of 375 surveyed freshman college students, 90 said they have passport. Suppose t

Question

of 375 surveyed freshman college students, 90 said they have passport. Suppose this constitutes a random sample. (a) Give a 90% Confidence interval for the true proportion of students who have a passport an interpret it in the context of this problem. (b) Give a 99% confidence interval for the true proportion of students who have a passport and interpret it in the context of this problem. (c) How big of a sample should we take if we wish to be within 5% of the true proportion of all citizens (not just college students) with 99% confidence? /assume we have no idea how many people actually have passports.

Explanation / Answer

a)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.24          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.022054478          
              
Now, for the critical z,              
alpha/2 =   0.05          
Thus, z(alpha/2) =    1.644853627          
Thus,              
Margin of error = z(alpha/2)*sp =    0.036276388          
lower bound = p^ - z(alpha/2) * sp =   0.203723612          
upper bound = p^ + z(alpha/2) * sp =    0.276276388          
              
Thus, the confidence interval is              
              
(   0.203723612   ,   0.276276388   ) [ANSWER]

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b)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.24          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.022054478          
              
Now, for the critical z,              
alpha/2 =   0.005          
Thus, z(alpha/2) =    2.575829304          
Thus,              
Margin of error = z(alpha/2)*sp =    0.056808571          
lower bound = p^ - z(alpha/2) * sp =   0.183191429          
upper bound = p^ + z(alpha/2) * sp =    0.296808571          
              
Thus, the confidence interval is              
              
(   0.183191429   ,   0.296808571   ) [ANSWER]

********************

c)

Note that      
      
n = z(alpha/2)^2 p (1 - p) / E^2      
      
where      
      
alpha/2 =    0.005  
As there is no previous estimate for p, we set p = 0.5.      
      
Using a table/technology,      
      
z(alpha/2) =    2.575829304  
      
Also,      
      
E =    0.05  
p =    0.5  
      
Thus,      
      
n =    663.4896601  
      
Rounding up,      
      
n =    664   [ANSWER]

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