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In hypothesis testing, we aim to determine whether there is enough statistical evidence to support a specific claim. For the mentioned exercise, the claim is that catalyst 2 results in a higher mean yield compared to catalyst 1. We set up two hypotheses:

• The null hypothesis (H_0) states that there is no difference in means, i.e., \( \mu_1 = \mu_2 \).
• The alternative hypothesis (H_a) claims that the mean yield of catalyst 2 is greater, i.e., \( \mu_2 > \mu_1 \).

To test this, a two-sample t-test is ideal because it compares the means of two independent groups. We calculate a test statistic from the sample data and compare it with a critical value from the t-distribution table at a significance level of \( \alpha = 0.01 \).

When the calculated test statistic exceeds the critical value, it tells us that the observed data is unlikely under the null hypothesis. Thus, we reject H_0 and accept H_a, suggesting that the mean yield of catalyst 2 is indeed higher.

A confidence interval provides a range of values within which we can be confident that the true difference in means lies, based on the sample data. For our data, we are asked to calculate a 99% confidence interval for the difference in mean yields between the two catalysts.

Here's how we determine it:
If the confidence interval contains zero, it suggests that there is no significant difference between the two means. The calculation is given by:
\[ (\bar{x}_2 - \bar{x}_1) \pm t_{\alpha/2, df} \cdot s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} \]where \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means, \( s_p \) is the pooled standard deviation, and \( t_{\alpha/2, df} \) is the critical t-value comparing a total confidence level of 99%.

In our example, since the confidence interval includes zero and positive values, it supports the idea that catalyst 2 might have a superior yield to catalyst 1. This acts as further evidence supporting the alternative hypothesis from the hypothesis testing section.

The pooled standard deviation is a crucial step when performing a two-sample t-test, especially when the assumption of equal variances is valid for the two groups being compared. It serves as a combined estimate of the two group standard deviations and is used to compute the t-statistic for hypothesis testing.

In mathematical terms, the pooled standard deviation is calculated using:\[ s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}} \]Here, \( s_1 \) and \( s_2 \) are the sample standard deviations, and\( n_1 \) and \( n_2 \) are the sample sizes of the two groups. It assumes both groups have the same underlying variance. Using this pooled measure helps standardize the variability when comparing the two sample means.

This aggregated variance measure permits accurate calculation of the standard error in the formula for the t-statistic and the confidence interval, ensuring precise and reliable statistical conclusions.