Question

An article included data from a survey of 2688 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 2688 people who participated in the survey, 1100 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. 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.)

(______,_______)

Answer 1

Solution :

Given that,

n = 2688

x = 1100

Point estimate = sample proportion = = x / n = 1100 / 2688 = 0.409

1 - = 1 - 0.409 = 0.591

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

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

= 1.96 * (((0.409*0.591) /2688 )

= 0.019

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.409-0.019< p <0.409+0.019

0.391< p <0.428

The 95% confidence interval for the population proportion p is :(0.391 , 0.428)