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September 21, 2014

September 21, 2014

Posted by **Michelle** on Thursday, September 11, 2008 at 1:29am.

In a recent national survey, the mean weekly allowance for a nine-year-old child from his or her parents was reported to be $3.65. A random sample of 45 nine-year-olds in northwestern Ohio revealed the mean allowance to be $3.69 with a standard deviation of .24. At the .05 level of significance, is there a difference in the mean allowances nationally and the mean allowances in northwestern Ohio for nine-year-olds?

- Statistics -
**MathGuru**, Thursday, September 11, 2008 at 8:53amTry a one-sample z-test for your problem.

Using the z-test formula to find the test statistic:

z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)

z = (3.69 - 3.65)/(.24/√45)

Finish the calculation. Use a z-table to determine the critical or cutoff value to reject or fail to reject the null hypothesis at .05 level of significance. This will be a two-tailed test because the problem is just asking if there is a difference. Keep that in mind when you are looking at the value in the table.

Once you have the value from the table, compare to the test statistic. If the test statistic exceeds the value from the table, reject the null and conclude a difference. If the test statistic does not exceed the value from the table, fail to reject the null and you cannot conclude a difference.

I hope this will help get you started on your problem.

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