# Z proportions

1. Sep 6, 2011

### Maybe_Memorie

1. The problem statement, all variables and given/known data

--------------------IG------IB-----
Non-Germinated ---405-----341
Germinated --------386-----180

Carry out the Z-test to compare the proportions of seeds germinating for the IG and IB columns and interpret the result.

2. Relevant equations

3. The attempt at a solution

I don't know what I'm doing wrong but I'm assuming it's something because my answer is too big for the Z-table

Z = (p1-p2)/root([(p1(1-p1)/n1)+[(p2(1-p2)/n2)

P1 = 386/791
P2 = 341/521

Then throw everything into the Z formula. What's wrong with this?

2. Sep 6, 2011

3. Sep 6, 2011

### Ray Vickson

Well, for one thing, you should use p1 = 386/791 and p2 = 180/521, because you want to compare similar proportions for columns 1B and 1G.

For z-values beyond the table you can use a simple asymptotic approximation: for Z a standard normal random variable we have Pr{Z > z} ~ exp(-z^2/2)/[z*sqrt(2*Pi)] for large z. For example, for z = 3 the exact value (to two decimal places) is Pr(Z > 3} = 1.35e-3 while the approximation is 1.48e-2, while for z = 4 the exact value is 3.17e-5 and the approximation is 3.35e-5. There are improved simple approximations available. For example, the approximation P{Z . z} ~ (1/z - 1/z^3)*exp(-z^2/2)/sqrt(2*Pi) gives much better accuracy.

RGV

4. Sep 6, 2011

### Maybe_Memorie

Ah, I see the mistake.

Our professor says we answers shouldn't go beyond the table, so is my formula incorrect, should I use the formula from here instead?
http://stattrek.com/ap-statistics-4/test-difference-proportion.aspx