Question

A website reported that 31​% of drivers 18 and older admitted to texting while driving in 2009. In a random sample of 800 drivers 18 years and older drawn in​ 2010, 204 of the drivers said they texted while driving. Complete parts a through c below. a. Construct a 99​% confidence interval to estimate the actual proportion of people who texted while driving in 2010. A 99​% confidence interval to estimate the actual proportion has a lower limit of nothing and an upper limit of nothing.

Answer 1

Solution :

Given that,

n = 800

x = 204

Point estimate = sample proportion = = x / n = 204 / 800 = 0.255

1 - = 1 - 0.255 = 0.745

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z_{0.005 = 2.576}

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

= 2.576 (((0.255 * 0.745) / 800)

= 0.040

A 99% confidence interval for population proportion p is ,

± E  

= 0.255 ± 0.040

= ( 0.215, 0.295 )

lower limit = 0.215

upper limit = 0.295