In: Statistics and Probability
A doctor wants to estimate the mean HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the mean HDL cholesterol within 4 points with 99% confidence assuming s=13.7 based on earlier studies? Suppose the doctor would be content with 90% confidence. How does the decrease in confidence affect the sample size required?
A) 99% confidence level requires_____ Subject. (round up to the nearest subject)
B) 90% confidence level requires_____ Subject. (round up to the nearest subject)
a)
Standard Deviation ,   σ =   
13.7          
       
sampling error ,    E =   4  
           
   
Confidence Level ,   CL=   99%  
           
   
          
           
   
alpha =   1-CL =   1%  
           
   
Z value =    Zα/2 =    2.576   [excel
formula =normsinv(α/2)]      
       
          
           
   
Sample Size,n = (Z * σ / E )² = (   2.576  
*   13.7   /   4   ) ²
=   77.831
          
           
   
          
           
   
So,Sample Size needed=      
78          
       
b)
Standard Deviation ,   σ =   
13.7          
       
sampling error ,    E =   4  
           
   
Confidence Level ,   CL=   90%  
           
   
          
           
   
alpha =   1-CL =   10%  
           
   
Z value =    Zα/2 =    1.645   [excel
formula =normsinv(α/2)]      
       
          
           
   
Sample Size,n = (Z * σ / E )² = (   1.645  
*   13.7   /   4   ) ²
=   31.738
          
           
   
          
           
   
So,Sample Size needed=      
32