Question

Suppose a survey revealed that 22% of 502 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  

Answer 1

Solution :

Given that,

n = 502

Point estimate = sample proportion = = 0.22

1 - = 1 - 0.22 = 0.78

a) At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

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

= 1.645 (((0.22 * 0.78) / 502)

= 0.030

A 90% confidence interval for population proportion p is ,

± E

= 0.22  ± 0.030

= ( 0.190, 0.250 )

b) At 98% confidence level

= 1 - 98%

=1 - 0.98 =0.02

/2 = 0.01

Z/2 = Z0.01 = 2.326

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

= 2.326 (((0.22 * 0.78) / 502)

= 0.043

A 98% confidence interval for population proportion p is ,

± E

= 0.22  ± 0.043

= ( 0.177, 0.263 )

c) The 98% confidence interval

d) The 90% confidence interval