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A paired sample t-test is a statistical method used to compare two related samples, typically used to determine whether there is a significant difference between two sets of paired data. In this context, the paired sample t-test helps us evaluate whether the impurity levels measured by two different analytical tests on the same steel alloy samples differ significantly. Each pair consists of the impurity level measured by Test 1 and Test 2 on the same specimen.

For the paired sample t-test, we follow these steps:

• Calculate the difference between the measurements for each pair. This difference measures the discrepancy in impurity levels between the two tests for each sample.
• Compute the mean of these differences, denoted as \( \bar{d} \).
• Calculate the standard deviation of the differences, \( s_d \).
• Formulate the null hypothesis (\( H_0: \mu_d = 0 \)), suggesting no difference between test means, and the alternative hypothesis (\( H_a: \mu_d eq 0 \)), suggesting a significant difference in means.
• Use the t-statistic given by \( t = \frac{\bar{d}}{s_d / \sqrt{n}} \), where \( n \) is the number of paired samples, and compare it to a critical value from the t-distribution.

By following these steps, we can determine if the impurity levels are statistically different based on the tests conducted.

Understanding the mean difference is crucial in hypothesis testing, especially when dealing with paired samples. The mean difference (\( \bar{d} \)) is the average of all differences calculated between paired measurements. In the context of impurity level testing, the mean difference shows the average change in impurity levels from one test to the other.

This concept becomes vital when you have a hypothesis about the direction or magnitude of a change. For instance, in part (b) of the exercise, the hypothesis is that Test 1 generates results that are a mean of 0.1 units lower than Test 2. The null hypothesis would be \( H_0: \mu_d = -0.1 \) (meaning no significant difference), while the alternative is \( H_a: \mu_d < -0.1 \) (indicating Test 1 results are significantly lower).

To test these claims, compute the t-statistic using \( t = \frac{\bar{d} + 0.1}{s_d / \sqrt{n}} \). This statistic tells us whether the observed mean difference is likely due to random chance or reflects a true difference in the impurity levels measured by the two tests.

Determining the appropriate sample size is a critical step in designing experiments, ensuring the study has enough power to detect a true effect. In the given scenario, part (c) requires us to determine whether testing eight specimens is enough to detect if Test 1's mean is 0.1 lower than Test 2's with a probability (or power) of at least 0.90.

To calculate this, use the following formula for paired t-tests:

• \[ n > \left(\frac{(Z_{1-\beta} + Z_{\alpha/2}) \cdot \sigma}{\Delta} \right)^2 \]
• Where \( Z_{1-\beta} \) is the z-value that corresponds to the desired power (0.90 in this case), \( Z_{\alpha/2} \) represents the z-value for the given significance level \( \alpha = 0.05 \).
• \( \sigma \) is the standard deviation of the differences.
• \( \Delta \) is the mean difference you wish to detect, which is 0.1 in this case.

If eight alloys do not meet this criterion, we must increase the sample size until this inequality holds true. Proper sample size ensures that the test results are reliable and minimizes the risk of type II errors (failing to detect a real difference when one exists).