QUICK HELP HURRY MAKRIMG PIEPLE AS BRIANLDIT

Step-by-step explanation:

From the given information:

Let assume:

population mean for catalyst 1

population mean for catalyst 2

Then:

Null hypothesis:

Alternative hypothesis:

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 =

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

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