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 for the proportion of students supporting the fee increase?
A)[0.767,0.815]
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)) / 1,100)= 0.87 +/- 0.024= [0.846, 0.894]Therefore, the confidence interval is [0.846, 0.894].

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