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. = 
~ N(0,1)
where, 
= sample proportion of students who are registered voters = 
= 0.60
n = sample of students = 500
p = true percent of students
<em>Here for constructing 90% confidence interval we have used One-sample z proportion test statistics.</em>
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 < < 1.6449) = 0.90
P( 
< 
< 
) = 0.90
P( 
< p < 
) = 0.90
<u>90% confidence interval for p</u> =[ , ]
= [ 
, 
]
= [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.
