Question

A research center project involved a survey of 843 internet users. It provided a variety of statistics on internet users.

(a) The sample survey showed that 90% of respondents said the internet has been a good thing for them personally. Develop a 95% confidence interval for the proportion of respondents who say the internet has been a good thing for them personally. (Round your answers to four decimal places.)

(b) The sample survey showed that 71% of internet users said the internet has generally strengthened their relationship with family and friends. Develop a 95% confidence interval for the proportion of respondents who say the internet has strengthened their relationship with family and friends. (Round your answers to four decimal places.)

(c) Fifty-seven percent of internet users have seen an online group come together to help a person or community solve a problem, whereas only 25% have left an online group because of unpleasant interaction. Develop a 95% confidence interval for the proportion of internet users who say online groups have helped solve a problem. (Round your answers to four decimal places.)

Answer 1

a)

sample proportion, = 0.9
sample size, n = 843
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.9 * (1 - 0.9)/843) = 0.0103

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0103
ME = 0.0202

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.9 - 1.96 * 0.0103 , 0.9 + 1.96 * 0.0103)
CI = (0.8798 , 0.9202)

b)

sample proportion, = 0.71
sample size, n = 843
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.71 * (1 - 0.71)/843) = 0.0156

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0156
ME = 0.0306

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.71 - 1.96 * 0.0156 , 0.71 + 1.96 * 0.0156)
CI = (0.6794 , 0.7406)

c)

sample proportion, = 0.57
sample size, n = 843
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.57 * (1 - 0.57)/843) = 0.0171

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0171
ME = 0.0335

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.57 - 1.96 * 0.0171 , 0.57 + 1.96 * 0.0171)
CI = (0.5365 , 0.6035)