Arandom sample of 384 people in a mid-sized city (city one) revealed 112 individuals who worked at more than one job. a second random sample of 432 workers from another mid-sized city (city two) found 91 people who work at more than one job. find a 99% confidence interval for the difference between the proportions of workers in the two cities who work at more than one job. select one: a. (0.003, 0.159)b. (0.021, 0.141)c. (-0.159, 0.004)d. (0.031, 0.131)e. sample sizes aren't large enough to justify using z-procedures

99% confidence interval is:

(0.00278 < P1 - P2< 0.15921)

Step-by-step explanation:

For calculating a confidence intervale for the difference between the proportions of workers in the two cities, we calculate the following:

             Where   : proportion sample of individuals who worked      

                                                     at more than one job in the city one

                           : Number of respondents in the city one

                            : proportion sample of individuals who worked      

                                                     at more than one job in the city two

                           : Number of respondents in the city two

Then

α = 0.01 and α/2 = 0.005

and

 and

The confidence interval is:

(0.00278 < P1 - P2< 0.15921)

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.

b