America\'s young people are heavy Internet users; 87% of Americans ages 12 to 17

Question

America's young people are heavy Internet users; 87% of Americans ages 12 to 17 are Internet users (The Cincinnati Enquirer, February 7, 2006). MySpace was voted the most popular website by 9% in a sample survey of Internet users in this age group. Suppose 1,750 youths participated in the survey. What is the margin of error (to 3 decimals), and what is the interval estimate (to 3 decimals) of the population proportion for which MySpace is the most popular website (to 4 decimals)? Use a 95% confidence level. For interval estimate Enter your answer using parentheses and a comma, in the form (n1,n2). Do not use commas in your numerical answer (i.e. use 1200 instead of 1,200, etc.)

Explanation / Answer

Margin of Error = 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
Sample Size(n)=1750
Sample proportion =0.09
Margin of Error = Z a/2 * ( Sqrt ( (0.09*0.91) /1750) )
= 1.96* Sqrt(0.00005)
=0.01341

Confidence Interval For Proportion
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
Sample Size(n)=1750
Sample proportion = x/n =0.09
Confidence Interval = [ 0.09 ±Z a/2 ( Sqrt ( 0.09*0.91) /1750)]
= [ 0.09 - 1.96* Sqrt(0) , 0.09 + 1.96* Sqrt(0) ]
= [ 0.077,0.103] ~ [7.7% , 10.3%]

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