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March 5, 2015

March 5, 2015

Posted by **Anonymous** on Sunday, October 3, 2010 at 7:07pm.

- statistics -
**MathGuru**, Monday, October 4, 2010 at 6:52pmTry a binomial proportion 2-sample z-test using proportions.

Hypotheses:

Ho: pA = pB

Ha: pA does not equal pB -->this is a two-tailed test (the alternate hypothesis does not show a specific direction).

The formula is:

z = (pA - pB)/√[pq(1/n1 + 1/n2)]

...where n represents the sample sizes, p is (x1 + x2)/(n1 + n2), and q is 1-p.

I'll get you started:

p = (73 + 56)/(5000 + 4500) = ? -->once you have the fraction, convert to a decimal (decimals are easier to use in the formula).

q = 1 - p

pA = 73/5000

pB = 56/4500

Convert all fractions to decimals. Plug those decimal values into the formula and find z. Compare z to the cutoff 0.05 for a two-tailed test (cutoff value is z = + or - 1.96). If the test statistic you calculated exceeds either the positive or negative cutoff z-value, reject the null and conclude a difference. If the test statistic does not exceed either the positive or negative cutoff z-value, do not reject the null (you cannot conclude a difference).

I hope this will help get you started.

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