A survey of an urban university showed that 750 of 1100 stud

A survey of an urban university showed that 750 of 1100 students sampled attended a home football game during the season. Using the 99% level of confidence, what is the confidence interval for the proportion of students attending a football game?

[0.767, 0.814]

[0.6550, 0.7050]

[0.6456, 0.7179

[0.6795, 0.6805]

Solution

large sample confidence intervals are used to find a region in which we are 100 (1-)% confident the true value of the parameter is in the interval.

For large sample confidence intervals for the proportion in this situation you have:

pHat ± z * sqrt( (pHat * (1-pHat)) / n)

where pHat is the sample proportion
z is the zscore for having % of the data in the tails, i.e., P( |Z| > z) =
n is the sample size = 1100

the z-score for a 0.99 Confidence Interval (CI) about the population proportion is:
P(|Z| > z) = 0.01
P(Z > z) = 0.005
z = -2.575829

the 100( 1 - 0.01 )% CI is:
( pHat - z * sqrt(pHat * (1 - pHat) / n) , pHat + z * sqrt(pHat * (1 - pHat) / n) )
( 0.6818182 - 2.575829 * 0.01404351 , 0.6818182 + 2.575829 * 0.01404351 )
( 0.6456445 , 0.7179919 )

with some difference in rounding your solution is answer [0.6550,0.7050]

this CI tells us that we can be 99% confident the true proportion of students attending the home football games is within this interval.