Question

for a class project, a political science student at a large university wants to estimate the percent of students that are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90% confidenct interval for the treu percent of students that are registered voters and interopret the confidence interval

Answer 1

90% confidence interval for the true percent of students that are registered voters is [0.56 , 0.64].

Explanation:

We are given that for a class project, a political science student at a large university wants to estimate the percent of students that are registered voters.

He surveys 500 students and finds that 300 are registered voters.

Firstly, the pivotal quantity for 90% confidence interval for the true proportion is given by;

P.Q. =
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where,
`\hat p` = sample proportion of students who are registered voters =
`(300)/(500)` = 0.60

n = sample of students = 500

p = true percent of students

Here for constructing 90% confidence interval we have used One-sample z proportion test statistics.

So, 90% confidence interval for the true proportion, p is ;

P(-1.6449 < N(0,1) < 1.6449) = 0.90 {As the critical value of z at 5%

level of significance are -1.6449 & 1.6449}

P(-1.6449 <
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` < 1.6449) = 0.90

P(
`-1.6449 * {\sqrt{(\hat p(1-\hat p))/(n) } }` <
`{\hat p-p}` <
`1.6449 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.90

P(
`\hat p-1.6449 * {\sqrt{(\hat p(1-\hat p))/(n) } }` < p <
`\hat p+1.6449 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.90

90% confidence interval for p =[
`\hat p-1.6449 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ,
`\hat p+1.6449 * {\sqrt{(\hat p(1-\hat p))/(n) } }`]

= [
`0.60-1.6449 * {\sqrt{(0.60(1-0.60))/(500) } }` ,
`0.60+1.6449 * {\sqrt{(0.60(1-0.60))/(500) } }` ]

= [0.56 , 0.64]

Therefore, 90% confidence interval for the true percent of students that are registered voters is [0.56 , 0.64].

Interpretation of the above confidence interval is that we are 90% confident that the true percent of students that are registered voters will lie between 0.56 and 0.64.