SOLUTION: in a random sample of 900 observations from a binomial population produced x = 378 successes. Find the probability of success? determine the 80% confidence interval for estimatin

Click here to see ALL problems on Probability-and-statistics

Question 844290: in a random sample of 900 observations from a binomial population produced x = 378 successes.
Find the probability of success?
determine the 80% confidence interval for estimating p.
How large of a sample is required to estimate p correct to within .025 with probability equal to .95.
Thanks in advance for all your help.
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
in a random sample of 900 observations from a binomial population produced
x = 378 successes.
Find the probability of success?
Ans: The sample proportion = 378/900 = 0.42
------------------------------------------------------------
determine the 80% confidence interval for estimating p.
ME = 1.2816*sqrt[0.42*0.58/900] = 0.0211
------
80%CI:: 0.42-0.0211 < p < 0.42+0.0211
================================================
How large of a sample is required to estimate p correct to within .025 with probability equal to .95.
Since E = z*sqrt(pq/n),
----
n = {z/E]^2*pq
-----
n = [1.96/0.025]^2*0.95*0.05 = 292 when rounded up
===================
Cheers,
Stan H.
==========================