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A confidence interval provides a range of values where we expect the true parameter to lie, with a certain level of confidence. In hypothesis testing, we often use a confidence interval to assess the validity of a hypothesis. For example, in testing if the means of two groups are equal, a confidence interval can help determine if the observed difference could realistically be zero. If zero lies within the interval, it suggests that our result could happen by chance if the null hypothesis is true. To calculate a one-sided confidence interval for the difference between two means, we use the formula:\[ CI = (\bar{x}_1 - \bar{x}_2) + z_{\alpha} \times \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \]Here, \(\bar{x}_1\) and \(\bar{x}_2\) are sample means, \(\sigma_1\) and \(\sigma_2\) are known variances, and \(n_1\) and \(n_2\) are sample sizes from each group. The \(z_{\alpha}\) value is chosen based on the desired level of confidence, such as 95%.Using this approach in the original exercise, we concluded that since zero is within our computed interval, we should not reject the null hypothesis.

The test statistic is a standardized value calculated from sample data during a hypothesis test. It helps determine the likelihood of observing the sample data if the null hypothesis is true. The formulation depends on the nature of the hypothesis and available information, like known variances.In our specific exercise, the test statistic \(z\) is calculated using:\[ z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]This formula checks how far the observed sample mean difference is from the hypothesized mean difference under the null hypothesis, measured in terms of standard errors.For the values given (\(\bar{x}_1 = 14.2\), \(\bar{x}_2 = 19.7\), etc.), the formula yielded \(z \approx -1.61\). Comparing this test statistic against the critical value from a normal distribution (-1.645 for 95% confidence), we determined there was not enough evidence to reject the null hypothesis.

The P-value quantifies the probability of obtaining test results at least as extreme as the observed data, assuming that the null hypothesis is correct. A smaller P-value indicates that the observed data are less likely under the null hypothesis, thus strengthening the evidence against it.In practice, the P-value is compared to the significance level \(\alpha\). If the P-value is less than or equal to \(\alpha\), we reject the null hypothesis. In our problem, we calculated the P-value to be approximately 0.0539, which is slightly higher than the \(\alpha = 0.05\) level. This larger value suggests insufficient evidence to reject the null hypothesis.Understanding the P-value helps students to interpret statistical significance and know when the null hypothesis might be considered true or questionable.

The power of a statistical test refers to its ability to correctly reject a false null hypothesis. It is the probability that the test will identify a true effect when one exists. Power informs us about the test's sensitivity and is calculated as \(1 - \beta\), where \(\beta\) is the probability of a Type II error (failing to reject a false null hypothesis).In the context of the exercise, power is determined for the specific scenario where the true difference is \(\mu_1 = \mu_2 - 4\). Using the modified test statistic, the power was calculated by evaluating the probability \(P(Z < -1.645 + 0.438)\). This yields a power of roughly 0.8849, implying a 88.49% chance of correctly rejecting the null hypothesis if the real mean difference is indeed 4 units.Plan for adequate power in hypothesis testing to minimize risks of making errors and drawing incorrect conclusions.