Comparing Seed Germination Proportions: Z-Test

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The discussion focuses on performing a Z-test to compare seed germination proportions between two groups, IG and IB. The initial calculations were incorrect due to using the wrong proportions for comparison, with the correct proportions being p1 = 386/791 and p2 = 180/521. Participants suggest using asymptotic approximations for Z-values that exceed standard Z-tables, providing formulas for better accuracy. There is also a mention of potential alternative resources for performing the Z-test correctly. The conversation emphasizes the importance of accurate proportion selection and understanding the limitations of Z-tables.
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Homework Statement



--------------------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.

Homework Equations





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?
 
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Any advice?
 
Maybe_Memorie said:

Homework Statement



--------------------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.

Homework Equations





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?

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
 
Ray Vickson said:
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

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
 
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