Question

As a medical social worker in a local hospital, you are asked to evaluate the success of a group instruction diabetic control program. You randomly assign twelve diabetic patients to the group program and twelve to the more traditional personal instruction program. Over a twelve-month period you collect data on the number of hospital admissions and emergency room visits for uncontrolled blood sugar for individuals in each group. These data are presented below. You are to analyze these data using an independent-samples test and then make a recommendation to the medical director as to whether to continue the program (Set α=.05).

      Number of admissions and emergency room visits

Group instruction   Personal Instruction

             n₁=12                            n₂=12

            ₁=1.4                           ₂=2.3

             S₁=1.2                           S₂=1.2

Answer 1

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 <   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   1.400                  
standard deviation of sample 1,   s1 =    1.2000                  
size of sample 1,    n1=   12                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   2.300                  
standard deviation of sample 2,   s2 =    1.2000                  
size of sample 2,    n2=   12                  
                          
difference in sample means =    x̅1-x̅2 =    1.4000   -   2.3   =   -0.90  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.2000                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.4899                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -0.9000   -   0   ) /    0.49   =   -1.837
                          
Degree of freedom, DF=   n1+n2-2 =    22                  
  
p-value =        0.0399   [ excel function: =T.DIST(t stat,df) ]               
Conclusion:     p-value <α , Reject null hypothesis                      

We conclude that the null hypothesis is rejected

hence there is evidence that mean for group instruction is less.

we should continue the program