Question

Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. 500 randomly selected adult residents in this city are surveyed, 481 responded yes, they own cell phones. Find a 92% confidence interval for the true proportion of adults residents in this city who have a cell phone. Then give a complete confidence statement. Round only the last calculation (both endpoints of the interval) to two decimal places

Answer 1

Solution :

Given,

n = 500 ....... Sample size

x = 481 .......no. of successes in the sample

Let denotes the sample proportion.

     = x/n   = 481 / 500 = 0.96

Our aim is to construct 92% confidence interval.

c = 0.92

= 1 - c = 1- 0.92 = 0.08

  /2 = 0.04 and 1- /2 = 0.96

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

= 1.751   

Now , the margin of error is given by

E = /2 *  

= 1.751 * [ 0.96 *(1 - 0.96)/500 ]

= 0.015

Now the confidence interval is given by

( - E)   ( + E)

( 0.96 - 0.015 )   ( 0.96 + 0.015 )

0.95   0.98  

Required 92% Confidence Interval is ( 0.95 , 0.98 )