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A college graduate school president is interested in knowing what proportion of applicants would like to be accepted into the statistics department. Of a simple random sample of 75 applicants, 12 requested the statistics department. Construct a 95% confidence interval for the true proportion of all applicants that prefer statistics.

User Danny Kim
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1 Answer

4 votes

Answer:

0.077 to 0.24

Explanation:


P=(12)/(75)=0.16

confidence level =95%=0.95

significance level =1-confidence level =1 -0.95= 0.05


z_(\alpha )/(2)=z_(0.05)/(2)=1.96 from the z table

standard error of P
SE=\sqrt{(P* \left ( 1-P \right ))/(n)}


\sqrt{(0.16* 0.84)/(75)} as n=75 given

=0.0423


E=z_(\alpha )/(2)*\sqrt(P* \left ( 1-P \right ))/(n)

=1.96×0.0423=0.0289

now confidence interval is given by (0.16-0.0289 ,0.16+0.0289)

=(0.077,0.24)

User Manuel Navarro
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