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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 90 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 alpha = 0.01.
(B) Find a 99% confidence interval on the difference in mean yields that can be used to test the claim in part (a). (e.g. 98.76).

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Answer:

Step-by-step explanation:

From the given information:

Let assume:

[tex]\mu_1 =[/tex] population mean for catalyst 1

[tex]\mu_2 =[/tex] population mean for catalyst 2

Then:

Null hypothesis: [tex]\mu_1 \ge \mu_2[/tex]

Alternative hypothesis: [tex]\mu_1 <\mu_2[/tex]

[tex]\alpha= 0.01[/tex]

By using MINITAB software to compute the 2 sample t-test, we have:

Two-Sample T_Test and CI

Sample    N     Mean     StDev      SE Mean

1               12     94.00      3.00       0.87

2              15      90.00     2.00      0.52

Difference = [tex]\mu_1(1)-\mu_2(2)[/tex]

Estimate of difference: -6.00

99% upper bound for difference: -3.603

T-test of difference: 0  (vs <): T-value = -6.22

P.value = 0.000

DF = 25

Both use Pooled  StDev = 2.4900

From above result

the test statistics = -6.00

p-value = 0.00

Decision Rule: To reject [tex]H_o\ \ if \ \ p \le \alpha[/tex]

Conclusion: Provided that p-value is < ∝, we reject [tex]H_o[/tex]. Hence, there is sufficient evidence to support the given claim.

b) From the MINITAB;

The 99% C.I on the difference in the mean yields that can be applied to test the claim in part (a) is:

[tex]\mathbf{\mu_1 -\mu_2 \le -3.60}[/tex]

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