Question

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 383 drivers and find that 296 claim to always buckle up. Construct a 87% confidence interval for the population proportion that claim to always buckle up.

Answer 1

n = Sample Size = 383

p = Sample proportion = 296/383 = 0.7728

q = 1 - p = 0.2272

SE =

=

= 0.13

ndf = 383 - 1 = 382

From Table, critiical values of t = 1.5174

So,

Confidence interval:

0.7728 (1.5174 X 0.0214)

= 0.7728 0.0325

= ( 0.7403, 0.8053)

So,

Confidence interval is:

0.7403 < P < 0.8053