Question

A survey of College students was conducted to see how many go to sporting events on campus. From the 1,365 students that were asked, 863 students said they attend events. Construct a 99% coincidence interval of the student that do not attend sporting events.

Answer 1

Solution:

Sample size = n = 1365

Number of students who attend events = 863

So ,

Number of students who do not attend events = 1365 - 863 = 502

Our aim is to confidence interval for proportion of the student that do not attend sporting events.

So , take x = 502

Let denotes the sample proportion

     = x/n   = 502/1365 = 0.368

Our aim is to construct 99% confidence interval.

c = 0.99

= 1 - c = 1- 0.99 = 0.01

  /2 = 0.005 and 1- /2 = 0.995

Search the probability 0.995 in the Z table and see corresponding z value

= 2.576   

Now , the margin of error is given by

E = /2 *  

= 2.576 * [0.368 *(1 - 0.368)/1365]

= 0.034

Now the confidence interval is given by

( - E)   ( + E)

(0.368 - 0.034)   (0.368 + 0.034)

0.334    0.402

i.e.

(0.334 , 0.402)