A sample survey by the Pew Internet and American Life Projec

A sample survey by the Pew Internet and American Life Project asked a random sample of adults about use of the Internet and about the type of community they lived in. Here is the two-way table:

Give 96 % confidence intervals (±0.001) for the proportions of adults who use the Internet

in rural communities: ____p_____ ?

in suburban communities: _____p_____?

in urban communities: ______p_______

Solution

a) In rural communities
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Rural(x)=415
Sample Size(n)=914
Sample proportion = x/n =0.454
Confidence Interval = [ 0.454 ±Z a/2 ( Sqrt ( 0.454*0.546) /914)]
= [ 0.454 - 2.05* Sqrt(0) , 0.454 + 2.05* Sqrt(0) ]
= [ 0.42,0.488]
in rural commnities [ 42.2% , 48.8% ] proportions of adults use the Internet

b)
In Suburban
Suburban(x)=1137
Sample Size(n)=1741
Sample proportion = x/n =0.653
Confidence Interval = [ 0.653 ±Z a/2 ( Sqrt ( 0.653*0.347) /1741)]
= [ 0.653 - 2.05* Sqrt(0) , 0.653 + 2.05* Sqrt(0) ]
= [ 0.63,0.676]
In Suburban commnities [ 63% , 67.6% ] proportions of adults use the Internet

c)
In Urban

Urban(x)=561
Sample Size(n)=962
Sample proportion = x/n =0.583
Confidence Interval = [ 0.583 ±Z a/2 ( Sqrt ( 0.583*0.417) /962)]
= [ 0.583 - 2.05* Sqrt(0) , 0.583 + 2.05* Sqrt(0) ]
= [ 0.55,0.616]

In urban commnities [ 55% , 61.6% ] proportions of adults use the Internet