Find the necessary confidence interval for the binomial proportion p. (Round your answers to three decimal places.) A 90% confidence interval for p, based on a random sample of n = 400 observations from a binomial population with x = 358 successes.

Accepted Solution

Answer:[0.875;0.925]Step-by-step explanation:Hello!You have a random sample of n= 400 from a binomial population with x= 358 success.Your variable is distributed X~Bi(n;ρ)Since the sample is large enough you can apply the Central Limit Teorem and approximate the distribution of the sample proportion to normal^Οβ‰ˆN(ρ;(ρ(1-ρ))/n)And the standarization is Z= ^ρ-ρ Β β‰ˆN(0;1) √(ρ(1-ρ)/n) The formula to estimate the population proportion with a Confidence Interval is[^ρ Β± [tex]Z_{1-\alpha/2}[/tex]*√(^ρ(1-^ρ)/n)]The sample proportion is calculated with the following formula:^ρ= x/n = 358/400 = 0.895 β‰… 0.90And the Z-value is [tex]Z_{1-\alpha/2} = Z_{0.95} = 1.648[/tex] β‰… 1.65[0.90 Β± 1.65 * √((0.90*0.10)/400)][0.875;0.925]I hope you have a SUPER day!