Posted by **Jenny** on Wednesday, December 19, 2007 at 9:37pm.

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]

- Statistics -
**economyst**, Thursday, December 20, 2007 at 11:40am
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)

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