in a random sample of 250 tv viewers in a large city, 190 had seen a certain controversial program. assuming the population is large compared to the sample size,
a.construct a 95% confidence interval for the true population proportion, using the estimated variance of p^
b/ construct a 95% confidence interval for the true population proportion, using the actual variance of p^
so far i got : p^= 190/250=0.76 and q^=1-p^=0.24
so (p^-p)/ Sp^~N(0,1)
so by using p^-Zc square root (p^q^/n) i got .707<μ<.813
so for part b how do we find the actual variance of p^
everyone please help
a.construct a 95% confidence interval for the true population proportion, using the estimated variance of p^
b/ construct a 95% confidence interval for the true population proportion, using the actual variance of p^
so far i got : p^= 190/250=0.76 and q^=1-p^=0.24
so (p^-p)/ Sp^~N(0,1)
so by using p^-Zc square root (p^q^/n) i got .707<μ<.813
so for part b how do we find the actual variance of p^
everyone please help