0
votes

The seeds of the garden pea are either yellow or green. A certain cross between pea plants produces progeny where 75% are plants with yellow seeds and 25% are plants with green seeds. What is the minimum number of progeny you would need to grow to have probability no less than 0.99 of obtaining at least 10 plants with green seeds?

I understand how to estimate a required sample size when I have data such as standard deviation, mean, correlation, etc., but I don't even know where to start to estimate it based on the percentage values with a certain probability.

So far I set up this code in SAS:

Proc power; 
onesamplefreq  test=Z method=normal
sides=1
alpha=.01
nullproportion=.5
proportion=.25
power=.99
ntotal= .;
run;

Running this program resulted in a sample size of 76, but I don't feel like this is correct. I don't know how to specify that I need at least 10 plants with green seeds, and I don't know how to set the nullproportion or if it matters.

1

1 Answers

0
votes

It is a Binomial distribution kind problem. Where chance of winning (green plant) is 25%. You want to win at least 10 times, so how many times you need to play (that is, how many seeds you need)?
Mean of binomial distribution will answer this question which is:

np = 10  
n*0.25 = 10  
n = 40  

So required seed is 40. This is purely probabilistic. But we need to consider Type I and Type II error. So sample size 76 seems reasonable to me.