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When conducting a test of hypotheses, you are essentially comparing what you observe in your data to what was expected under a specific hypothesis. The hypothesis typically comes in two forms: the null hypothesis (\(H_0\) which usually states that there is no effect or no difference) and the alternative hypothesis (\(H_a\) or \(H_1\) which suggests that there is an effect or a difference). In the context of the given exercise, the null hypothesis is that the true mean \( \mu = 4 \) and the alternative hypothesis is that \( \mu eq 4 \).
The importance of distinguishing these hypotheses lies in the direction of your research question and in setting the stage for the statistical test you'll perform, which in this case involves using a t-distribution.

The significance level, often denoted by \( \alpha \), is the threshold for determining whether a statistical result is significant enough to reject the null hypothesis. Typically a 5% or 0.05 significance level is used, meaning there is a 5% risk of concluding that there is a difference when there is none (Type I error).
It sets the bar for how extreme the test statistic needs to be in order to say that the results are not just due to random chance. In the exercise, a 5% significance level indicates that you require strong evidence against the null hypothesis to reject it.

The concept of degrees of freedom (df) is crucial in the context of making statistical estimates. It refers to the number of values in the final calculation of a statistic that are free to vary. Degrees of freedom are often associated with the sample size. In hypothesis testing, df is used to determine the critical value of the test statistic.
In the given exercise example, with a sample size (\( n \) of 15, the degrees of freedom would be \( n - 1 = 14 \). Knowing the degrees of freedom, you can reference the t-distribution table to find the critical values required for finding the significance of the test statistic.

The test statistic calculation is a standardized value that is calculated from sample data during a hypothesis test. It's a critical figure as it allows you to make a decision about the hypothesis. To compute a t-test statistic, you subtract the null hypothesis value from the sample mean, then divide by the standard error of the mean. That formula looks like this: \[ t = \frac{{\bar{x} - \mu}}{{s/ \sqrt{n}}} \]
In the problem, after plugging in the provided numbers, you determined the test statistic based on the sample average, the hypothesized mean, the sample standard deviation, and the sample size.

A two-tailed test is a statistical hypothesis test in which the area of rejection is on both sides of the sampling distribution. This means you are checking for the possibility of an effect in two directions—both higher and lower than the hypothesized value.
In relation to the task at hand, the null hypothesis is tested against an alternative hypothesis that just says the mean is not equal to 4 (\( \mu eq 4 \)), without specifying whether it is greater or less. Therefore, you look at both tails of the t-distribution when seeking evidence to reject the null. The calculated t-score is then compared against the critical values for both ends of the distribution.