Respuesta :
Answer:
a) 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally:
[tex]0.8801\leq \pi \leq 0.9199[/tex]
b) 95% confidence interval for the proportion of respondents who say the Internet has strengthened their relationship with family and friends:
[tex]0.6386\leq \pi \leq 0.7014[/tex]
c) 95% confidence interval for the proportion of Internet users who say online groups have helped solve a problem:
[tex]0.5267\leq \pi \leq 0.5933[/tex]
Explanation:
a) We have a sample proportion of 90% (p=0.90).
The sample size is n=857.
For a 95% CI, the z-value is z=1.96.
The standard deviation for the proportion is:
[tex]\sigma_p=\sqrt{\frac{p(1-p)}{n}}= \sqrt{\frac{0.9*0.1}{857}}=0.0102[/tex]
Then the upper and lower limit of the 95% CI is:
[tex]LL=p-z\cdot \sigma_p=0.9-1.96*0.0102=0.9-0.0199=0.8801\\\\UL=p+z\cdot \sigma_p=0.9+1.96*0.0102=0.9+0.0199=0.9199[/tex]
b) We have a sample proportion of 67% (p=0.67).
The sample size is n=857.
For a 95% CI, the z-value is z=1.96.
The standard deviation for the proportion is:
[tex]\sigma_p=\sqrt{\frac{p(1-p)}{n}}= \sqrt{\frac{0.67*0.33}{857}}=0.0160[/tex]
Then the upper and lower limit of the 95% CI is:
[tex]LL=p-z\cdot \sigma_p=0.67-1.96*0.0160=0.67-0.0314=0.6386\\\\UL=p+z\cdot \sigma_p=0.67+1.96*0.0160=0.67+0.0314=0.7014[/tex]
c) We have a sample proportion of 56% (p=0.56).
The sample size is n=857.
For a 95% CI, the z-value is z=1.96.
The standard deviation for the proportion is:
[tex]\sigma_p=\sqrt{\frac{p(1-p)}{n}}= \sqrt{\frac{0.56*0.44}{857}}=0.0170[/tex]
Then the upper and lower limit of the 95% CI is:
[tex]LL=p-z\cdot \sigma_p=0.56-1.96*0.0170=0.56-0.0333=0.5267\\\\UL=p+z\cdot \sigma_p=0.56+1.96*0.0170=0.56+0.0333=0.5933[/tex]
