Question

the national association of colleges and employers sponsors the graduating student and alumni survey. part of the survey gauges student optimism in landing a job after graduation.according to one years survey results, published in American demographics, among the 1500 respondents,733 said that they expected difficulty finding a job.construct a 90% confidence interval for the proportion of students who expected difficulty finding a job. interpret your result.

Answer 1

Solution :

Given that,

n = 1500

x = 733

Point estimate = sample proportion = = x / n = 733/1500=0.489

1 - = 1 - 0.489=0.511

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645} ( Using z table )

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

= 1.645 (((0.489*0.511) /1500 )

E= 0.0212

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.489- 0.0212< p <0.489+ 0.0212

0.4678< p < 0.5102

(0.4678, 0.5102)

The 90% confidence interval for the population proportion p is : 0.4678, 0.5102