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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.

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.

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.

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.

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.