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}=(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{\frac{\hat{p}(1-\hat{p})}{n}}\\\\=0.29\pm(2.33)\sqrt{((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{\frac{\hat{p}(1-\hat{p})}{n}}`

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

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