Posted by **Meep** on Wednesday, October 31, 2012 at 11:21am.

A local McDonald's manager will return a shipment of hamburger buns if more than 10% of the buns are crushed. A random sample of 81 buns finds 13 crushed buns. A 5% significance test is conducted to determine if the shipment should be accepted. The p value for this situation is:

0.0348

0.0500

0.0700

0.0436

0.0218

- AP Statistics -
**MathGuru**, Wednesday, October 31, 2012 at 8:07pm
You can try a proportional one-sample z-test for this one since this problem is using proportions.

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

z = .16 - .10 -->test value (13/81 is .16) minus population value (.10) divided by

√[(.10)(.90)/81] --> .90 represents 1-.10 and 81 is sample size.

Finish the calculation. Use a z-table to determine the p-value for the z-test statistic.

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