Question

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 404 drivers and find that 301 claim to always buckle up. Construct a 91 % confidence interval for the population proportion that claim to always buckle up. Use interval notation, for example, [1,5]

Answer 1

Solution :

Given that,

Point estimate () = x / n = 301 / 404 = 0.745

1 - = 0.255

Z/2 = 1.695

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

= 1.695 * (((0.745 * 0.255) / 404)

= 0.037

A 91% confidence interval for population proportion p is ,

- E < p < + E

0.745 - 0.037 < p < 0.745 + 0.037

0.708 < p < 0.782

(0.708, 0.782)