A pollster wants to construct a 95 confidence interval for t

A pollster wants to construct a 95% confidence interval for the proportion of adults who believe that economic conditions are getting better. A Gallup poll taken in July 2010 estimates this proportion to be 0.33. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.062? I also need to know what command is being used to solve this in the TI-84 calculator

Solution

Note that      
      
n = z(alpha/2)^2 p (1 - p) / E^2      
      
where      
      
alpha/2 =    0.025  
       
      
Using a table/technology,      
      
z(alpha/2) =    1.959963985  
      
Also,      
      
E =    0.062  
p =    0.33  
      
Thus,      
      
n =    220.9538359  
      
Rounding up,      
      
n =    221   [ANSWER]

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Unfortunately, your ti84 can only help finding the z score here, but you will still need the formula, so we do it this way.