Question

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 410 drivers and find that 282 claim to always buckle up. Construct a 95% confidence interval for the population proportion that claim to always buckle up.

Answer 1

Solution :

Given that,

n = 410

x = 282

Point estimate = sample proportion = = x / n = 282/410 = 0.688

1 - = 1 -0.688 = 0.312

At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

Z/2 = Z_{0.025 = 1.960}

Z_/2 = 1.960

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

= 1.960 * ((0.688*(0.312) /410 )

= 0.045

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.688 - 0.045< p < 0.688+ 0.045

0.643 < p < 0.733  

(0.643,0.733 )