Posted by Sheila on Friday, April 20, 2012 at 7:20pm.
Try a proportional one-sample z-test for this one since this problem is using proportions.
Here's a few hints to get you started:
Null hypothesis:
Ho: p ≥ .90 -->meaning: population proportion is greater than or equal to .90
Alternate hypothesis:
Ha: p < .90 -->meaning: population proportion is less than .90
Using a formula for a proportional one-sample z-test with your data included:
z = .88 - .90 -->test value (352/400 = .88) minus population value (.90) divided by
√[(.90)(.10)/400] --> .10 represents 1-.90 and 400 is sample size.
Finish the calculation. Remember if the null is not rejected, then there is no difference. If you need to find the p-value for the test statistic, check a z-table. The p-value is the actual level of the test statistic. Otherwise, check a z-table for a one-tailed test at .05 level of significance. Compare to your test statistic. If the test statistic exceeds the value from the table, reject the null and conclude p < .90. If the test statistic does not exceed the value from the table, do not reject the null.
I hope this will help.
Thanks MathGuru! Makes a ton more sense now!
MathGuru,
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statistic. Otherwise, check a z-table for a one-tailed test at .05 level of significance. Compare to your test statistic. If the test statistic exceeds the value from the table, reject the null and conclude p < .90. If the test statistic does not exceed the value from the table, do not reject the null.
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I think you got the opposite.
Decision rule should be--> Accept Ho if test statistic >= -1.645