Question

1. A national survey of 900 women golfers was conducted to learn how women golfers view their treatment at golf courses in United States. The survey found that 396 of the women golfers were satisfied with the availability of tee times.

Suppose one wants to develop a 99% confidence interval estimate for the proportion of the population of women golfers satisfied with the availability of tee times.

Answer 1

sample success x =		396
sample size          n=		900
sample proportion p̂ =x/n=		0.440
std error se= √(p*(1-p)/n) =		0.0165
for 99 % CI value of z=		2.58
margin of error E=z*std error   =		0.043
lower bound=p̂ -E                       =		0.397
Upper bound=p̂ +E                     =		0.483

from above 99% confidence interval for population proportion =(0.397,0.483)