Question

6. Suppose a researcher believes that college faculty vote at a higher rate than college students.She collects data from 200 college faculty and 200 college students using simple random sampling. If 167 of the faculty and 138 of the students voted in the 2008 Presidential election, find a 90% confidence interval for the difference between the proportion of faculty and the proportion of students.

Answer 1

Solution:

90% Confidence interval for difference in proportions is given by (p1-p2) +_ 1.645 *S.E(p1-p2)

( Critical value of Z at 10% level of significance is 1.645)

p1=167/200= 0.835

p2= 138/200= 0.69

p1-p2= 0.145

q1=1-p1, q2=1- p2

S. E(p1-p2) =sqrt(p1q1/n1 + p2q2/n2)

= sqrt(0.1378/200+ 0.2139/200) = 0.0419

90% confidence interval= ( 0.145- 0.0689, 0.145+ 0.0689)

= (0.0761, 0.2139)

Therefore, based on the data provided, the 90% confidence interval for the difference between the population proportions is ( 0.076,0.214) which indicates that we are 90% confident that the true difference between population proportions is contained by the interval (0.076, 0.214) .

answered by: anonymous