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%

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