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 283 claim to always buckle up. Construct a 91% confidence interval for the population proportion that claim to always buckle up.

Answer 1

Solution :

Given that,

n = 410

x = 283

Point estimate = sample proportion = = x / n = 283/410=0.690

1 -   = 1-0.690 =0.31

At 91% confidence level the z is ,

= 1 - 91% = 1 - 0.91 = 0.09

/ 2 = 0.09 / 2 = 0.045

Z/2 = Z0.045 = 1.695   ( Using z table )

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

= 1.695 (((0.69*0.31) /410 )

E = 0.039

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.69-0.039 < p < 0.69+0.039

0.651< p <0.729

91% confidence interval for the population proportion that claim to always buckle up (0.651 to 0.729)