Question

A major metropolitan newspaper selected a simple random sample of 544 readers from their list of 100,000 subscribers. They asked whether the paper should increase its coverage of local news, and 33% agreed that they should. What is the upper bound on the 99% confidence interval for the proportion of readers who would like more coverage of local news? Round your answer to 3 decimal places.

Answer 1

Given,

n = 544 ....... Sample size

Let denotes the sample proportion.

     = 33% = 0.33  

Our aim is to construct 99% confidence interval.

c = 0.99

= 1- c = 0.01

  /2 = 0.005

= 2.576 (use z table)

Now , the margin of error is given by

E = /2 *  

= 2.576 * [(0.33*(1 - 0.33)/544]

= 0.052

Now the confidence interval is given by

( - E)   ( + E)

(0.33 - 0.052 )   (0.33 + 0.052 )

0.278 0.382  

Upper Bound = 0.382