Question

University dean is interested in determining the proportion of students who receive some sort offinancial aid. rather than examine the records for all students, the dean randomly selects 200students and finds that 118 of them are receiving financial aid. use a 98% confidence interval toestimate the true proportion of students on financial aid. express the answer in the form p± e andround to the nearest thousandth.

Answer 1

The confidence interval is
$0.59\pm0.081$

We first find p, our sample proportion.  118/200 = 0.59.

Next we find the z-score associated with this level of confidence:
Convert 98% to a decimal: 98% = 98/100 = 0.98
Subtract from 1:  1-0.98 = 0.02
Divide by 2:  0.02/2 = 0.01
Subtract from 1:  1-0.01 = 0.99

Using a z-table (http://www.z-table.com) we see that this value is associated with a z-score of 2.33.

The margin of error (ME) is given by
$ME=z\sqrt{\frac{p(1-p)}{n}} \\ \\=2.33\sqrt{\frac{0.59(1-0.59)}{200}}=2.33\sqrt{\frac{0.2419}{200}}\approx0.081$

This gives us the confidence interval 
$0.59\pm0.081$