/*! 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 65 The computer age has allowed tea... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The computer age has allowed teachers to use electronic tutorials to motivate their students to learn. Issues in Accounting Education published the results of a study that showed that an electronic tutorial, along with intentionally induced peer pressure, was effective in enhancing preclass preparations and in improving class attendance, test scores, and course evaluations when used by students studying tax accounting. Suppose a similar study is conducted at your school using an electronic study guide (ESG) as a tutor for students of accounting principles. For one course section, the students were required to use a new ESG computer program that generated and scored chapter review quizzes and practice exams, presented textbook chapter reviews, and tracked progress. Students could use the computer to build, take, and score their own simulated tests and review materials at their own pace before they took their formal in-class quizzes and exams composed of different questions. The same instructor taught the other course section, used the same textbook, and gave the same daily assignments, but he did not require the students to use the ESG. Identical tests were administered to both sections, and the mean scores of all tests and assignments at the end of the year were tabulated: $$\begin{array}{lccc}\text { Section } & n & \text { Mean Score } & \text { Std. Dev. } \\\\\hline \text { ESG }|1\rangle & 38 & 79.6 & 6.9 \\\\\text { No ESG }|2| & 36 & 72.8 & 7.6 \\\\\hline\end{array}$$ Do these results show that the mean scores of tests and assignments for students taking accounting principles with an ESG to help them are significantly greater than the mean scores of those not using an ESG? Use a 0.01 level of significance. a. Solve using the \(p\) -value approach.

Short Answer

Expert verified
The final decision regarding whether the ESG group mean test scores are significantly greater than those of the No ESG group is based on comparing the p-value to the significance level (0.01). The exact p-value and decision can only be determined by computing the t-test statistic using the given data.

Step by step solution

01

Formulate the Hypotheses

The null hypothesis (\(H_0\)) represents a statement of no effect or no change and the alternative hypothesis (\(H_1\)) represents a statement of the effect or change. In this case, the statements can be written as: \n\nThe null hypothesis (\(H_0\)): \(\mu_1 = \mu_2\) or \(\mu_1 - \mu_2 = 0\), where \(\mu_1\) and \(\mu_2\) are the mean test scores of the students in the ESG group and the No ESG group respectively. \n\nThe alternative hypothesis (\(H_1\)): \(\mu_1 > \mu_2\) or \(\mu_1 - \mu_2 > 0\). This indicates that the mean test score for the ESG group is greater than the mean test score for the No ESG group. \n\nTo test these hypotheses, a paired t-test can be conducted.
02

Calculate the Test Statistic

We use the formula for the test statistic in a two-sample t-test:\[ t = \frac{(\bar{X_1}-\bar{X_2}) - (\mu_1 - \mu_2)}{\sqrt{(s_1^2/n_1) + (s_2^2/n_2)}},\]where, \(\bar{X_1}\) and \(\bar{X_2}\) are the sample means,\(s_1\) and \(s_2\) are the sample standard deviations, and \(n_1\) and \(n_2\) are the sample sizes for the ESG group and the No ESG group respectively. \(\mu_1\) and \(\mu_2\) are the hypothetical population means (in this case, \(\mu_1 - \mu_2\) is presumed to be 0 in our null hypothesis). Given the data, the test statistic can now be calculated.
03

Determine the p-value

The p-value is the probability, under the null hypothesis, of obtaining a test statistic at least as extreme as the one computed from the sample data. By comparing the calculated t-statistic to the t-distribution with \(n_1 + n_2 - 2\) degrees of freedom, we can find the p-value. If the calculated p-value is less than the significance level (0.01), then we reject the null hypothesis in favor of the alternative hypothesis.
04

Making a Decision

If the calculated p-value in step 3 is equal to or greater than 0.01, we fail to reject the null hypothesis as it suggests that there is no significant difference in the mean test scores of the two groups. If the calculated p-value is less than 0.01, we reject the null hypothesis, implying that the mean test scores in the ESG group are significantly different (and greater as per alternative hypothesis) than those in the No ESG group.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electronic Study Guides (ESG)
Electronic Study Guides (ESG) are digital tools designed to assist students in their study process. They provide a variety of resources to support learning, such as quizzes, practice exams, and chapter summaries. With the ability to track progress, students can identify areas where they need more practice and adjust their study strategy accordingly. ESGs engage students more interactively, allowing a customized learning experience.

For accounting education, an ESG can effectively facilitate pre-class preparations and improve overall course outcomes. It encourages a self-paced study environment, enabling students to build their knowledge base before taking formal assessments.
  • Custom quizzes and exams allow for targeted practice.
  • Immediate feedback helps reinforce learning.
  • Progress tracking motivates continuous improvement.
With these benefits, ESGs can play a significant role in enhancing student performance in accounting education.
Hypothesis Testing in Statistics
Hypothesis testing in statistics is a method used to determine if there is enough evidence to infer that a condition is true for the entire population. This process starts with formulating two hypotheses: the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)).

The null hypothesis typically states that there is no effect or difference, while the alternative suggests a significant effect or difference exists. In our context with the ESG study, the hypotheses are:
  • \(H_0\): The mean scores of students with ESG (\( \mu_1 \)) equal those without ESG (\( \mu_2 \)).
  • \(H_1\): The mean scores for students using ESG (\( \mu_1\)) are greater than those without (\( \mu_2\)).
Hypothesis testing involves comparing sample data against these hypotheses.
A t-test is conducted using the sample means and standard deviations to decide whether the null hypothesis can be rejected. If the calculated p-value is less than the stated significance level (in this case, 0.01), it indicates that the mean score increase for ESG users is statistically significant.
Impact of Technology on Learning
Technology has transformed education by providing innovative tools and resources that enhance the learning experience. In accounting education, tools like electronic study guides (ESGs) can significantly impact student performance.

By introducing technology into the learning process, students benefit from:
  • Interactive learning experiences, which can make complex subjects more understandable.
  • Increased access to diverse resources, enabling them to explore topics more thoroughly.
  • Flexible learning environments, allowing students to learn at their own pace and aligning study sessions with their schedules.
The use of ESGs in the study presented showed improvements in test scores and attendance, providing empirical evidence of technology's potential to enhance educational outcomes in accounting. As technology continues to evolve, it will likely offer even more ways to engage and support learners, ultimately improving student performance across various educational fields.

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

At a large university, a mathematics placement exam is administered to all students. Samples of 36 male and 30 female students are randomly selected from this year's student body and the following scores recorded: $$\begin{array}{lllllllllll}\hline \text { Male } & 72 & 68 & 75 & 82 & 81 & 60 & 75 & 85 & 80 & 70 \\\& 71 & 84 & 68 & 85 & 82 & 80 & 54 & 81 & 86 & 79 \\\& 99 & 90 & 68 & 82 & 60 & 63 & 67 & 72 & 77 & 51 \\\& 61 & 71 & 81 & 74 & 79 & 76 & & & & \\\\\hline \text { Female } & 81 & 76 & 94 & 89 & 83 & 78 & 85 & 91 & 83 & 83 \\\& 84 & 80 & 84 & 88 & 77 & 74 & 63 & 69 & 80 & 82 \\\& 89 & 69 & 74 & 97 & 73 & 79 & 55 & 76 & 78 & 81 \\ \hline\end{array}$$ a. Describe each set of data with a histogram (use the same class intervals on both histograms), the mean, and the standard deviation. b. Construct \(95 \%\) confidence interval for the mean score for all male students. Do the same for all female students. c. Do the results found in part b show that the mean scores for males and females could be the same? Justify your answer. Be careful! d. Construct the \(95 \%\) confidence interval for the difference between the mean scores for male and female students. e. Do the results found in part d show that the mean scores for male and female students could be the same? Explain. f. Explain why the results in part b cannot be used to draw conclusions about the difference between the two means.

To test the effect of a physical-fitness \- course on one's physical ability, the number of sit-ups that a person could do in 1 minute, both before and after the course, was recorded. Ten randomly selected participants scored as shown in the following table. Can you \- conclude that a significant amount of improvement took place? Use \(\alpha=0.01\) and assume normality.$$\begin{array}{lllllllllll}\text { Before } & 29 & 22 & 25 & 29 & 26 & 24 & 31 & 46 & 34 & 28 \\\\\text { After } & 30 & 26 & 25 & 35 & 33 & 36 & 32 & 54 & 50 & 43 \\\\\hline\end{array}$$a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

The article "Car rental prices can change in a Theartbeat" (USA Today, March 14, 2007) reported that mates for car rentals can change many times during a day. The national average rate for the January-March quarter Eof 2007 was \(\$ 52.71,\) yet in some cities, car rentals can cost more than \(\$ 100\) a day. A similar study of two major cities found the following results: $$\begin{array}{lccc}\text { City } & \text { n } & \text { Average Daily Rate } & \text { Standard Deviation } \\\\\hline \text { Boston } & 10 & 95.94 & 7.50 \\ \text { New York Cily } & 16 & 127.75 & 15.83 \\ \hline\end{array}$$ Set a \(95 \%\) confidence interval on the difference in average daily rates between the two major East Coast cities of Boston and New York City. Assume normality for the sampled populations and that the samples were selected randomly.

State the null hypothesis, \(H_{o}\), and the alternative hypothesis, \(H_{a},\) that would be used to test these claims: a. There is no difference between the proportions of men and women who will vote for the incumbent in next month's election. b. The percentage of boys who cut classes is greater than the percentage of girls who cut classes. c. The percentage of college students who drive old cars is higher than the percentage of noncollege people of the same age who drive old cars.

Determine the critical values that would be used for the following hypothesis tests (using the classical approach) about the difference between two means with population variances unknown. a. \(\quad H_{a}: \mu_{1}-\mu_{2} \neq 0, n_{1}=26, n_{2}=16, \alpha=0.05\) b. \(\quad H_{a}: \mu_{1}-\mu_{2}<0, n_{1}=36, n_{2}=27, \alpha=0.01\) c. \(\quad H_{a}: \mu_{1}-\mu_{2}>0, n_{1}=8, n_{2}=11, \alpha=0.10\) d. \(\quad H_{a}: \mu_{1}-\mu_{2} \neq 10, n_{1}=14, n_{2}=15, \alpha=0.05\)

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.