Getting a Job. The National Association of Colleges and Employers sponsors the Graduating Students and Alumni...

Getting a Job.

The National Association of Colleges and Employers sponsors the Graduating Students and Alumni Survey. Part of the survey gauges student optimism in landing a job after graduation. According to one year’s survey results among the 1218 respondents, 733 said they expected difficulty finding a job. a. Use the data to find and interpret a 95% confidence interval for the proportion of students who expect difficulty finding a job. b. What is the margin of error?

Solutions

Expert Solution

Solution :

n = 1218

x = 733

= x / n = 733 / 1218 =0.602

1 - = 1 - 0.602 = 0.398

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.960

Margin of error = E = Z / 2 * $\sqrt$ (( * (1 - )) / n)

= 1.960 * ( $\sqrt$ ((0.602 * 0.398) / 1218)

= 0.027

A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.602 - 0.027 < p < 0.602 + 0.027

0.574 < p < 0.629

orchestra answered 1 year ago

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