BookmarkSubscribeSubscribe to RSS Feed

Re: What formula does JMP use to calculate single-proportion sample size?

Can someone tell me how JMP calculates single-proportion sample size? That is, what is the underlying formula?

Most importantly why is it different from the "usual" published formula?

for example that used on the online calculator:

and detailed here:

For example (JMP8)
with p0 = 0.1, p1 = 0.4, alpha = 0.05, 1-sided test, power = 0.8

JMP8 gives sample size = 7 (actually 6.183 - rounded up to whole number 7)

"usual" formula gives sample size = 10 (actually 9.116 - rounded up to whole number 10)


I've changed the example so it was relevant to JMP8 (which only does 1-sided one-proportion sample size tests - or is there another way to do 2-sided?)

Message was edited by: reedme
I am very interested in seeing a response from SAS to this poster's question. I also have seen inconsistencies in the several calculations for proportions data, especially when dealing with small p (<1%). I think JMP may be using the normal approximation, which may be okay for proportions you would see in marketing studies, but makes the software relatively useless for Six Sigma. At this point, given these inconsistencies I am seriously considering switching my group over to Minitab.

Super User


Jun 23, 2011

JMP 8.0.1 does an Exact calculation. The 2 references you give use a normal approximation calculation. (The JMP docs say JMP is using a normal approximation but they are being updated.)

When I use 8.0.1 to calculate the sample size for the example you give, I get sample size = 13 (not 7). This agrees with the Exact calculation performed at this excellent power and sample size page: