A doctor wants to estimate the HDL cholestrol of all 20 to 2

A doctor wants to estimate the HDL cholestrol of all 20 to 29 year females. How many subjects are neede to estimate the HDL chrolestrol with 3 points with 99% confidence assuming o=12.8? Suppose the doctor would be content with 95% confidence. How does the decrease in confidence affect the sample size required?

Solution

a)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.005  
      
Using a table/technology,      
      
z(alpha/2) =    2.575829304  
      
Also,      
      
s = sample standard deviation =    12.8  
E = margin of error =    3  
      
Thus,      
      
n =    120.7846066  
      
Rounding up,      
      
n =    121   [ANSWER]

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b)

As the sample size is directly proportional to the square of the critical z (which decreases with decreasing confidence level), then n would decrease. [ANSWER, SAMPLE SIZE DECREASES]

To verify,

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.025  
      
Using a table/technology,      
      
z(alpha/2) =    1.959963985  
      
Also,      
      
s = sample standard deviation =    12.8  
E = margin of error =    3  
      
Thus,      
      
n =    69.93162369  
      
Rounding up,      
      
n =    70   (it decreased from 121 to 70)

A doctor wants to estimate the HDL cholestrol of all 20 to 29 year females. How many subjects are neede to estimate the HDL chrolestrol with 3 points with 99% c
A doctor wants to estimate the HDL cholestrol of all 20 to 29 year females. How many subjects are neede to estimate the HDL chrolestrol with 3 points with 99% c

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