Question

A website reported that 33​% of drivers 18 and older admitted to texting while driving in 2009. In a random sample of 200 drivers 18 years and older drawn in​ 2010, 58 of the drivers said they texted while driving. Complete parts a through c below. a. Construct a 98​% confidence interval to estimate the actual proportion of people who texted while driving in 2010. A 98​% confidence interval to estimate the actual proportion has a lower limit of 0.215 and an upper limit of 0.365. ​(Round to three decimal places as​ needed.) b. What is the margin of error for this​ sample? ​(Round to three decimal places as​ needed.)

Answer 1

Answer with explanation:

Given : In a random sample of 200 drivers 18 years and older drawn in​ 2010, 58 of the drivers said they texted while driving.

i.e. $\hat{p}=\dfrac{58}{200}=0.29$

n= 200

Significance level : $\alpha: 1-0.98=0.02$

Critical value : $z_{\alpha/2}=2.33$

a) The confidence interval for population proportion is given by :-

$\hat{p}\ \pm\ z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.29\pm(2.33)\sqrt{\dfrac{(0.29)(1-0.29)}{200}}\\\\\approx0.29\pm0.075\\\\=(0.29-0.075,0.29+0.075)=(0.215,\ 0.365 )$

Hence, a 98​% confidence interval to estimate the actual proportion has a lower limit of 0.215 and an upper limit of 0.365.

b). The margin of error for this​ sample : $E=z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

i.e. $(2.33)\sqrt{\dfrac{(0.29)(1-0.29)}{200}}=0.0747599662252\approx0.075$

Hence, the margin of error for this​ sample is 0.075.