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

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

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 < < 1.6449) = 0.90

P( < < ) = 0.90

P( < p < ) = 0.90

90% confidence interval for p =[ , ]

 = [ , ]

 = [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.

how is that middle school

step-by-step explanation:

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