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When performing hypothesis testing with two sample groups, we often assume that the variances of the two populations are equal. This assumption allows us to compute the pooled variance, a key concept that helps us get a more stable estimate of variance when the sample sizes are equal or nearly equal.
The pooled variance is calculated using the formula:

• \( s_p^2 = \frac{{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}}{{n_1 + n_2 - 2}} \)

This formula averages the variances, weighted by the degrees of freedom from each sample. In our context, it combines variance data from two sample groups to provide a comprehensive view.
Why use pooled variance? It enhances the precision of estimations by incorporating more data (from both groups), giving us a better understanding of the data's overall variability. By using pooled variance, you effectively have more information to base your hypothesis test on, which increases the reliability of the conclusions drawn.

The t-distribution is central to hypothesis testing when dealing with small sample sizes, typically under 30. It's used to understand the variability of the sample data. Unlike a normal distribution, the t-distribution takes into account this variability with a shape that is slightly more spread out and clumped towards the center for small sample sizes.
What makes the t-distribution special is its shape, which depends on degrees of freedom. As the degrees of freedom increase, the t-distribution begins to resemble a normal distribution more closely.
In our exercise, we calculate the test statistic using the t-distribution:

• \( t = \frac{{\bar{x}_1 - \bar{x}_2}}{{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}} \)

The t-distribution helps determine the probability of observing a test statistic as extreme as, or more extreme than, the observed value if the null hypothesis is true. Knowing this probability is what ultimately guides decisions in hypothesis testing, such as whether to reject the null hypothesis.

A confidence interval provides a range of values in which you can be reasonably sure the true parameter (like a mean difference) lies. In the context of hypothesis testing, it allows us to test hypotheses without relying solely on a critical value or \( P \)-value.
In our problem, a one-sided confidence interval is constructed for the difference in means:

• \((\bar{x}_1 - \bar{x}_2) \pm t_{crit} \cdot s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\)

This interval is checked against the value proposed by the null hypothesis. If the interval doesn't include the null hypothesis value (usually zero for differences), we reject the null hypothesis.
Using confidence intervals adds context and depth to hypotheses, showing not just whether to reject a hypothesis but also the precision and potential range of the actual effect.

The power of a test is the probability that it correctly rejects a false null hypothesis. A test's power depends on several factors:

• The true difference between population means (effect size)
• The sample size (larger samples increase power)
• The chosen significance level \( \alpha \)
• The variability in the data (less variance increases power)

In essence, power measures a test's sensitivity to detect an actual effect or difference when it exists. In our exercise, where the alternative hypothesis proposes a mean shifted by 3 units, the power was found to be approximately 0.92, indicating a very high probability of correctly rejecting a false null hypothesis.
To achieve desired power levels (like 0.95 for high confidence), you may need to adjust sample sizes or acceptive more significant effects.