A survey of an urban university (population of 25,450) showed that 870 of 1100 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 for the proportion of students supporting the fee increase?
A) [0.767, 0.814]
B) [0.759, 0.822]
C) [0.771, 0.811]
D) [0.714, 0.866]

Question

Quizplus · Accepted Answer

The confidence interval is calculated as:sample proportion +/- margin of errorwhere the margin of error is:z * sqrt((sample proportion * (1 - sample proportion)) / sample size)where z is the z-score corresponding to the desired confidence level. For a 95% confidence level, z = 1.96.Plugging in the values, we get:0.87 +/- 1.96 * sqrt((0.87 * (1 - 0.87)) / 1100)= 0.87 +/- 0.024= [0.846, 0.894]Therefore, the confidence interval for the proportion of students supporting the fee increase is [0.846, 0.894].

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