I need help with tri-inequality.

Question

Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 388 with 78% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

Answer 1

z=1.645 for 90%

mean= ,78,n=388,sd=sqr(pq=sqr(.78(.22))=about .41

MOE= z(sd)/sqrn= 1.645(.41)/sqr388=about .0346

.745<p<.815

alternative calculation

mean of binomial distribution= np = 388(.78)= 302.64

variance = npq = mean times(1-.78)= 302.64(.22)= 66.58

standard deviation = sqrVar= sqr66.58 = about 8.16

8.16/388 = about .021

MOE = zsd/sqrn = 1.645(.021)/sqr388 = about .035

CI = .78+/-.035= (.745,.815)

.745 < p < .815