Question

The Pew Research Center Internet Project, conducted on the 25th anniversary of the Internet, involved a survey of 857 Internet users (Pew Research Center website, April 1, 2014). It provided a variety of statistics on Internet users. For instance, in 2014, 87% of American adults were Internet users. In 1995 only 14% of American adults used the Internet.
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 4 decimals).

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 4 decimals)

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 4 decimals).

d. Compare the margin of error for the interval estimates in parts (a), (b), and (c). How is the margin of error related to sample proportion (to 2 decimals)?

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:

$0.5267\leq \pi \leq 0.5933$

d) margin of error for the interval estimates in parts (a) e=0.02, (b) e=0.05, and (c) e=0.06.

Step-by-step 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:

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

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:

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

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:

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

d)

For a), the relative margin of error is:

$e=\frac{z\sigma_p}{p} =\frac{0.0199}{0.9}= 0.02$

For b), the relative margin of error is:

$e=\frac{z\sigma_p}{p} =\frac{0.0314}{0.67}= 0.05$

For c), the relative margin of error is:

$e=\frac{z\sigma_p}{p} =\frac{0.0333}{0.56}= 0.06$