Question

In a test of the effectiveness of garlic for lowering​ cholesterol, 45 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes ​(beforeminus​after) in their levels of LDL cholesterol​ (in mg/dL) have a mean of 3.2 and a standard deviation of 17.3. Construct a 90​% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL​ cholesterol?

Answer 1

sample size ,    n =    45          
Degree of freedom, DF=   n - 1 =    44   and α =    0.1  
t-critical value =    t α/2,df =    1.6802   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =        17.3000          
                  
std error , SE = Sd / √n =    17.3000   / √   45   =   2.5789
margin of error, E = t*SE =    1.6802   *   2.5789   =   4.3332
                  
mean of difference ,    D̅ =   3.200          
confidence interval is                   
Interval Lower Limit= D̅ - E =   3.200   -   4.3332   =   -1.133
Interval Upper Limit= D̅ + E =   3.200   +   4.3332   =   7.533
                  
so, confidence interval is (   -1.1332   < µd <   7.5332   )  

since, confidence interval contain 0, so, it suggest that there is no significant effect of  garlic in reducing LDL​ cholesterol