Question

A website reported that 31% of drivers 18 and older admitted to texting while driving in 2009. In a random sample of 400 drivers 18 years and older drawn in 2010, 99 of the drivers said they texted while driving.

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 ___ and an upper limit of ___. ?(Round to three decimal places as? needed.)

Answer 1

Solution :

Given that,

n = 400

x = 99

= x / n = 99 / 400 = 0.247

1 - = 1 - 0.247 = 0.753

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * (((0.247 * 0.753) / 400) = 0.056

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.247 - 0.056 < p < 0.247 + 0.056

0.191 < p < 0.303

Lower limit = 0.191

Upper limit = 0.303