Statistics question.

Question

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.

When 332 college students are randomly selected and surveyed, it is found that 113 own a car. Find a 99% confidence interval for the true proportion of all college students who own a car.

Answer 1

The point estimate for p = r/n = 113/332 = 0.340

where r = number of successful trials = 113, and n = 332

 

To justify the approximation to normal approximation, we make sure that np>5 and nq>5

 

np = 332*0.340 = 113 >5

nq = 332*(1-0.340) = 219 >5

 

Z_critical for a 99% confidence level = 2.575

 

E = Z_critical * √p(1-p)/n = 2.575*√[(0.340)(1-0.340)/332]

 

E = 0.026

 

99% confidence interval is p ± E = 0.340 ± 0.026

 

0.314<p<0.366