Question

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 391 drivers and find that 280 claim to always buckle up. Construct a 96% confidence interval for the population proportion that claim to always buckle up. Use interval notation

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 280 / 391 = 0.716

1 - = 1 - 0.716 = 0.284

Z_/2 = 2.054

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

= 2.054 * (((0.716 * 0.284) / 391)

Margin of error = E = 0.047

A 96% confidence interval for population proportion p is ,

- E < p < + E

0.716 - 0.047 < p < 0.716 + 0.047

0.669 < p < 0.763