91Ó°ÊÓ

Hypothesis testing is a crucial part of statistics for determining whether there's a significant difference between two group means. The Two-sample z-test is specifically used when comparing the means from two different groups. In this scenario, the key question is whether the mean volumes filled by two machines are the same or different.

Start by setting up two hypotheses:

• Null Hypothesis \( H_0 \): The two group means are equal, \( \mu_1 = \mu_2 \).
• Alternative Hypothesis \( H_a \): The two group means are not equal, \( \mu_1 eq \mu_2 \).

You calculate the z-statistic using the formula:\[ z = \frac{ (\bar{x}_1 - \bar{x}_2) }{ \sqrt{ \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} } } \]

This involves the mean difference from your samples, \( \sigma_1 \) and \( \sigma_2 \) as standard deviations of the two groups, and \( n_1 \) and \( n_2 \) as the sizes of your samples. Finally, compare the calculated \( z \)-value to the critical \( z \)-value corresponding to your significance level (\( \alpha = 0.05 \)) to decide whether to reject or fail to reject the null hypothesis.

A confidence interval gives a range of values for the difference between two means, providing a better sense of how different or similar the means could be. In this exercise, a 95% confidence interval offers insights into the potential difference in filling volumes between two machines.

The formula for a confidence interval on the difference between two means is as follows:\[ (\bar{x}_1 - \bar{x}_2) \pm z_{\alpha/2} \sqrt{ \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} } \]

Here, \( \bar{x}_1 - \bar{x}_2 \) is the observed sample mean difference. The value of \( z_{\alpha/2} \) is known from z-tables (for a 95% confidence level, it is 1.96). Substitute the known values to get the interval.

If the confidence interval includes zero, it suggests that there's no significant difference between the machine outputs at a 5% significance level. If zero is outside this interval, you can infer a significant difference exists.

The power of a statistical test measures its ability to correctly reject a false null hypothesis, essentially reflecting the test's effectiveness. It is crucial because it tells you the likelihood that the test detects an actual difference when it truly exists.

In this situation, you are tasked with finding out the power of the test when the true mean difference is 0.04. The power is calculated as \( 1 - \beta \), where \( \beta \) represents the probability of a Type II error. This is mapped onto the standard normal distribution's cumulative distribution function:\[ \beta = \Phi \left( -z_{\alpha/2} + \frac{\delta}{\sqrt{ \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} }} \right) \]

With \( \delta = 0.04 \) as the true mean difference, substituting known values into the equation will provide the test power. A high power value (e.g., 0.8 or 80%) indicates that the test has a good chance of detecting a true difference.

Sample size calculation is a vital step to ensure that your test has sufficient power to detect a true difference when it occurs. It ties together significance level (\(\alpha\)), power (\(1 - \beta\)), and effect size (\(\delta\), the smallest difference you want to detect).

For our exercise, to achieve a power of 0.95 when \( \beta = 0.05 \) and the true mean difference \( \delta \) is 0.04, you use the formula:

\[ n = \left( \frac{(z_{\alpha/2} + z_{\beta}) \sqrt{ \sigma_1^2 + \sigma_2^2 }}{\delta} \right)^2 \]

The values \( z_{\alpha/2} \) and \( z_{\beta} \) are obtained from z-tables for the given \( \alpha \) and \( \beta \). Substituting the known values guides the determination of the minimal sample size each group should have, ensuring the test's reliability.