Question

Please Answer the following regarding probability:

6. A 2011 Gallup poll found that 76% of Americans believe that high achieving high school students should be recruited to become teachers. This poll was based on a random sample of 1002 Americans.

a) Find a 90% confidence interval for the proportion of Americans who would agree with this.

b) Interpret your interval in this context.

Please show all work.

Please typed answers only, no handwriting.

Answer 1

Solution:

We have

Sample size(n)=1002

Sample proportion(p^)=76%

=76/100

=0.76

(a) Here Significance level(alpha)=1-confidence interval

=1-90/100

=1-0.90

=0.1

z(alpha/2)=0.1/2

=0.05

Now to find the 90% confidence interval:

The 90% Confidence Interval = P^P^(1-P^)/n

=0.760.76(1-0.76)/1002

=0.76(0.76*0.24)/1002

=0.76(0.1824)/1002

=0.760.00018

=0.760.0134

=(0.76-0.0134,0.76+0.0134)

=(0.7466,0.7734)

(b) The Gallup Poll is 90% confident that 74 to 77% of americans believe that high achieving high school students should be recruited to become teachers.