Construct a 95% confidence interval for the population proportion who claim they always buckle up.

Question

Suppose that insurance companies did a survey. They randomly surveyed 430 drivers and found that 330 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up. NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

 

x:330

n:430

p':0.7674

 

Construct a 95% confidence interval for the population proportion who claim they always buckle up.

Answer 1

the error margin is:

 

    1.96 * sqrt(  p(1-p)/n )  =

 

  1.96* sqrt(  0.7674 * 0.2326 / 430 )

 

   = 0.0204

 

 

 ( 0.7674 - 0.0204 ,   0.7674 + 0.0204) =

 

 ( 0.747 ,  0.7878)