Question

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.

Answer 1

Answer:

0.077 to 0.24

Step-by-step explanation:

$P=\frac{12}{75}=0.16$

confidence level =95%=0.95

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

$z_\frac{\alpha }{2}=z_\frac{0.05}{2}=1.96$  from the z table

standard error of P $SE=\sqrt{\frac{P\times \left ( 1-P \right )}{n}}$

$\sqrt{\frac{0.16\times 0.84}{75}}$ as n=75 given

=0.0423

$E=z_\frac{\alpha }{2}\times\sqrt\frac{P\times \left ( 1-P \right )}{n}$

=1.96×0.0423=0.0289

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

=(0.077,0.24)