Suppose a survey revealed that 21 of 504 respondents said th

Suppose a survey revealed that 21% of 504 respondents said they had in the past sold unwanted gifts over the Internet. (a) Use the information to construct a 90% confidence interval for the population proportion who sold unwanted gifts over the Internet, rounding your margin of error to the nearest hundredth. (Round your answers to two decimal places.)
( , )

(b) Use the information to construct a 98% confidence interval for the population proportion who sold unwanted gifts over the Internet, rounding your margin of error to the nearest hundredth. (Round your answers to two decimal places.)
( , )
(c) Which interval has a higher probability of containing the true population proportion?

the 90% confidence interval the 98% confidence interval    

(d) Which interval is narrower? the 90% confidence interval the 98% confidence interval

Solution

(a) Given a=1-0.9=0.1, Z(0.05) = 1.645 (from standard normal table)

So the lower bound is

p-Z*sqrt(p*(1-p)/n) = 0.21- 1.645*sqrt(0.21*(1-0.21)/504) =0.18

So the upper bound is

p+Z*sqrt(p*(1-p)/n) = 0.21+ 1.645*sqrt(0.21*(1-0.21)/504) =0.24

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(b) Given a=1-0.98=0.02, Z(0.01) = 2.33 (from standard normal table)

So the lower bound is

p-Z*sqrt(p*(1-p)/n) = 0.21- 2.33*sqrt(0.21*(1-0.21)/504) =0.17

So the upper bound is

p+Z*sqrt(p*(1-p)/n) = 0.21+ 2.33*sqrt(0.21*(1-0.21)/504) =0.25

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(c)the 98% confidence interval    

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(d)the 90% confidence interval