91Ó°ÊÓ

The article "An Alternative Vote: Applying Science to the Teaching of Science" (The Economist, May 12,2011 ) describes an experiment conducted at the University of British Columbia. A total of 850 engineering students enrolled in a physics course participated in the experiment. These students were randomly assigned to one of two experimental groups. The two groups attended the same lectures for the first 11 weeks of the semester. In the twelfth week, one of the groups was switched to a style of teaching where students were expected to do reading assignments prior to class and then class time was used to focus on problem solving, discussion and group work. The second group continued with the traditional lecture approach. At the end of the twelfth week, the students were given a test over the course material from that week. The mean test score for students in the new teaching method group was 74 and the mean test score for students in the traditional lecture group was 41 . Suppose that the two groups each consisted of 425 students and that the standard deviations of test scores for the new teaching method group and the traditional lecture method group were 20 and 24 , respectively. Can you conclude that the mean test score is significantly higher for the new teaching method group than for the traditional lecture method group? Test the appropriate hypotheses using a significance level of \(\alpha=0.01\).

Step by step solution

01

Formulate the Hypotheses

We want to compare the mean test scores between the new teaching method and the traditional lecture method. We set up the null hypothesis (\(H_0\)) and alternative hypothesis (\(H_a\)) as follows: \(H_0: \mu_1 - \mu_2 = 0\) - There is no difference in mean test scores between the new teaching method and the traditional lecture method. \(H_a: \mu_1 - \mu_2 > 0\) - The mean test score for the new teaching method is significantly higher than the traditional lecture method.

02

Determine the Test Statistic

We will perform a two-sample t-test using the provided means (\(\bar{x}_1\) and \(\bar{x}_2\)), standard deviations (s1 and s2), and sample sizes (n1 and n2). The test statistic formula for a two-sample t-test is: *t = \(\frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{s^2_1}{n_1}+ \frac{s^2_2}{n_2}}}\)* Using the given data: - \(\bar{x}_1 = 74\) - \(\bar{x}_2 = 41\) - \(s_1 = 20\) - \(s_2 = 24\) - \(n_1 = n_2 = 425\) Plugging in the values into the formula: *t = \(\frac{(74 - 41) - (0)}{\sqrt{\frac{20^2}{425} + \frac{24^2}{425}}}\)*

03

Calculate the Test Statistic

Calculate the test statistic: *t = \(\frac{33}{\sqrt{\frac{400}{425} + \frac{576}{425}}}\)* *t ≈ 19.81*

04

Determine the Critical Region

We need to find the t-value that corresponds to the significance level (\(\alpha=0.01\)) and the degrees of freedom (df). The degrees of freedom for a two-sample t-test can be calculated using: *df = \(n_1 + n_2 - 2 = 425 + 425 - 2 = 848\)* Using a t-table or a calculator, we find the critical t-value for a one-tailed test with 848 degrees of freedom and \(\alpha=0.01\) is: *Critical t-value ≈ 2.33*

05

Make a Conclusion

Since the calculated test statistic (t ≈ 19.81) is greater than the critical t-value (2.33), we reject the null hypothesis (\(H_0\)). Therefore, we can conclude that the mean test score is significantly higher for the new teaching method group than for the traditional lecture method group at a significance level of \(\alpha=0.01\).

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.

Two-Sample T-Test

When comparing two groups, like different teaching methods, a two-sample t-test helps determine if there's a significant difference in their means.
This test considers both sample size and variability to see if observed differences are due to chance.

• Purpose: It tests the difference between means from two independent groups.
• Assumptions: Each group should have a normal distribution, and data should have similar variances.
• Calculation: The formula for the two-sample t-test is:
  \[ t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{s^2_1}{n_1}+ \frac{s^2_2}{n_2}}} \]
  where \(\bar{x}_1\) and \(\bar{x}_2\) are sample means, and \(s_1\) and \(s_2\) are standard deviations.

With these parameters, statisticians can evaluate if observed differences are statistically significant.

Null Hypothesis

In hypothesis testing, the null hypothesis \((H_0)\) asserts that there's no effect or difference between the groups.
It's a claim that any observed difference is due to chance.

• Example: In comparing teaching methods, \(H_0\) is that both methods yield the same mean test score.
• Formulation: For the exercise, it's set as \(H_0: \mu_1 - \mu_2 = 0\).
• Importance: It provides a baseline for testing significance and helps strengthen claims by being proven false.

Rejecting \(H_0\) indicates significant differences between groups.

Alternative Hypothesis

The alternative hypothesis \((H_a)\) suggests that there is a meaningful difference between the groups.
For our example, it supports the notion that one teaching method is superior.

• Example: In the teaching method study, \(H_a\): \(\mu_1 - \mu_2 > 0\) suggests the new teaching approach results in higher scores.
• Role: It opposes the null hypothesis and aims to show a link or effect.
• Outcome: If statistical evidence is strong enough, \(H_a\) is accepted, supporting the claim of difference.

The alternative hypothesis ensures researchers can validate changes or improvements with evidence.

Significance Level

The significance level \((\alpha)\) helps determine how strong the evidence must be to reject the null hypothesis.
It's a threshold for deciding if an observed effect is statistically significant.

• Value: Common choices are 0.05, 0.01, etc. In this exercise, \(\alpha = 0.01\).
• Interpretation: An \(\alpha\) of 0.01 means there's a 1% risk of concluding a difference exists when it doesn't.
• Application: If the calculated test statistic exceeds the critical value from a t-table at \(\alpha\), we reject \(H_0\).

This benchmark helps ensure reliability and confidence in research findings.