in a survey of 612 males ages 1864 398 say they have gone to

in a survey of 612 males ages 18-64 398 say they have gone to the dentist in the past year. construct 90% and 95% confidence intervals for the population proportion

Solution

A)

90% CONFIDENCE:

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.650326797          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.019276194          
              
Now, for the critical z,              
alpha/2 =   0.05          
Thus, z(alpha/2) =    1.644853627          
Thus,              
Margin of error = z(alpha/2)*sp =    0.031706518          
lower bound = p^ - z(alpha/2) * sp =   0.61862028          
upper bound = p^ + z(alpha/2) * sp =    0.682033315          
              
Thus, the confidence interval is              
              
(   0.61862028   ,   0.682033315   ) [ANSWER]

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B)

95% CONFIDENCE:

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.650326797          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.019276194          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.037780646          
lower bound = p^ - z(alpha/2) * sp =   0.612546151          
upper bound = p^ + z(alpha/2) * sp =    0.688107443          
              
Thus, the confidence interval is              
              
(   0.612546151   ,   0.688107443   ) [ANSWER]