Question

An article included data from a survey of 2996 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 2996 people who participated in the survey, 1500 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 professionais. 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.) Interpret the interval. % confident that the proportion of all hiring managers - Select this confidence interval. onals who Select social networking sites to research job applicants is You may need to use the appropriate table in Appendix A to answer this question Need Help? Read Tak te Tuter -78 POINTS PODSTATS 9.0.022. MY NOTES In a survey of 1000 randomly selected adults in the United States, participants were asked what the most favorite and what their least favorite subject was when they were in school (Associated Press, August 17, 2005). In what might seem like a contradiction, math was chosen ore often than any other subject in both categories Math was chosen by 230 of the 1000 as the favorite subject, and it was also chosen by 374 of the 1000 as the least favorite subject (a) Construct a 95% confidence interval for the proportion of U.S. adults for whom math was the favorite subject in school (Give the answers to four decimal places) (b) Construct a 95% confidence interval for the proportion of U.S. adults for whom math was the least favorite subject. Give the answers to four decimal places.)

Answer 1

Sample size = n = 2996 Sample number of events = x = 1500 Sample proportion = p = x/n = (1500/2996) = 0.5007 (point estimate of population proportion) Stanadard Error = P(1-) 2 = sqrt(0.5007*(1-0.5007)/2996) = 0.0091 V n Confidence Level = 95 Significance Level = a = (100-95)% = 0.05 Critical value = z* = 1.96 [Using Excel =ABS(NORMINV(0.05/2,0,1))] Margin of Error = z* (standard error )= 1.96*0.0091 = 0.0179 Lower Limit =Ộ - Margin of Error = 0.5007 -0.0179 = 0.483 Upper Limit =p + Margin of Error = 0.5007 + 0.0179 = 0.519 Ans:(0.483,0.519)

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