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October 25, 2014

October 25, 2014

Posted by **Sheila** on Friday, April 20, 2012 at 7:20pm.

- Statistics -
**MathGuru**, Friday, April 20, 2012 at 7:38pmTry 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.

- Statistics -
**Sheila**, Friday, April 20, 2012 at 8:33pmThanks MathGuru! Makes a ton more sense now!

- Statistics -
**Srinivas**, Monday, August 13, 2012 at 7:49pmMathGuru,

/****

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 think you got the opposite.

Decision rule should be--> Accept Ho if test statistic >= -1.645

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