Question

Do teenage boys worry more than teenage girls? This is one of the questions addressed by the authors of the article "The Relationship of Self-Esteem and Attributional Style to Young People's Worries" (Journal of Psychology [1987]: 207-215). A scale called the Worries Scale was administered to a group of teenagers, and the results are summarized in the accompanying table. \begin{tabular}{lccc} & & Sample & Sample \\ Gender & \(n\) & Mean Score & sd \\ \hline Girls & 108 & \(62.05\) & \(9.5\) \\ Boys & 78 & \(67.59\) & \(9.7\) \\ & & & \\ \hline \end{tabular} Is there sufficient evidence to conclude that teenage boys score higher on the Worries Scale than teenage girls? Use a significance level of \(\alpha=.05\).

Short answer

The answer really depends on the t-statistic calculated in step 2. If the calculated t-statistic is higher than 1.653, then we can conclude that there is sufficient evidence to say that the teenage boys score higher on the Worries Scale than teenage girls. If it is lower, we cannot conclude the same. You'll need to measure the values into the formulas correctly in order to achieve a valid result.

Step-by-step solution

Formulation of Null and Alternative Hypotheses

The first step in hypothesis testing is to formulate the null and alternative hypotheses. The null hypothesis (\(H_0\)) is the claim that there is no difference between the mean scores of boys and girls, i.e., the mean score of boys equals the mean score of girls. The alternative hypothesis (\(H_1\)) is the claim that there is a difference, in this case, that the mean score of boys is higher than that of girls. So, \(H_0: \mu_{boys} = \mu_{girls}\) and \(H_1: \mu_{boys} > \mu_{girls}\) (\(\mu\) refers to the population mean).

Calculate the Test Statistic

Next, calculate the t-statistic using the formula: \( t = (\bar{x}_{boys} - \bar{x}_{girls}) / \sqrt{ (s_{b}^2/n_{b}) + (s_{g}^2/n_{g}) } \) where \(\bar{x}\) denotes the sample mean, \(s\) denotes the standard deviation, and \(n\) denotes the sample size. In this scenario, the values would be as follows: \(\bar{x}_{boys} = 67.59\), \(\bar{x}_{girls} = 62.05\), \(s_{boys} = 9.7\), \(s_{girls} = 9.5\), \(n_{boys} = 78\), and \(n_{girls} =108\).

Determine the Critical t-value

The critical value of t is determined by the chosen significance level, and the degrees of freedom, which is the sum of the two sample sizes minus 2 (\(n_{boys} + n_{girls} - 2\)). For a significance level of \(alpha = .05\) with a one-tailed test (because the problem involves a 'greater than' hypothesis), degrees of freedom = 184 (78 +108 -2), you get a critical value of approximately 1.653.

Compare Test Statistic with Critical t-value and Make Decision

The decision rule is as follows: if the calculated t-statistic is greater than the critical value, reject the null hypothesis. If it is less, fail to reject the null hypothesis. Therefore, compare the calculated t-statistic with 1.653 decide whether to reject or accept \(H_0\).

Conclusion

Interpret results in the context of the problem. If you rejected the null hypothesis, you can conclude that there is sufficient evidence that teenage boys worry more than teenage girls according to the Worries Scale. If you failed to reject the null hypothesis, you can conclude that there is not enough evidence to suggest that one worries more than the other based on the given sample data.

Key concepts

Null and Alternative Hypotheses

When we talk about null and alternative hypotheses in statistics, we delve into the heart of hypothesis testing. The null hypothesis, denoted as \( H_0 \), represents the default statement that there is no effect or no difference. In the context of the exercise, the null hypothesis posits that there is no difference in the mean Worries Scale scores between teenage boys and girls.

On the other hand, the alternative hypothesis, denoted as \( H_1 \), contradicts the null hypothesis, suggesting that there is an effect or a difference. In our study, the alternative hypothesis claims that teenage boys have a higher mean Worries Scale score than girls. To sum it up, for the given problem, the null hypothesis is \( H_0: \mu_{boys} = \mu_{girls} \) and the alternative hypothesis is \( H_1: \mu_{boys} > \mu_{girls} \). Establishing these hypotheses frames the direction of our statistical test.

Test Statistic Calculation

The test statistic is a standardized value that is calculated from sample data during a hypothesis test. Its purpose is to help us decide whether to support or reject the null hypothesis. To compute this statistic, we use the formula:
\[ t = (\bar{x}_{boys} - \bar{x}_{girls}) / \sqrt{ (s_{b}^2/n_{b}) + (s_{g}^2/n_{g}) } \]
This formula compares the difference in sample means to the variability of the scores. The numerator represents the observed difference between sample means, while the denominator encompasses the standard error of the difference. A higher absolute value of the test statistic indicates greater evidence against the null hypothesis. For the Worries Scale problem, plugging in the sample means, standard deviations, and sizes leads us to our calculated t-statistic.

Critical Value Determination

Determining the critical value is essential in the process of hypothesis testing. It is the threshold against which the test statistic is compared to decide whether the null hypothesis should be rejected. The critical value depends on the predetermined significance level and the distribution of the test statistic.

In this case, we're using a t-distribution to find our critical value. Since we are performing a one-tailed test with a significance level \( \alpha = .05 \), we look to the t-distribution table to find the critical value that corresponds to an area of 0.05 in the upper tail of the distribution with the appropriate degrees of freedom (184 in this scenario). A t-distribution is skewed based on the number of degrees of freedom but tends towards a normal distribution as the degrees of freedom increase.

Significance Level

The significance level, denoted as \( \alpha \), is a threshold set by the researcher indicating the probability of rejecting the null hypothesis when it is actually true. Commonly, a level of 0.05 is used, which offers a good balance between the risks of making a Type I error (false positive) and not detecting a true effect (Type II error).

In the current problem, we're using a significance level of \( \alpha = .05 \). This represents a 5% risk of concluding that a difference exists when there is none. Based on this level, and the critical value, we will decide whether the evidence from our sample is strong enough to reject the null hypothesis.

t-test Analysis

The t-test is a statistical procedure used to determine whether there is a significant difference between the means of two groups. In the Worries Scale problem, we’re conducting an independent samples t-test because we're comparing the scores of two separate groups—teenage boys and girls.

Analysis involves comparing the calculated t-statistic to the critical t-value. If the t-statistic falls into the critical region (exceeding the critical value), it suggests that there's a statistically significant difference between the group means at the chosen significance level. Thus, the null hypothesis would be rejected in favor of the alternative hypothesis—that teenage boys score higher on the Worries Scale than teenage girls.