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

Answer:

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

Step-by-step 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. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample proportion of students who are registered voters = $\frac{300}{500}$ = 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 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.6449) = 0.90

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

P( $\hat p-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

90% confidence interval for p =[$\hat p-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$]

 = [ $0.60-1.6449 \times {\sqrt{\frac{0.60(1-0.60)}{500} } }$ , $0.60+1.6449 \times {\sqrt{\frac{0.60(1-0.60)}{500} } }$ ]

 = [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.