5 A polling organization interviewed by phone 400 randomly s

5. A polling organization interviewed by phone 400 randomly selected adults in New York City about their opinion on random drug testing for taxi drivers and found that 38% favored such a regulation.

a. Find the standard error of the proportion.

b. Find and interpret the 95% confidence interval for the population proportion.

c. Find and interpret the 99% confidence interval for the population proportion.

Solution

a)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.38          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.024269322   [ANSWER]

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B)      
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
              
lower bound = p^ - z(alpha/2) * sp =   0.332433003          
upper bound = p^ + z(alpha/2) * sp =    0.427566997          
              
Thus, the confidence interval is              
              
(   0.332433003   ,   0.427566997   ) [ANSWER]

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c)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.38          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.024269322          
              
Now, for the critical z,              
alpha/2 =   0.005          
Thus, z(alpha/2) =    2.575829304          
Thus,              
              
lower bound = p^ - z(alpha/2) * sp =   0.317486369          
upper bound = p^ + z(alpha/2) * sp =    0.442513631          
              
Thus, the confidence interval is              
              
(   0.317486369   ,   0.442513631   ) [ANSWER]