Posted by **kay** on Monday, September 2, 2013 at 1:49am.

It has been reported that 40% of children in the United States live in homes where both parents work to make ends meet. A school administrator claimed that more than 40% of children in his school district lived in homes where both parents work to make ends meet. In a sample of 300 children, 138 of them lived in homes where both parents work to make ends meet.

1) State the correct null and alternative hypotheses that the school administrator wishes to test.

2) Calculate the test statistic that would be used for conducting the desired test.

3) Find the p-value of this test.

4) Assuming á = 0.01, what is your conclusion?

- university of phoenix -
**MathGuru**, Tuesday, September 3, 2013 at 7:14pm
You can 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 = .40 -->meaning: population proportion is equal to .40

Alternative hypothesis:

Ha: p > .40 -->meaning: population proportion is greater than .40 (this is a one-tailed test).

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

z = .46 - .40 -->test value (138/300 is .46) minus population value (.40) divided by

√[(.40)(.60)/300] --> .60 represents 1-.40 and 300 is sample size.

Finish the calculation. 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. Compare that value to 0.01 to determine whether or not to reject the null. If you reject the null, you can conclude that p > .40.

I hope this will help.

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