LARGE SAMPLE ESTIMATION OF A POPULATION PROPORTION A medical researcher wants to

Question

LARGE SAMPLE ESTIMATION OF A POPULATION PROPORTION

A medical researcher wants to investigate the amount of time it takes for patients' headache pain to be relieved after taking a new prescription painkiller. She plans to use statistical methods to estimate the mean of the population of relief times. She believes that the population is normally distributed with a standard deviation of 22 minutes. How large a sample should she take to estimate the mean time to within 5 minutes with 92% confidence?

Explanation / Answer

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.04  
      
Using a table/technology,      
      
z(alpha/2) =    1.750686071  
      
Also,      
      
s = sample standard deviation =    22  
E = margin of error =    5  
      
Thus,      
      
n =    59.3364973  
      
Rounding up,      
      
n =    60   [answer]

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