In 2001, 10% of all U.S. families had incomes below the poverty level, as reported by the U.S. Census Bureau in American Community Survey. During that same year, of 360 randomly selected Georgia families, 42 had incomes below the poverty level. At the 10% significance level, do the data provide sufficient evidence to conclude that, in 2001, the percentage of families with incomes below the poverty level among those living in Georgia was greater than than among all U.S. families?

Procedure: Select an answer One mean Z Hypothesis Test One variance χ² Hypothesis Test One proportion Z Hypothesis Test One mean T Hypothesis Test

Assumptions: (select everything that applies)

Step 1. Hypotheses Set-Up:

$\displaystyle {H}_{{0}}:$  Select an answer σ² p μ =	, where Select an answer σ² μ p is the Select an answer population mean population variance population proportion and the units are Select an answer subjects 100%
$\displaystyle {H}_{{a}}:$ Select an answer p μ σ² ? > < ≠	, and the test is Select an answer Two-Tail Right-Tail Left-Tail

Step 2. The significance level $\displaystyle \alpha=$ %

Step 3. Compute the value of the test statistic: Select an answer χ²₀ z₀ t₀ f₀ = (Round the answer to 3 decimal places)

Step 4. Testing Procedure: (Round the answers to 3 decimal places)

CVA	PVA
Provide the critical value(s) for the Rejection Region:	Compute the P-value of the test statistic:
left CV is and right CV is	P-value is

Step 5. Decision:

CVA	PVA
Is the test statistic in the rejection region?	Is the P-value less than the significance level?
? yes no	? yes no

Conclusion: Select an answer Do not reject the null hypothesis in favor of the alternative. Reject the null hypothesis in favor of the alternative.

Step 6. Interpretation:

At 10% significance level we Select an answer DO NOT DO have sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.