Question

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all​ students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. Use a 99​% confidence interval to estimate the true proportion of students on financial aid.

Answer 1

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 = $\frac{118}{200} \\=0.59$

$q-1-o=0.41\\std error = \sqrt{pq/n} =0.0348$

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

=$(0.500,1.090)$