Question

An article included data from a survey of 2769 hiring managers and human resource professionals. The article noted that many employers are using social networks to screen job applicants and that this practice is becoming more common. Of the 2769 people who participated in the survey, 1200 indicated that they use social networking sites (such as Facebook, MySpace, and LinkedIn) to research job applicants. For the purposes of this exercise, assume that the sample is representative of hiring managers and human resource professionals.

a) Construct a 95% confidence interval for the proportion of hiring managers and human resource professionals who use social networking sites to research job applicants. (Round your answers to three decimal places.)

(_________ , __________)

b) Interpret the interval.

We are __________ % confident that the proportion of all hiring managers and human resource professionals who use social networking sites to research job applicants is within this confidence interval.

Answer 1

Solution :

Given that,

a) n = 2769

x = 1200

Point estimate = sample proportion = = x / n = 1200/2769 = 0.433

1 - = 1 - 0.433 = 0.567

Z_/2 = 1.96

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 * ((0.433*(0.567) / 2769)

= 0.018

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.433 - 0.018 < p < 0.433+ 0.018

0.415< p < 0.452

( 0.415 , 0.452)

The 95% confidence interval for the population proportion p is : 0.415 and 0.452

b) We are 95% confident that the proportion of all hiring managers and human resource professionals who use social networking sites to research job applicants is within this confidence interval.