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.

[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

^ρ≈N(ρ;(ρ(1-ρ))/n)

And the standarization is

Z= ^ρ-ρ  ≈N(0;1)

√(ρ(1-ρ)/n)

The formula to estimate the population proportion with a Confidence Interval is

[^ρ ± *√(^ρ(1-^ρ)/n)]

The sample proportion is calculated with the following formula:

^ρ= x/n = 358/400 = 0.895 ≅ 0.90

And the Z-value is ≅ 1.65

[0.90 ± 1.65 * √((0.90*0.10)/400)]

[0.875;0.925]

I hope you have a SUPER day!

overpriced

step-by-step explanation:

what level of math? also i don't need to be paid if i can you.