Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1, resulting in an average yield of 84 and a sample standard deviation of 3. Fifteen batches were prepared using catalyst 2, and they resulted in an average yield of 92 with a standard deviation of 2. Assume that yield measurements are approximately normally distributed with the same standard deviation.
(a) Is there evidence to support the claim that catalyst 2 produces higher mean yield than catalyst 1? Use a = 0.01. Yes.
(b) Find a 99% confidence interval on the difference in mean yields that can be used to test the claim in part (a). Round your answer to two decimal places (e.g. 98.76).

To address the given problem, we will perform a hypothesis test to determine if there is evidence that catalyst 2 produces a higher mean yield than catalyst 1 and find the 99% confidence interval for the difference in mean yields.

### Part (a): Hypothesis Testing

We will conduct a two-sample t-test for the difference in means.

1. **Define the hypotheses:**
   \[
   H_0: \mu_1 = \mu_2 \quad \text{(the mean yields are the same)}
   \]
   \[
   H_a: \mu_2 > \mu_1 \quad \text{(catalyst 2 has a higher mean yield than catalyst 1)}
   \]

2. **Calculate the test statistic:**

The formula for the test statistic for the difference in means with unequal sample sizes but assuming equal population variances is:
\[
t = \frac{\bar{X}_2 - \bar{X}_1}{\sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}
\]
where \(s_p^2\) is the pooled sample variance, calculated as:
\[
s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}
\]

Given:
- \(\bar{X}_1 = 84\), \(s_1 = 3\), \(n_1 = 12\)
- \(\bar{X}_2 = 92\), \(s_2 = 2\), \(n_2 = 15\)

First, calculate the pooled variance:
\[
s_p^2 = \frac{(12 - 1)3^2 + (15 - 1)2^2}{12 + 15 - 2} = \frac{11 \cdot 9 + 14 \cdot 4}{25} = \frac{99 + 56}{25} = \frac{155}{25} = 6.2
\]

Next, calculate the test statistic:
\[
t = \frac{92 - 84}{\sqrt{6.2 \left(\frac{1}{12} + \frac{1}{15}\right)}} = \frac{8}{\sqrt{6.2 \left(\frac{5}{60}\right)}} = \frac{8}{\sqrt{6.2 \cdot \frac{1}{12}}} = \frac{8}{\sqrt{0.5167}} = \frac{8}{0.7189} \approx 11.13
\]

3. **Determine the critical value and compare:**

For a one-tailed test with \(\alpha = 0.01\) and degrees of freedom \(df = n_1 + n_2 - 2 = 25\), the critical value from the t-distribution table is approximately 2.485.

Since \( t = 11.13 \) is much greater than 2.485, we reject the null hypothesis.

**Conclusion:**
There is strong evidence to support the claim that catalyst 2 produces a higher mean yield than catalyst 1.

### Part (b): Confidence Interval

To find the 99% confidence interval for the difference in means, use the formula:
\[
(\bar{X}_2 - \bar{X}_1) \pm t_{\alpha/2, \, df} \times \sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
\]

Using the same pooled variance \(s_p^2 = 6.2\) and \(t_{\alpha/2, \, df} = 2.787\) for 99% confidence (two-tailed):

\[
\text{Margin of error} = 2.787 \times \sqrt{6.2 \left(\frac{1}{12} + \frac{1}{15}\right)} = 2.787 \times 0.7189 \approx 2.00
\]

Thus, the 99% confidence interval is:
\[
(92 - 84) \pm 2.00 = 8 \pm 2.00 = (6.00, 10.00)
\]

**Conclusion:**
The 99% confidence interval for the difference in mean yields between catalyst 2 and catalyst 1 is (6.00, 10.00).

Would you like more details or have any questions? Here are some related questions you might want to explore:

1. How do we interpret the results of a hypothesis test?
2. What are the assumptions underlying the t-test for comparing two means?
3. How do we calculate the pooled variance in a t-test?
4. What is the difference between a one-tailed and a two-tailed test?
5. How do we determine the appropriate critical value for a given significance level?

**Tip:** Always check the assumptions of the statistical test you are using to ensure the validity of the results.

Explore a detailed solution to compare mean yields of two catalysts using hypothesis testing and confidence intervals. Analyze the statistical evidence supporting which catalyst produces higher yields with a 99% confidence interval. Essential for understanding statistical inference in batch chemical processes.

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