Question

An insurance company checks police records on 573 accidents selected at random and notes that teenagers were at the wheel in 83 of them. Complete parts ​a) through ​d).

​a) Construct the​ 95% confidence interval for the percentage of all auto accidents that involve teenage drivers.

​95%

CIequals=left parenthesis nothing % comma nothing % right parenthesis%,%

​(Round to one decimal place as​ needed.)

Answer 1

Solution :

Given that,

n = 573

x = 83

Point estimate = sample proportion = = x / n = 83/573=0.145

1 - = 1-0.145=0.855

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )   

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

= 1.96 (((0.145*0.855) / 573)

E = 0.0288

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.145 -0.0288 < p <0.145 + 0.0288

0.1162< p < 0.1738

The 95% confidence interval for the population proportion p is : 11.6% and17.4%