How do I state the distribution to use for this test?

Question

"Untitled," by Stephen Chen

"I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1%.

So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right?"

Conduct a hypothesis test at the 5% level.

Note: If we are using a Student's t-distribution for the problem, we may assume that the underlying population is normally distributed. (In general, we must first prove that assumption, though.)"

1. I need help stating the distribution to use for the test (I have to round it to four decimal places)

P' ~ N (__?__, __?__)

Answer 1

You should use the binomial distribution. If you want to approximate it, you should use the Poisson, not the Normal. That is because np = 50*.01=.5 << 5

if p =.01, P(x = 0) = .99^50 = .605006

P(x = 1) = C(50,1).99^49*.01 = .305559

Thus,P(x <= 1) = .910565

Then, the p-value at 2 is 1- .910565 = .089435

As p > α =.05, you are unable to prove that p has increased.

Incidentally, with the Poisson, P(X <= 1) with λ =.5 is e^-.5(1 + .5) = .909796; this is fairly close to .910565, the exact probability.