In: Statistics and Probability
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
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