Question 368:

Asked on September 5, 2008

Tags: Binomial Probability , Normal Approximation , 2-proportion test

1

Answer:

No answer provided yet.The samples are large enough that we can use the Normal Approximation to the Binomial to answer the difference between two proportions. You want to know if the observed difference in proportions 120/200 - 240/500  = .12 is greater than chance.

You divide this difference by the square-root of a denominator which accounts for the chance which will provide you with a z-score. The denominator is the square root of:

1         1
--   +  --     * PQ
n₁       n₂

Where P = (x1 + x2)/(n1+n2)  Q is 1-P. The x's are just the number of tea-drinkers and the n's are the sample size.

P = (120+240)/(200+500) = .514
Q = 1-.514 = .486
PQ = .486*.513 = .2498

1/n1 + 1/n2 = 1/200 + 1/500 = .007

So multiply .2498 * .007= 0.00175

Now the square root of this is SQRT(0.00175) = .04183

So the equation is the observed difference .12/.04183 = 2.87

That result is the z-score which is your test-statistic. You now look this value up using the z-score to percentile calculator using the 2-sided area. You should get 0.413% or a p-value of .00413, meaning there is about a  .04% chance the difference is due to chance. 

See also question 314