Question

(2 points) Over half of all American teens (ages 12 to 17 years) have an online profile, mainly on Facebook. A random sample of 550 teens with profiles found that 328 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

Answer 1

a)

Level of Significance,   α = 1-0.999 = 0.001          
Number of Items of Interest,   x =   328          
Sample Size,   n =    550          
                  
Sample Proportion ,    p̂ = x/n =    0.596          
z -value =   Zα/2 =    3.291   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0209          
margin of error , E = Z*SE =    3.291   *   0.0209   =   0.0688
                  
Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.596   -   0.0688   =   0.5275
Interval Upper Limit = p̂ + E =   0.596   +   0.0688   =   0.6652
                  
99.9%   confidence interval is (   0.528   < p <    0.665   )

b)

Level of Significance,   α =    0.001          
Number of Items of Interest,   x +2 = 328 +2 =   330          
Sample Size,   n+4 = 550+4 = 554          
                  
Sample Proportion ,    p̂ = (x+2)/(n+4) =    0.596          
z -value =   Zα/2 =    3.291   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0209          
margin of error , E = Z*SE =    3.291   *   0.0209   =   0.0686
                  
Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.596   -   0.0686   =   0.5271
Interval Upper Limit = p̂ + E =   0.596   +   0.0686   =   0.6643
                  
99.9%   confidence interval is (   0.527   < p <    0.664   )