A national health survey weighed a sample of 200 boys aged 6-11 and found that 29 of them were overweight. They weighed a sample of 220 girls aged 6-11 and found that 34 of them were overweight. Can you conclude that the proportion of overweight kids aged 6-11 among boys is different from the proportion of overweight kids aged 6-11 among girls? Use level of significance 1%.

Procedure: Select an answer Two means T (non-pooled) Hypothesis Test Two means T (pooled) Hypothesis Test Two paired means Z Hypothesis Test Two proportions Z Hypothesis Test Two means Z Hypothesis Test Two variances F Hypothesis Test

Assumptions: (select everything that applies)

Step 1. Hypotheses Set-Up:

$\displaystyle {H}_{{0}}:$ Select an answer p₁-p₂ μ₁-μ₂ σ₁²/σ₂² μ =	, where Select an answer μ's are p's are μ is σ's are the Select an answer difference between population means population means population standard deviations population proportions and the units are Select an answer pieces 100% people counts
$\displaystyle {H}_{{a}}:$ Select an answer p₁-p₂ μ₁-μ₂ μ σ₁²/σ₂² ? < ≠ >	, and the test is Select an answer Right-Tail Two-Tail Left-Tail

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

Step 3. Compute the value of the test statistic: Select an answer z₀ χ²₀ f₀ t₀ = (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 1% significance level we Select an answer DO NOT DO have sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.