Answer:
(0.500, 1.090)
Step-by-step explanation:
Given that a university dean is interested in determining the proportion of students who receive some sort of financial aid.
A sample of (random) 200 students taken and 118 were found out receiving financial aid.
Sample proportion p = [tex]\frac{118}{200} \\=0.59[/tex]
[tex]q-1-o=0.41\\std error = \sqrt{pq/n} =0.0348[/tex]
For margin of error we use Z critical value as this is proportion and also sample size is large. np and nq both are greater than 5.
So margin of error = 2.58*std error = 0.090
Confidence interval
=[tex](0.500,1.090)[/tex]
