A survey of an urban university (population of 25,450) showed that 870 of 1,100 students sampled supported a fee increase to fund improvements to the student recreation center. Using the 95% level of confidence, what is the confidence interval?
A) [0.767, 0.814]
B) [0.759, 0.822]
C) [0.771, 0.811]
D) [0.714, 0.866]
E) None of the choices are correct

Question

Quizplus · Accepted Answer

The correct answer is A because the confidence interval is [0.767, 0.814]. This can be calculated using the formula:CI = p +/- z * sqrt((p * (1-p)) / n)where p is the sample proportion, z is the z-score for the desired confidence level, and n is the sample size.In this case, p = 870/1100 = 0.791, z = 1.96 (for a 95% confidence level), and n = 1100. Plugging these values into the formula gives us:CI = 0.791 +/- 1.96 * sqrt((0.791 * (1-0.791)) / 1100)CI = 0.791 +/- 1.96 * sqrt((0.791 * 0.209) / 1100)CI = 0.791 +/- 1.96 * sqrt(0.164 / 1100)CI = 0.791 +/- 1.96 * sqrt(0.000149)CI = 0.791 +/- 1.96 * 0.0122CI = 0.791 +/- 0.024CI = [0.767, 0.814]

A Survey of an Urban University (Population of 25,450) Showed