Question

Suppose a survey revealed that 19% of 494 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.)

(_________ , __________ )

Answer 1

Solution :

Given that,

n = 494

Point estimate = sample proportion = = 0.19

1 - = 1 - 0.19 = 0.81

a) At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 (((0.19 * 0.81) / 494)

= 0.03

A 90% confidence interval for population proportion p is ,

± E

= 0.19 ± 0.03

= (0.16, 0.22)

b) At 98% confidence level

= 1 - 98%

=1 - 0.98 =0.02

/2 = 0.01

Z/2 = Z_{0.01 = 2.326}

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.326 (((0.19 * 0.81) / 494)

= 0.04

A 98% confidence interval for population proportion p is ,

± E

= 0.19 ± 0.04

= (0.15, 0.23)