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April 1, 2015

April 1, 2015

Posted by **Torri** on Tuesday, April 23, 2013 at 5:23pm.

- Statistics -
**MathGuru**, Tuesday, April 23, 2013 at 7:20pmTry a z-test for proportions (two samples).

Here is one such formula:

z = (p1 - p2)/(√pq)[√(1/n1 + 1/n2)]

p1 = 41/100 = .41

p2 = 37/100 = .37

p = (x1 + x2)/(n1 + n2) = (41 + 37)/(100 + 100) = 78/200 = .39

q = 1 - p = 1 - .39 = .61

Substitute into the formula:

z = (.41 - .37)/[√(.39)(.61)][√(1/100 + 1/100) = .58 (rounded)

Testing at the 1% significance level for a one-tailed test, you will fail to reject the null. You cannot conclude a difference.

Double check these calculations.

I hope this helps.

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