Question 1177068
This has a half-interval of z(0.975)*sqrt(p*(1-p)/n)
or 1.96* sqrt (0.032*0.968/225)
=0.0230
so 95%CI=(0.009, 0.055) proportion of people with dizziness
The calculator will give (0.008, 0.54) because 3.2% of 225 is 7.2, and we have to use integers here or 7, which will give a slightly different interval.