Please help me answer the below: The Pew Research Center Internet Project conduc

Question

Please help me answer the below:

The Pew Research Center Internet Project conducted a survey of 1,057 Internet users. This survey provided a variety of statistics on them.

If required, round your answers to four decimal places.

(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. _______ to _______ (b) The sample survey showed that 67% 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. ______to _________ (c) Fifty-six 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. ________to __________

Explanation / Answer

a)

CI for 95%

n = 1057

p = 0.9

z-value of 95% CI = 1.9600

SE = sqrt(p*(1-p)/n) = sqrt(0.9*0.1/1057) = 0.00923

ME = z*SE = 1.96*0.00923 = 0.01809

Lower Limit = p - ME = 0.9 - 0.01809 = 0.88191

Upper Limit = p + ME = 0.9 + 0.01809 = 0.91809

95% CI (0.8819 , 0.9181 )

b)

CI for 95%

n = 1057

p = 0.67

z-value of 95% CI = 1.9600

SE = sqrt(p*(1-p)/n) = sqrt(0.31*0.69/1057) = 0.01446

ME = z*SE = 1.96*0.01446 = 0.02835

Lower Limit = p - ME = 0.67 - 0.02835 = 0.64165

Upper Limit = p + ME = 0.67 + 0.02835 = 0.69835

95% CI (0.6417 , 0.6983 )

c)

CI for 95%

n = 1057

p = 0.56

z-value of 95% CI = 1.9600

SE = sqrt(p*(1-p)/n) = sqrt(0.56*0.44/1057) = 0.01527

ME = z*SE = 1.96*0.01527 = 0.02992

Lower Limit = p - ME = 0.56 - 0.02992 = 0.53008

Upper Limit = p + ME = 0.56 + 0.02992 = 0.58992

95% CI (0.5301 , 0.5899 )

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