Question

Yield of Chemical Process Two catalysts were analyzed to determine how they affect the mean yield of chemical process. Specifically, catalyst 1 is currently used; but catalyst 2 is acceptable. Because catalyst 2 is cheaper, it should be adopted, if it does not change the process yield. Two tests were run in the pilot plant. Eight samples of each type of catalyst were randomly and independently selected from the pilot plant, and the chemical process yield were recorded. The data can be found in the excel file Catalyst with variables name Type and Yield. At the 5% significance level, is there any difference in the mean yield of two catalysts of a chemical process? (19 points total)

1. Label the Parameters: (3 points)
2. State null and alternative hypotheses in parameter notation: (2 points)

H₀:

H_a:

3. Propose an appropriate hypothesis test. Explain! (4 points)
4. Verify the required conditions for the proposed hypothesis test: (3 points)

Randomization assumption: 

Normality assumption:

Equal population variance assumption:                               

                       Note: Attach R codes and outputs in the following space:  

5. Report a test statistic, df, and p-value: (3 points)

     Test statistic = 
df = 

p-value =

                         Note: Attach R codes and outputs in the following space:

6. Make a decision of hypothesis test: (2 points)
7. Make a conclusion of hypothesis test in context: (2 points)

Type	Yield
Catalyst 1	91.5
Catalyst 1	94.18
Catalyst 1	92.18
Catalyst 1	95.39
Catalyst 1	91.79
Catalyst 1	89.07
Catalyst 1	94.72
Catalyst 1	89.21
Catalyst 2	89.19
Catalyst 2	90.95
Catalyst 2	90.46
Catalyst 2	93.21
Catalyst 2	97.19
Catalyst 2	97.04
Catalyst 2	91.07
Catalyst 2	92.75

Answer 1

a) The average catalyst 1 is mu1
The average catalyst 2 is mu2

b)
H0: There is no significance difference in the mean yield of two catalysts of a chemical process
H1: There is significance difference in the mean yield of two catalysts of a chemical process

c) t - test for difference of means is used

d) Normality assumption:

Test for difference of variances:

Since p-value > alpha 0.05 so we conclude that the two population variances are equal

e) Test for difference of means:

since p-value = 0.7289 which is > alpha 0.05 so we accept H0

Thus we conclude that There is no significance difference in the mean yield of two catalysts of a chemical process