The distribution of sample proportions is reasonably normally distributed if the number of success and the number of failures are at least 10. If p is the population proportion and n is the sample size, then we want np 10 (the number of success) and n (1p) 10 (the number of failures). Similarly, the distribution of difference in the sample proportions is reasonably normally distributed if the number of success and the number of failures are at least 10 for each population. In other words, if p1 and p2 are the population proportions and n1 and n2 are the sample sizes, then we want n1 p1 10, n2 p2 10 (the number of success) and n1 (1p1) 10, n2 (1p2) 10 (the number of failures).

First question : Why is this condition necessary? In other words, what goes wrong if we use the the sampling distribution is normal when the number of success or the number of failures is not at least 10?

Second question : Go to the following applet http://www.rossmanchance.com/applets/ConfSim.html, under method, choose proportions, binomial and Wald. Let us choose the sample size n= 15, and a population proportion p= 0.1 (in the applet you will see instead of p so choose = 0.1). Choose intervals to be 500 and the confidence interval to be 95%. Hit "Sample". What is the percentage of the intervals containing 0.1? (Include a screenshot.)