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 99% level of confidence,what is the confidence interval for the proportion of students supporting the fee increase?
A)[0.751,0.829]
B)[0.759,0.823]
C)[0.767,0.814]
D)[0.771,0.811]

Question

Quizplus · Accepted Answer

To calculate the confidence interval for the proportion of students supporting the fee increase, we need to use the formula:

CI = p ± Z*sqrt(p*(1-p)/n)

where:

p = proportion of students supporting the fee increase
Z = critical value for the given level of confidence (99% in this case)
n = sample size

Plugging in the values from the problem, we get:

p̂ = 870/1100 = 0.791
Z = 2.58 (using a table or calculator)
n = 1100

CI = 0.791 ± 2.58*sqrt(0.791*(1-0.791)/1100)
   = 0.791 ± 0.024
   = [0.767, 0.815]

So the best choice is B, which has the closest range to [0.767, 0.815].

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