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The t-statistic is a key component in hypothesis testing, specifically when dealing with small sample sizes. It helps determine if there is a statistically significant difference between means of two groups. In this context, we use it to evaluate the difference in performance between two types of coatings.

To compute the t-statistic, the formula is:\[t = \frac{\underline{d} - \mu_{鈧攠}{s / \sqrt{n}}\]Where:

• \(\underline{d}\) is the sample mean difference.
• \(\mu_{鈧攠\) is the mean difference under the null hypothesis.
• \(s\) is the standard deviation of differences.
• \(n\) is the sample size.

This formula considers the variance within the sample and the deviation from the hypothesized mean, allowing us to see how many standard deviations the sample mean is away from the hypothesized population mean.

In this exercise, with a sample mean difference \(\underline{d} = 8\), the null hypothesis mean \(\mu_{鈧攠 = 0\), a standard deviation \(s = 11.03\), and a sample size of 15, our t-statistic is approximately 2.08. This number tells us how extreme our sample statistic is in relation to the hypothesized population parameter.

The null hypothesis is a fundamental part of hypothesis testing. It acts as a starting assumption that there is no effect or difference, and any observed effect is due to random chance. In this exercise, the null hypothesis \((H鈧�)\) states that there is no significant difference between the corrosion preventing abilities of coatings A and B, implying that the mean of differences between the coatings is zero.

Formally, the null hypothesis is represented as:

• \(H鈧�: \mu_{鈧攠 = 0\)

This statement assumes that any observed difference in the mean maximum pits between the two coatings could simply be a random occurrence rather than a true difference in their effectiveness.

Testing the null hypothesis involves calculating a statistic, like the t-statistic, and comparing it against a critical value to determine if the assumed equality (no difference) should be rejected or not. If the t-statistic falls outside the range defined by the critical values, we would reject the null hypothesis in favor of the alternative, which suggests that there is a significant difference. However, in our case, as the calculated t-statistic is less than the critical value, we fail to reject the null hypothesis.

The significance level in hypothesis testing is a threshold that determines whether to reject the null hypothesis. It's usually denoted by \(\alpha\) and represents the probability of rejecting the null hypothesis when it is actually true. In simpler terms, it's the risk we take to make a Type I error.

In this exercise, the significance level is set at 0.05, meaning there is a 5% risk of concluding that a difference exists when there is none. This level was chosen before performing the test to ensure that any conclusions are based on a pre-determined threshold of confidence.

With a two-tailed test, as in this problem, this significance level implies that the critical region, where we would reject the null hypothesis, is split on both ends of the distribution. The critical t-value thus divides 5% into 2.5% on each tail. Therefore, a calculated t-statistic that falls beyond these critical values would suggest that the null hypothesis should be rejected.

Setting a significance level of 0.05 is common practice, as it balances the risk of Type I error with the power to detect an actual effect. However, it's important to choose a level that is suitable for the specific context and acceptable to the stakeholders involved.