Question

A newsgroup is interested in constructing a 99% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 557 randomly selected Americans surveyed, 408 were in favor of the initiative. Round answers to 4 decimal places where possible.

a. With 99% confidence the proportion of all Americans who favor the new Green initiative is between ____ and ____.

Answer 1

Solution :

Given,

n = 557 ....... Sample size

x = 408 .......no. of successes in the sample

Let denotes the sample proportion.

     = x/n   = 408 / 557 = 0.732  

Our aim is to construct 99% confidence interval.

c = 0.99

= 1 - c = 1- 0.99 = 0.01

  /2 = 0.005 and 1- /2 = 0.995

Search the probability 0.995 in the Z table and see corresponding z value

= 2.576   

Now , the margin of error is given by

E = /2 *  

= 2.576 * [ 0.732 *(1 - 0.732)/557]

= 0.0483

Now the confidence interval is given by

( - E)   ( + E)

( 0.732 - 0.0483 )   ( 0.732 + 0.0483 )

0.6837   0.7803

Required 99% Confidence Interval is ( 0.6837 , 0.7803 )

The 99% confidence the proportion of all Americans who favor the new Green initiative is between 0.6837 and 0.7803.