Suppose a survey revealed that 21% of 501 respondents said they had in the past

Question

Suppose a survey revealed that 21% of 501 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.)

Explanation / Answer

Given n=501, p=0.21

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(a) a=0.1, Z(0.05)=1.645 (check standard normal table)

So 90% CI is
p ± Z*[p*(1-p)/n]

-- > 0.21 ± 1.645*sqrt(0.21*(1-0.21)/501)

--> ( 0.18, 0.24)

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(b)

a=0.02, Z(0.01)=2.33 (check standard normal table)

So 98% CI is
p ± Z*[p*(1-p)/n]

-- > 0.21 ± 2.33*sqrt(0.21*(1-0.21)/501)

--> ( 0.17, 0.25)

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