Question 1150274
90% CI has z=1.645
p hat is 451/600=0.752
half interval for a 1 sample proportion is z*sqrt(p*(1-p)/n); sqrt (0.751*0.249/600)=0.01765
z*se=0.029
90% CI=(0.723, 0.781)
We don't know the true value, but we can be 90% confident it is in that interval.