The National Association of Colleges and Employers sponsors the Graduating Stude

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 year’s survey results, published in American Demographics, among the 1218 respondents, 733 said that they expected finding difficulty finding a job. Note: In this problem, round your proportion to four decimal places. a)Use these data to find and interpret a 95% confidence interval for the proportion of students who expect difficulty finding a job.

Explanation / Answer

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.60180624          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.014026584          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.0274916          
lower bound = p^ - z(alpha/2) * sp =   0.57431464          
upper bound = p^ + z(alpha/2) * sp =    0.62929784          
              
Thus, the confidence interval is              
              
(   0.57431464   ,   0.62929784   ) [ANSWER]

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