Question

Suppose a consumer advocacy group would like to conduct a survey to find the proportion p of consumers who bought the newest generation of an MP3 player were happy with their purchase. The advocacy group took a random sample of 1000 consumers who recently purchased this MP3 player and found that 6 HUNDRED were happy with their purchase. Find a 90% confidence interval (CI)l for p y entering the Lower Limit and the Upper Limit of the CI?

Answer 1

from the given data of information

n= 1000

x=600

=600/1000=x/n

=0.6

point estimate is 0.6

1 - =1-0.6 =0.4

at 90% confidence level z is

= 1-90% = 1-0.90

=0.01

/2 =0.01/2 =0.005

= = 1.645

margin of error E =*

=1.645*

=0.025

at 90%confidence interval of p is

-E<P<+E

0.6-0.025<P<0.6+0.025

0.575<P<0.625

(0.575,0.625)

LOWER LIMIT IS 0.575

UPPER LIMIT IS 0.625