Question

. In a survey of 600 community college students, 470 indicated that they have read a book for personal enjoyment during the school year. Determine a 95% confidence interval for the proportion of community college students who have read a book for personal enjoyment during the school year.

Answer 1

C.I. at 95% is as C.I.[0.7504 to 0.8162]

Explanation:

Given:

Total number n=600

and Required No of students=x=470

To Find:

Determine a 95% confidence interval .

Solution:

Using normal distribution table for Z,

when 95 % C.I. Z=1.96

Now calculate ,

Population proportion,p=x/n

p=470/600

p=0.7833

Now calculate margin of error

MOE=Z*Sqrt[p(1-p)/n]

MOE=1.96*Sqrt[(0.7833*0.21667)/600]

=1.96*Sqrt(0.00028286)

=1.96*0.01681

=0.0329

So

The Boundaries will be as follows:

1)p-MOE

=0.7833-0.0329

=0.7504

2)p+MOE

=0.7833+0.0329

=0.8162