Two catalysts may be used in a batch chemical process Twelve

Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1, resulting in an average yield of 86 and a sample standard deviation of 3. Fifteen batches were prepared using catalyst 2, and they resulted in an average yield of 89 with a standard deviation of 2. Assume that yield measurements are approximately normally distributed with the same standard deviation. Is there evidence to support a claim that catalyst 2 produces a higher mean yield than catalyst 1? Use a =0.01. Find a 95% confidence interval on the difference in mean yields.

Solution

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   >=   0  
Ha:   u1 - u2   <   0  
At level of significance =    0.01          
As we can see, this is a    left   tailed test.      
Calculating the means of each group,              
              
X1 =    86          
X2 =    89          
              
Calculating the standard deviations of each group,              
              
s1 =    3          
s2 =    2          
              
Thus, the pooled standard deviation is given by              
              
S = sqrt[((n1 - 1)s1^2 + (n2 - 1)(s2^2))/(n1 + n2 - 2)]               
              
As n1 =    12   , n2 =    15  
              
Then              
              
S =    2.48997992          
              
Thus, the standard error of the difference is              
              
Sd = S sqrt (1/n1 + 1/n2) =    0.964365076          
              
As ud = the hypothesized difference between means =    0   , then      
              
t = [X1 - X2 - ud]/Sd =    -3.110855084          
              
Getting the critical value using table/technology,              
df = n1 + n2 - 2 =    25          
tcrit =    -   2.485107175      
              
Getting the p value using technology,              
              
p =    0.002309561          
              
As t < -2.4851, and P < 0.01, we   REJECT THE NULL HYPOTHESIS.      

Thus, there is significant evidence that catalyst 2 produces a higher mean yield than catalyst 1. [CONCLUSION]
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B)      
              
For the   0.95   confidence level, then      
              
alpha/2 = (1 - confidence level)/2 =    0.025          
t(alpha/2) =    2.059538553          
              
lower bound = [X1 - X2] - t(alpha/2) * Sd =    -4.986147053          
upper bound = [X1 - X2] + t(alpha/2) * Sd =    -1.013852947          
              
Thus, the confidence interval is              
              
(   -4.986147053   ,   -1.013852947   ) [ANSWER]