Social Networking Sites In 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.7 Find and interpret a 95% confidence interval for each of the following proportions:
(a) Proportion of US adults who use the Internet regularly.
(b) Proportion of US adult Internet users who use a social networking site.
(c) Proportion of all US adults who use a social networking site. Use the confidence interval to estimate whether it is plausible that 50% of all US adults use a social networking site.

(a). ( 0.776 ,0.809).

(b). (0.567 , 0.613).

(c). 0.600.

Step-by-step 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.

your answer is the third one

Step-by-step explanation:

10 you would take 50/ .2

step-by-step explanation:

its kind of blurry. can you put a clearer pic?

step-by-step explanation: