Assume that a sample is used to estimate a population proportion p. find the 99.5% confidence interval for a sample of size 385 with 70.9% successes. enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

Step-by-step explanation:

The confidence interval for population proportion (p) is given by :-

Given : Significance level :

Critical value :

Sample size : n= 385

Sample proportion :

Then , the 99.5% confidence interval for population proportion is given by :-

Hence, the 99.5% confidence interval for population proportion :

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18