Question

1. Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car.

When designing a study to determine this population proportion, what is the minimum number of drivers you would need to survey to be 95% confident that the population proportion is estimated to within 0.04? (Round your answer up to the nearest whole number.)

2. According to a poll, 75% of California adults (377 out of 506 surveyed) feel that education is one of the top issues facing California. We wish to construct a 90% confidence interval for the true proportion of California adults who feel that education is one of the top issues facing California.

Find a 90% confidence interval for the population proportion. (Round your answers to three decimal places.)
( ? , ? )

Answer 1

1. Solution :

Given that margin of error E = 0.04

=> if we don't know, assume that p = q = 0.5

=> for 95% confidence level, Z = 1.96

=> Sample size n = p*q*(Z/E)^2

= 0.5*0.5*(1.96/0.04)^2

= 600.25

= 600 (nearest whole number)

2. Solution :

Given that p = 0.75 , n = 506

=> q = 1 - p = 0.25

=> for 90% confidence level, Z = 1.645

=> A 90% confidence interval of the population proportion is

=> p +/- Z*sqrt(p*q/n)

=> 0.75 +/- 1.645*sqrt(0.75*0.25/506)

=> (0.7183,0.7817)

=> (0.718,0.782) (rounded)