Suppose that the efficacy of a certain drug is 0.59. Consider the sampling distribution (sample size n = 182) for the proportion of patients cured by this drug

Question

a. What is the mean of this distribution? 

b. What is the standard error of this distribution? Round the answer to three decimal places.   

Answer 1

Hi Bailey,

This is a binomial distribution, which differs from the normal distributions we worked with previously. The way to identify the binomial distribution is most fundamentally, do we have two binary outcomes of success and failure? It also requires a fixed sample size and statistical independence, but I won’t overcomplicate. Here, we do indeed have a binary outcome.

Once we identify the binomial distribution, the formulas to compute mean and standard deviation of a sample are relatively simple.

mu-p=p

sigma-p=sqrt[(p(1-p)/n]

Let’s break this down:

Mu-p=mean of sampling distribution

sigma-p=standard deviation of sampling distribution

p=population probability of success

1-p=population probability of failure

n=sample size

So, for our problem:

a) mu=p=0.59

b) sigma-p=[sqrt(p(1-p))/n]

p=0.59

1-p=0.41

n=182

Substituting:

sigma-p=[sqrt(0.59)(0.41)/182]

sigma-p=0.036

I hope this helps.