Question

Question 3 View Policies Current Attempt in Progress What Do People Do on Facebook? In a survey of 2255 US adults, we learn that 970 of the respondents use the social networking site Facebook.+ Of the 970 Facebook users, the survey shows that on an average day: 15% update their status 22% comment on another's post or status 20% comment on another user's photo 26% "like" another user's content 10% send another user a private message. 1Hampton, K., Goulet, L., Rainie, L., and Purcell, K., "Social Networking Sites and Our Lives." Pew Research Center, June 16, 2011. (a) For each of the bulleted activities, find a 90% confidence interval for the proportion of Facebook users engaging in that activity on an average day. Round your answers to three decimal places. Bulleted Activities Confidence Interval Update status: Comment on another's post or status: i Comment on another user's photo: "Like" another user's content: Send another user a private message: i e Textbook and Media

Answer 1

Formula for confidence interval for single population proportion : p

PZ/2® P(1-P)/n

$\alpha$ for 90% confidence level = (100-90)/100 =0.10

$\alpha$/2 = 0.10/2 =0.05

Z_$\alpha$/2 = Z_0.05 = 1.6449

Sample size : n = 970

Bulleted Activities

Update Status :

Sample Proportion of facebook users engaging in updating their status on an average day : $\widehat{p}$ = 15/100 =0.15

20.05 x V (1-P/n = 0.15+20.05 0.15(1 – 0.15)/970 = 0.15+0.018859 = (0.131141, 0.168859)

90% confidence interval for the proportion of Facebook users  engaging in updating their status on an average day :(0.131141 to 0.168859)

Comment on other's post or status

Sample Proportion of facebook users engaging in Comment on other's post or status on an average day : $\widehat{p}$ = 22/100 =0.22

20.05 x V (1-P/n = 0.22–20.05 X V0.22(1 – 0.22)/970 = 0.22=0.021878 = (0.198122, 0.241878)

90% confidence interval for the proportion of Facebook users  engaging in Comment on other's post or status on an average day :(0.198122 to 0.241878)

Comment of another user's photo

Sample Proportion of facebook users engaging in Comment on another user's photo on an average day : $\widehat{p}$ = 20/100 =0.20

20.05 x V (1-P/n = 0.2 – 20.05 X 10.2(1 – 0.2)/970 = 0.2 + 0.021126 = (0.178874,0.221126)

90% confidence interval for Proportion of facebook users engaging in Comment on another user's photo on an average day

(0.178874 to 0.221126)

"Like" Another user's content

Sample Proportion of facebook users engaging in "Like" Another user's content on an average day : $\widehat{p}$ = 26/100 =0.26

20.05 * P(1 - p/n = 0.26+20.05 X V0.26(1 -0.26)/970 = 0.26+0.023166 = (0.236834, 0.283166)

90% confidence interval for Proportion of facebook users engaging in "Like" Another user's content on an average day

(0.236834 to 0.283166)

Send another user a private message

Sample Proportion of facebook users engaging in Send another user a private message on an average day : $\widehat{p}$ = 10/100 =0.10

P 20.05 VP(1 - p/n = 0.1 20.05 X 10.1(1 -0.1)/970 = 0.1 +0.015844 = (0.084156,0.115844)

90% confidence interval for Proportion of facebook users engaging in Send another user a private message on an average day

(0.084156 to 0.115844)

90% confidence interval for the proportion of Facebook users engaging in that activity on an average day.

Bulleted Activities	Confidence Interval
Update Status	0.131	to	0.169
Comment on another's post status	0.198	to	0.242
Comment on another's user's photo	0.179	to	0.221
"Like" another user's content	0.237	to	0.283
Send Another user a private message	0.084	to	0.116