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 98% confidence interval to estimate the true proportion of students on financial aid.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 118 / 200 = 0.590

1 - = 1 - 0.590 = 0.41

Z/2 = 2.326

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.326 * (((0.590 * 0.41) / 200)

Margin of error = E = 0.081

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.590 - 0.081 < p < 0.590 + 0.081

0.509 < p < 0.671

The 98% confidence interval for the population proportion p is : (0.509 , 0.671)