Two catalysts in a batch chemical process, are being considered for their effect

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: 80.49 79.30 79.37 79.77 78.13 79.62 80.55 78.99 On the other hand, you found the following for the yield using catalyst 2: 82.63 80.59 81.64 81.22 81.62 80.79 82.54 81.49 80.17 80.57 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 greater than 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.

Explanation / Answer

Null hypothesis: H0 : µ1 = µ2

Alternative hypothesis: H1 : µ1 < µ2 => µ1 - µ2 < 0

Using R software, the test was conducted.

> c1= c(80.49,79.30,79.37,79.77,78.13,79.62,80.55,78.99)

> c2=c(82.63,80.59,81.64,81.22,81.62,80.79,82.54,81.49,80.17,80.57)

> var.test(c1,c2) #To test if C1 and C2 have equal variance

        F test to compare two variances

data: c1 and c2

F = 0.91337, num df = 7, denom df = 9, p-value = 0.9261

alternative hypothesis: true ratio of variances is not equal to 1

95 percent confidence interval:

0.2176217 4.4053744

sample estimates:

ratio of variances

         0.9133685

Since, the p-value(0.92) > 0.05, we can conclude that C1 and C2 have equal variance.

> t.test(c1,c2,conf.level=0.975,var.equal=T,alternative="less")

        Two Sample t-test

data: c1 and c2

t = -4.6822, df = 16, p-value = 0.0001249

alternative hypothesis: true difference in means is less than 0

97.5 percent confidence interval:

       -Inf    -0.9842146

sample estimates:

mean of x mean of y

79.5275   81.3260

97.5% confidence interval is given by: ( -Inf , -0.9842146)

From the difference, it is evident that the difference is less than zero. Therefore, we can conclude that mean of C2 is greater than mean of C1.

Since the p-value(0.00012) < 0.025, we reject the null hypothesis and conclude that mean of C2 is greater than mean of C1.

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