A poll conducted in 2013 found that 52 of US adult Twitter u

A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news on Twitter.40. The standard error for this estimate was 2.4%, and a normal distribution may be used to model the sample proportion. Construct a 99% confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter, and interpret the confidence interval in context.

Solution

Note that              
              
Lower Bound = p^ - z(alpha/2) * se              
Upper Bound = p^ + z(alpha/2) * se              
              
where              
              
alpha/2 = (1 - confidence level)/2 =    0.005          
              
Thus,              
p^ = sample mean proportion =    0.52          

z(alpha/2) = critical z for the confidence interval =    2.575829304          
se = standard error =    0.024          
              
              
Thus,              
              
Lower bound =    0.458180097          
Upper bound =    0.581819903          
              
Thus, the confidence interval is              
              
(   0.458180097   ,   0.581819903   ) [ANSWER]