Question

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).

Answer 1

Explanation:

From the given information:

Let assume:

`\mu_1 =` population mean for catalyst 1

`\mu_2 =` population mean for catalyst 2

Then:

Null hypothesis:
`\mu_1 \ge \mu_2`

Alternative hypothesis:
`\mu_1 <\mu_2`

`\alpha= 0.01`

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 =
`\mu_1(1)-\mu_2(2)`

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
`H_o\ \ if \ \ p \le \alpha`

Conclusion: Provided that p-value is < ∝, we reject
`H_o`. 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:

`\mathbf{\mu_1 -\mu_2 \le -3.60}`