Question

Two catalysts are being analyzed to determine how they affect the mean yield of a chemical process. Specifically, catalyst 1 is currently used. Because catalyst 2 is cheaper, it should be adopted, if it does not change the process yield. A test is run in the pilot plant and results in the data shown in the Table below. Both populations are assumed normal.

Observation #	Catalyst 1	Catalyst 1
1	91.5	89.19
2	94.18	90.95
3	92.18	90.46
4	95.39	93.21
5	91.79	97.19
6	89.07	97.04
7	94.72	91.07
8	89.21	92.75

a) Assuming equal variances, using a hypothesis test at 5% alpha level, show if there is any difference in the mean yields.

b) Assuming unequal variances, using a hypothesis test at 5% alpha level, show if catalyst 2 yields greater mean than catalyst 1.

c) Find a 95% confidence interval for the mean of catalyst 1 minus that of catalyst 2 assuming equal variances.

Answer 1

a)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   Catalyst 1                  
mean of sample 1,    x̅1=   92.26                  
standard deviation of sample 1,   s1 =    2.39                  
size of sample 1,    n1=   8                  
                          
Sample #2   ---->   Catalyst 2                  
mean of sample 2,    x̅2=   92.73                  
standard deviation of sample 2,   s2 =    2.98                  
size of sample 2,    n2=   8                  
                          
difference in sample means =    x̅1-x̅2 =    92.2550   -   92.7   =   -0.48  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    2.70                  
std error , SE =    Sp*√(1/n1+1/n2) =    1.3504                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -0.4775   -   0   ) /    1.35   =   -0.354
                          
Degree of freedom, DF=   n1+n2-2 =    14                  

p-value =        0.7289   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value>α , Do not reject null hypothesis                      
                          
There is not enough evidence to conclude that there is any difference in the mean yields

b)

Ho :   µ1 - µ2 =   0          
Ha :   µ1-µ2 <   0          
                  
Level of Significance ,    α =    0.05          
                  
Sample #1   ---->   Catalyst 1          
mean of sample 1,    x̅1=   92.26          
standard deviation of sample 1,   s1 =    2.385018718          
size of sample 1,    n1=   8          
                  
Sample #2   ---->   Catalyst 2          
mean of sample 2,    x̅2=   92.733          
standard deviation of sample 2,   s2 =    2.98          
size of sample 2,    n2=   8          
                  
difference in sample means = x̅1-x̅2 =    92.255   -   92.7325   =   -0.4775
                  
std error , SE =    √(s1²/n1+s2²/n2) =    1.3504          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -0.4775   /   1.3504   ) =   -0.3536

p-value =        0.3647   [ excel function: =T.DIST(t stat,df) ]
Conclusion:     p-value>α , Do not reject null hypothesis      
          
There is not enough evidence to conclude that  catalyst 2 yields greater mean than catalyst 1.

c)

Degree of freedom, DF=   n1+n2-2 =    14              
t-critical value =    t α/2 =    2.1448   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    2.7009              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    1.3504              
margin of error, E = t*SE =    2.1448   *   1.3504   =   2.8964  
                      
difference of means =    x̅1-x̅2 =    92.2550   -   92.733   =   -0.4775
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -0.4775   -   2.8964   =   -3.374
Interval Upper Limit=   (x̅1-x̅2) + E =    -0.4775   +   2.8964   =   2.419