Question

in a study of cell phone use and brakn hemispheric dominance, an internet survey was emailed to 2319 subjects randomly selected from an online group involved with ears. 1022 surveys were returned. construct a 95% confidence interval for the proportion of returned surveys

A. find the best point estimate of the populatiom proportion p

B. identify the value of the margin of error E

C. construct the confidence interval

Answer 1

Solution:

Given,

n = 2319 ....... Sample size

x = 1022 .......no. of successes in the sample

A) Let denotes the sample proportion.

     = x/n   = 1022/ 2319 = 0.441

B)

Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025 and 1- /2 = 0.975

Search the probability 0.975 in the Z table and see corresponding z value

= 1.96

Now , the margin of error is given by

E = /2 *  

= 1.96 * [(0.441 * 0.441)/2319]

=  0.0202

margin of error =  0.0202

C)

Now the confidence interval is given by

( - E)   ( + E)

(0.441 - 0.0202)   (0.441 + 0.0202)

0.4208   0.4612

Interval is (0.4208 , 0.4612)