Answer Happy

Consider the case where X is the binomial random variable with parameters n = the number of trials/sample size and p = the probability of success. Then for a single sample of size n (or, equivalently, a single run of n trials), x would represent the # of “successes" for that sample/run. So, for that single sample/run of n trials, we can approximate the # of "successes" probability of success as the sample proportion p =. For all such samples/runs of n trials, we can then define the variable Ể = to be the collection of all possible sample proportions for samples of size n/runs of n trials. 72 - For large values of n, the binomial random variable, X, is approximately normally distributed with mean Mx = np and standard deviation Ox = Vnp(1 - p). Assuming that n is sufficiently large for the normal approximation to apply, do the following.

- 1. In most cases, the margin of error for a sample statistic/point estimate, (which approximates the population parameter, e) is the critical value from the distribution of times the standard a error, 03. Using this format, (1) define the 2-tailed margin of error for P as an estimate of p and (2) write out the formula for the 2-tailed (1 - a) x 100% confidence interval for the population proportion p centered at P. Please be sure to show all of your work and to clearly explain your logic throughout. If you reference results from the GW 9 discussion, please be sure to clearly identify the result and the question it is from. It has been shown that replacing p with p in both the margin of error and the endpoints of the confidence interval (which contain the parameter we are estimating, p), results in a confidence interval that will capture the value of the true population parameter, p, approximately (1 - a) x 100% of the time. In your final step for the margin of error and the final step of finding the endpoints for the confidence interval, replace p with p.