Question

College administrators want to know if students would be willing to pay a new fee if most of it would be budgeted for Student Activities. A random sample of 632 students resulted in 516 that would be willing to pay the new fee. Construct a 90% Confidence Interval estimate for the proportion of students willing to pay the new fee.

Answer 1

Solution :

Given that,

n = 632

x = 516

Point estimate = sample proportion = = x / n = 516/632=0.816

1 -   = 1- 0.816 =0.184

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 *((0.816*0.184) / 632)

E = 0.0254

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.816 0.0254< p <0.816+ 0.0254

0.7906< p < 0.8414

The 90% confidence interval for the population proportion p is : 0.7906, 0.8414