Answer:
(a). ( 0.776 ,0.809).
(b). (0.567 , 0.613).
(c). 0.600.
Explanation:
Okay, we are given the following set of values or data or parameters;
=> "A survey of 2255 randomly selected US adults (age 18 or older)"
=> "1787 of them use the Internet regularly. Of the Internet users, 1054 use a social networking site".
=> Also, "95% confidence interval for each of the following proportions"
Therefore, we are going to make use of one (major ) mathematical formula in solving this particular Question and it is given below;
Confidence Interval = p +/- z* × [ √p( 1 - p) / n].
(a).
Where p = 1787/2255 = 0.793.
95% confidence Interval = z* = 1.96.
= 0.793 +/- 1.96 × [√0.793 ( 1 - 0.793)/ 2255] .
= 0.793 +/- 0.0167.
= ( 0.776 ,0.809).
(b). Where p = 1054/ 1787 = 0.5900
95% confidence Interval = z* = 1.96.
= 0.5900 +/- 1.96 × [√0.5900 ( 1 - 0.5900)/ 1787]
= 0.5900 +/- 0.0228.
= (0.567 , 0.613).
(c). 1054/1787 = 0.59 = 0.600.