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March 29, 2015

March 29, 2015

Posted by **Anonymous** on Tuesday, March 26, 2013 at 2:37pm.

Ho: p

Ha: p

z =

- statistics -
**MathGuru**, Tuesday, March 26, 2013 at 6:21pmTry a proportional one-sample z-test for this one since this problem is using proportions.

Null hypothesis:

p ≤ .33

Alternate hypothesis:

p > .33

Using a formula for a proportional one-sample z-test with your data included, we have:

z = .37 - .33 -->test value minus population value

divided by

√[(.33)(.67)/782]-->.67 = 1 - .33; 782 = sample size

Finish the calculation. Use a z-table to determine the critical value for a one-tailed test at .05 level of significance. Determine whether or not to reject the null and conclude a difference (p > .33).

I hope this will help get you started.

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