91Ó°ÊÓ

The article "Religion and Well-Being Among Canadian University Students: The Role of Faith Groups on Campus" (Journal for the Scientific Study of Religion [1994]: \(62-73\) ) compared the self-esteem of students who belonged to Christian clubs and students who did not belong to such groups. Each student in a random sample of \(n=169\) members of Christian groups (the affiliated group) completed a questionnaire designed to measure self-esteem. The same questionnaire was also completed by each student in a random sample of \(n=124\) students who did not belong to a religious club (the unaffiliated group). The mean self-esteem score for the affiliated group was \(25.08\), and the mean for the unaffiliated group was \(24.55 .\) The sample standard deviations weren't given in the article, but suppose that they were 10 for the affiliated group and 8 for the unaffiliated group. Is there evidence that the true mean self- esteem score differs for affiliated and unaffiliated students? Test the relevant hypotheses using a significance level of \(.01\).

Step by step solution

01

Formulate Hypotheses

The null hypothesis (\(H0\)) is that the true mean self-esteem score for both groups is the same: \(μ1 = μ2\). The alternative hypothesis (\(Ha\)) is that the means are not equal: \(μ1 ≠ μ2\).

02

Calculate the Test Statistic

The test statistic for comparing two means is calculated as follows: \[Z = \frac{{(\overline{X}_1 - \overline{X}_2) - (μ1 - μ2)}}{{\sqrt{\frac{{σ1^2}}{n1} + \frac{{σ2^2}}{n2}}}}\]. Here, \(\overline{X}_1\) and \(\overline{X}_2\) are the sample means, \(n1\) and \(n2\) are the sample sizes, \(σ1^2\) and \(σ2^2\) are the variances for the two groups. Substituting from the problem, we have \[Z = \frac{{(25.08 - 24.55) }}{{\sqrt{\frac{{10^2}}{169} + \frac{{8^2}}{124}}}}\]. After the calculation, obtain the Z score.

03

Determine the Critical Value and Make Decision

At a significance level of 0.01, the critical values for a two-tailed test are -2.58 and 2.58. If your Z score falls outside of this range, we reject the null hypothesis. Apply this rule to your computed Z score.

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.

Null Hypothesis

In hypothesis testing, the null hypothesis, often denoted as \(H_0\), represents a statement that there is no effect or no difference in the case being assessed. It is the hypothesis that researchers usually aim to test. In the given exercise, the null hypothesis is that the mean self-esteem scores of students belonging to Christian clubs (affiliated) and those who do not belong to such groups (unaffiliated) are the same. This can be mathematically expressed as \(μ_1 = μ_2\).

• It assumes that any observed difference in the sample data is due to random chance.
• The null hypothesis acts as a neutral statement that stands until evidence strongly suggests otherwise.

The goal in testing the null hypothesis is to collect statistical evidence to assess its validity. If the evidence in the data is strong, the null hypothesis may be rejected in favor of the alternative hypothesis.

Alternative Hypothesis

In contrast to the null hypothesis, the alternative hypothesis, symbolized as \(H_a\), proposes that there is a meaningful effect or difference. It states what the researcher is typically trying to prove or ascertain. In our exercise, the alternative hypothesis is that the mean self-esteem scores of the affiliated and unaffiliated students are not equal, represented as \(μ_1 eq μ_2\).

• The alternative hypothesis captures the idea that they differ significantly, essentially implying \(μ_1 eq μ_2\).
• It is the assertion that will be considered only if the null hypothesis is deemed unlikely.

Acceptance of the alternative hypothesis suggests that differences in means are statistically significant and not a product of random chance.

Significance Level

The significance level, represented by \(\alpha\), is a threshold that helps determine when a result is remarkable enough to reject the null hypothesis. It represents the probability of rejecting the null hypothesis when it is actually true (a Type I error). In this exercise, a significance level of 0.01 is used.

• A significance level of 0.01 means that there is only a 1% risk of concluding that a difference exists when there is none.
• This strict threshold is chosen to reduce the likelihood of a false positive and helps ensure that the findings are robust and reliable.

The smaller the significance level, the stronger the evidence must be to reject the null hypothesis.

Test Statistic

The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used to decide whether to reject the null hypothesis. In the exercise, the test statistic for comparing two means is calculated using Z-scores. The formula used is: \[Z = \frac{{(\overline{X}_1 - \overline{X}_2) - (μ_1 - μ_2)}}{{\sqrt{\frac{{σ_1^2}}{n_1} + \frac{{σ_2^2}}{n_2}}}}\] Where:

• \(\overline{X}_1\) and \(\overline{X}_2\) are the sample means of the groups.
• \(n_1\) and \(n_2\) are the sizes of the samples.
• \(σ_1^2\) and \(σ_2^2\) are the sample variances.

The Z-score allows us to determine how far away our sample statistic is from the null hypothesis. If the Z-score falls beyond our critical value range, it suggests the sample provides sufficient evidence to reject the null hypothesis.