Question

The Pew Research Center Internet Project conducted a survey of 907 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.
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(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.
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(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.
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(d)	Compare the margin of error for the interval estimates in parts (a), (b), and (c). How is the margin of error related to the sample proportion?
	The margin of error - Select your answer -increasesdecreasesItem 7 as p gets closer to .50.

Answer 1

Answer

The Pew Research Center Internet Project conducted a survey of 907 Internet users.

(a) values are given

p bar=(90/100)= 0.9 (sample proportion)

z=1.96 (z critical value of 95% )

n=907 (sample size)

using the formula

put the values in the formula

=0.8804,0.9195

The respondents who say the Internet has been a good thing for them personally.=0.8804,0.9195

(b)  values are given

p bar=0.67

z critical value=1.96

n=907

using the formula

put the values in the formula

0.6395,0.7005

The respondents who say the Internet has strengthened their relationship with family and friends.

= 0.6395,0.70050.

(c) values are given

p bar=(56-25)= 31/100=0.31

z critical value=1.96

n=907

using the formula

put the values in the formula

=0.2801,0.3399

The Internet users who say online groups have helped solve a problem

=0.2801,0.3399

(d)	Compare the margin of error for the interval estimates in parts (a), (b), and (c).

(a) margin of error=(upper limit-lower limit)/2

=(0.9195-0.8804)/2=0.0195

(b) margin of error=(upper limit-lower limit)/2=(0.7005-0.6395)/2=0.0305

(c)margin of error=(upper limit-lower limit)/2=(0.3399-0.2801)/2=0.0598

We can see that the margin of error increases as p gets close to 0.5