Question

Two catalysts in a batch chemical process, are being considered for their effect on the output of a process reaction. Based on literature, data with different samples with catalyst 1 were found to result to the following yield:

78.53

78.17

81.05

81.77

78.13

81.24

81.84

79.63

78.59

79.74

79.84

On the other hand, you found the following for the yield using catalyst 2:

82.05

81.57

81.71

81.29

81.79

81.47

81.75

81.50

81.75

If you can assume that the distribution characterizing yield is approximately normal and the samples are independent:

Perform hypothesis testing using 97.5 % confidence to test whether the average yield using catalyst 2 is not equal to the average yield using catalyst 1

Report your findings with the same data and hypothesis, but now using significance testing (P-value approach) only for the last test (test on difference of means). Use the same confidence level as previously for any other test.

Answer 1

n1 = 11

= 79.8664

s1^2 = 2.0187

n2 = 9

= 81.6533

s2^2 = 0.0492

Claim: The average yield using catalyst 2 is not equal to the average yield using catalyst 1.

The null and alternative hypothesis is

For doing this test first we have to check the two groups have population variances are equal or not.

The null and alternative hypothesis is

Test statistic is

F = largest sample variance / Smallest sample variances

F = 2.0187 / 0.0492 = 40.99

Degrees of freedom => n1 - 1 , n2 - 1 => 11 - 1 , 9 - 1 => 10 , 8

Critical value = 4.295 ( Using f table)

Critical value > test statistic so we fail to reject null hypothesis.

Conclusion: The population variances are equal.

So we have to use here pooled variance.

Test statistic is

Degrees of freedom = n1 + n2 - 2 = 11 + 9 - 2 = 18

Critical value = 2.445 ( Using t table)

| t | > critical value we reject null hypothesis.

Conclusion: The average yield using catalyst 2 is not equal to the average yield using catalyst 1.

P-value approach:-

P-value = 2* P(T < - 3.72) = 0.0016

P-value < 0.025 we reject null hypothesis.

Conclusion: The average yield using catalyst 2 is not equal to the average yield using catalyst 1.