, part a, support this conclusion? Explain

A random sample of 175 measurements possessed a mean x¯=8.2 and a standard deviation s = .79.

a. Test H0:μ=8.3 against Ha:μ8.3Use a=0.05

7.83 A random sample of n observations is selected from a normal population to test the null hypothesis that σ2=25 . Specify the rejection region for each of the following combinations of Ha,αand n.

a.Ha:σ225;α=0.5;n=16

b. Ha:σ2>25;α=.01;n=23

c. Ha:σ2>25;α=.10;n=15

d. Ha:σ2<25;α=.01;n=13

e. Ha:σ225;α=.10;n=7

f.Ha:σ2<25;α=.05;n=25

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.

Sign-up for free