Posted by Tyler on Sunday, November 18, 2012 at 10:40pm.
You can try a proportional one-sample z-test for this one since this problem is using proportions.
Null hypothesis:
p ≤ .50
Alternate hypothesis:
p > .50
Using a formula for a proportional one-sample z-test with your data included, we have:
z = .52 - .50 -->test value minus population value
divided by
√[(.50)(.50)/1132]
Finish the calculation. Determine the p-value using the z-score you calculate. (The p-value is the actual level of the test statistic.) Compare to the level of significance for a one-tailed test, which is 0.05. Determine whether or not to reject the null and conclude a difference (p > .50). For part c, redo the same process using .53 instead of .52.
(Hint: One null hypothesis will be rejected; one will not.)
I hope this will help get you started.
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