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home / study / questions and answers / math / statistics and probability / over half of all american teens (ages 12 to 17 ... Your question has expired and been refunded. We were unable to find a Chegg Expert to answer your question. Question Over half of all American teens (ages 12 to 17 years) have an online profile, mainly on Facebook. A random sample of 461 teens with profiles found that 357 included photos of themselves. (a) Give the 99.9% large-sample confidence interval for the proportion p of all teens with profiles who include photos of themselves. Large-sample Interval: to (b) Give the plus four 99.9% confidence interval for p . (The plus-four interval always pulls the results away from 0% or 100%, whichever is closer. Even though the condition for using the large-sample interval is met, the plus four interval is more trustworthy.) Plus four Interval: to

Solution

a)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.774403471          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.019467016          
              
Now, for the critical z,              
alpha/2 =   0.0005          
Thus, z(alpha/2) =    3.290526731          
Thus,              
Margin of error = z(alpha/2)*sp =    0.064056737          
lower bound = p^ - z(alpha/2) * sp =   0.710346734          
upper bound = p^ + z(alpha/2) * sp =    0.838460208          
              
Thus, the confidence interval is              
              
(   0.710346734   ,   0.838460208   ) [ANSWER]

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b)

Here now, x = 357+2 = 359 and n = 461+4 = 465.

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.772043011          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.019454529          
              
Now, for the critical z,              
alpha/2 =   0.0005          
Thus, z(alpha/2) =    3.290526731          
Thus,              
Margin of error = z(alpha/2)*sp =    0.064015649          
lower bound = p^ - z(alpha/2) * sp =   0.708027362          
upper bound = p^ + z(alpha/2) * sp =    0.836058659          
              
Thus, the confidence interval is              
              
(   0.708027362   ,   0.836058659   ) [ANSWER]