Question

The Pew Research Center Internet Project conducted a survey of 857 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

Answer 1

Answer:

a) 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally:

$0.8801\leq \pi \leq 0.9199$

b) 95% confidence interval for the proportion of respondents who say the Internet has strengthened their relationship with family and friends:

$0.6386\leq \pi \leq 0.7014$

c) 95% confidence interval for the proportion of Internet users who say online groups have helped solve a problem:

$0.5267\leq \pi \leq 0.5933$

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:

$\sigma_p=\sqrt{\frac{p(1-p)}{n}}= \sqrt{\frac{0.9*0.1}{857}}=0.0102$

Then the upper and lower limit of the 95% CI is:

$LL=p-z\cdot \sigma_p=0.9-1.96*0.0102=0.9-0.0199=0.8801\\\\UL=p+z\cdot \sigma_p=0.9+1.96*0.0102=0.9+0.0199=0.9199$

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:

$\sigma_p=\sqrt{\frac{p(1-p)}{n}}= \sqrt{\frac{0.67*0.33}{857}}=0.0160$

Then the upper and lower limit of the 95% CI is:

$LL=p-z\cdot \sigma_p=0.67-1.96*0.0160=0.67-0.0314=0.6386\\\\UL=p+z\cdot \sigma_p=0.67+1.96*0.0160=0.67+0.0314=0.7014$

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:

$\sigma_p=\sqrt{\frac{p(1-p)}{n}}= \sqrt{\frac{0.56*0.44}{857}}=0.0170$

Then the upper and lower limit of the 95% CI is:

$LL=p-z\cdot \sigma_p=0.56-1.96*0.0170=0.56-0.0333=0.5267\\\\UL=p+z\cdot \sigma_p=0.56+1.96*0.0170=0.56+0.0333=0.5933$