I believe the answer is 'none of the above'
The formula for the confidence interval is:
P +or- Z(99) * sqrt(P*Q/n)
P is the estimated probability of attendence = 750/1100 = .6818
Q = 1-P
n = sample size = 1100
Z(99) is the number of standard deviations away from the mean that 99% of the population fall = 2.575
So Z(99)*sqrt(P*Q/n) = .03616.
So the range should be (.6456, .7180)
Thank you!!
A survey of an urban university (population of 25,450) showed that 750 of 1100 students sampled attended a home football game during the season. What inferences can be made about student attendance at football games?
Using the 99% level of confidence, what is the confidence interval?
a. [0.767, 0.814]
b. [0.0.6550, 0.7050]
c. [0.6659, 0.6941]
d. [0.0.6795, 0.6805]
1 answer
2 answers