Answer:
a) 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally:
![0.8801\leq \pi \leq 0.9199](https://tex.z-dn.net/?f=0.8801%5Cleq%20%5Cpi%20%5Cleq%200.9199)
b) 95% confidence interval for the proportion of respondents who say the Internet has strengthened their relationship with family and friends:
![0.6386\leq \pi \leq 0.7014](https://tex.z-dn.net/?f=0.6386%5Cleq%20%5Cpi%20%5Cleq%200.7014)
c) 95% confidence interval for the proportion of Internet users who say online groups have helped solve a problem:
![0.5267\leq \pi \leq 0.5933](https://tex.z-dn.net/?f=0.5267%5Cleq%20%5Cpi%20%5Cleq%200.5933)
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:
![\sigma_p=\sqrt{\frac{p(1-p)}{n}}= \sqrt{\frac{0.9*0.1}{857}}=0.0102](https://tex.z-dn.net/?f=%5Csigma_p%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%3D%20%5Csqrt%7B%5Cfrac%7B0.9%2A0.1%7D%7B857%7D%7D%3D0.0102)
Then the upper and lower limit of the 95% CI is:
![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](https://tex.z-dn.net/?f=LL%3Dp-z%5Ccdot%20%5Csigma_p%3D0.9-1.96%2A0.0102%3D0.9-0.0199%3D0.8801%5C%5C%5C%5CUL%3Dp%2Bz%5Ccdot%20%5Csigma_p%3D0.9%2B1.96%2A0.0102%3D0.9%2B0.0199%3D0.9199)
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:
![\sigma_p=\sqrt{\frac{p(1-p)}{n}}= \sqrt{\frac{0.67*0.33}{857}}=0.0160](https://tex.z-dn.net/?f=%5Csigma_p%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%3D%20%5Csqrt%7B%5Cfrac%7B0.67%2A0.33%7D%7B857%7D%7D%3D0.0160)
Then the upper and lower limit of the 95% CI is:
![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](https://tex.z-dn.net/?f=LL%3Dp-z%5Ccdot%20%5Csigma_p%3D0.67-1.96%2A0.0160%3D0.67-0.0314%3D0.6386%5C%5C%5C%5CUL%3Dp%2Bz%5Ccdot%20%5Csigma_p%3D0.67%2B1.96%2A0.0160%3D0.67%2B0.0314%3D0.7014)
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:
![\sigma_p=\sqrt{\frac{p(1-p)}{n}}= \sqrt{\frac{0.56*0.44}{857}}=0.0170](https://tex.z-dn.net/?f=%5Csigma_p%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%3D%20%5Csqrt%7B%5Cfrac%7B0.56%2A0.44%7D%7B857%7D%7D%3D0.0170)
Then the upper and lower limit of the 95% CI is:
![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](https://tex.z-dn.net/?f=LL%3Dp-z%5Ccdot%20%5Csigma_p%3D0.56-1.96%2A0.0170%3D0.56-0.0333%3D0.5267%5C%5C%5C%5CUL%3Dp%2Bz%5Ccdot%20%5Csigma_p%3D0.56%2B1.96%2A0.0170%3D0.56%2B0.0333%3D0.5933)