Question

A market survey is conducted at local shopping complexes.
A random sample of 1250 consumers are asked whether they consume supplements and 850 of them say “YES”.
Construct an approximate 96% confidence interval for the true proportion of consumers who say “YES”.
Give your answer correct to 3 decimal places

Answer 1

Solution:

Given,

n = 1250 ....... Sample size

x = 850 .......no. of successes in the sample

Let denotes the sample proportion.

     = x/n   = 850/1250 = 0.68

Our aim is to construct 96% confidence interval.

c = 0.96

= 1- c = 1- 0.96 = 0.04

  /2 = 0.02

= 2.054 ...(use z table)

Now , the margin of error is given by

E = *  

= 2.054 * [0.68*(1 - 0.68)/1250]

= 0.027

Now the confidence interval is given by

( - E)   ( + E)

(0.68 - 0.027)   (0.68 + 0.027)

0.653 0.707

i.e.

(0.653 , 0.707)