A Pew Internet survey reported that 61% of Facebook users have taken a voluntary

Question

A Pew Internet survey reported that 61% of Facebook users have taken a voluntary break from Facebook of several weeks or more at one time or another. The survey contacted 1006 adults living in the United States by landline and cell phone. There were 338 Internet users who don't use Facebook. Of these, 64 reported that they have used Facebook in the past.

(a) Find (±0.0001) SEpˆ the standard error of pˆ?

SEpˆ =

(b) Give the 95% confidence interval for p in the form of estimate plus or minus the margin of error (±0.0001)?

±

(c) Give the confidence interval as an interval of percents (±.0.1)?

% to %

Explanation / Answer

a)

We get the standard error of p, sp:              
              
sp = sqrt[p(1 - p) / n] =    0.026530099   [ANSWER]

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B)  

First, we get the point estimate of the proportion, p^,              
              
p^ = x / n =    0.189349112          
              
Also, the margin of error is

E = z(alpha/2) * sp = 0.051998039

Thus, our confidence interval is

0.189349112 +/- 0.051998039. [ANSWER]

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c)

              
z(alpha/2) =    1.959963985          

Thus,              
              
lower bound = p^ - z(alpha/2) * sp   0.137351074          
upper bound = p^ + z(alpha/2) * sp =    0.241347151          
              
Thus, the confidence interval is              
              
(13.7351074% , 24.1347151%) [ANSWER]

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