Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1, resulting in an average yield of 85 and a sample standard deviation of 3. Fifteen batches were prepared using catalyst 2, and they resulted in an average yield of 89 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 (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).

We are testing whether catalyst 2 produces a higher yield. We always test for equality, that's why the null has an equals sign. If catalyst 2 has a higher yield, the difference would result in a negative a number (take a number, and subtract a larger number than that number).

The problem just asks if there is evidence to support the claim that catalyst 2 produces a higher yield, it doesn't give a significance level, so we will use the same significance level for part b, which is 99%. We will be using a t-score since both samples sizes are less than 30. For this, the degress of freedom are 2 less than the sum of the two sample sizes. in this case 25. This is a left tailed test since we are testing for a difference, claiming that µ2 is greater, so if the test statistic is too low a value, we will reject the null.

The corresponding t-score for a significance level of 99% with a one tailed test is t < -2.485

See attached photo 1 for the hypothesis test calculations...

Our test statistic is t = -4.148

-4.148 < -2.485, so we reject the null hypothesis. This means that there is sufficient evidence to support the claim that catalyst 2 produces a higher yield that catalyst 1.

We must now build a confidence interval. See attached photo 2 for the construction of the confidence interval

joaobezerra joaobezerra · Answer 2 · 2021-11-16T18:09:42+01:00

Using the t-distribution, it is found that:

a) Since the lower bound of the confidence interval found in item a is above 0, there is evidence to support the claim that catalyst 2 produces higher mean yield than catalyst 1.

b) The 99% confidence interval on the difference in mean yields is (1.19, 6.81).

The best way to solve this question is finding the confidence interval in item b, and then using it to answer item a.

The standard errors are:

[tex]s_{1} = \frac{3}{\sqrt{12}} = 0.866[/tex]

[tex]s_{2} = \frac{2}{\sqrt{15}} = 0.5164[/tex]

The distribution of the differences has:

[tex]\overline{x} = \mu_{2} - \mu_{1} = 89 - 85 = 4[/tex]

[tex]s = \sqrt{s_{2}^2 + s_{1}^2} = \sqrt{0.866^2 + 0.5164^2} = 1.0083[/tex]

Item b:

The confidence interval is:

[tex]overline{x} \pm ts[/tex]

The critical value, using a t-distribution calculator, for a two-tailed 99% confidence interval, with 12 + 15 - 2 = 25 df, is t = 2.7874.

Then:

[tex]\overline{x} - ts = 4 - 2.7874(1.0083) = 1.19[/tex]

[tex]\overline{x} + ts = 4 + 2.7874(1.0083) = 6.81[/tex]

The 99% confidence interval on the difference in mean yields is (1.19, 6.81).

Item a:

Since the lower bound of the confidence interval found in item a is above 0, there is evidence to support the claim that catalyst 2 produces higher mean yield than catalyst 1.

A similar problem is given at https://brainly.com/question/15682365