A sample of birth weights of 34 girls was taken. Below are the results (in g):

3636.7	2713.6	3383.6	4315.3	3928.6
2338.8	3183.5	2982.3	3621.9	3022.7
2947.3	4022.8	3413.3	3728.8	3703.6
3808.7	3550	3938.5	3906.9	3374.3
2656.5	3735.8	4244.7	2568.6	3362.1
3552	3484.2	3150.5	3384.7	3508.4
3816.5	3469.9	2333.6	4426.7

(Note: The average and the standard deviation of the data are respectively 3447.5 g and 527.52 g.)

Use a 5% significance level to test the claim that the standard deviation of birthweights of girls is different from the standard deviation of birthweights of boys, which is 540 g.

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

Assumptions: (select everything that applies)

Step 1. Hypotheses Set-Up:

$\displaystyle {H}_{{0}}:$ Select an answer σ² μ p =	, where ? σ μ p is the Select an answer population standard deviation population mean population proportion and the units are ? lbs kg g mg
$\displaystyle {H}_{{a}}:$ Select an answer p μ σ² ? < ≠ >	, and the test is Select an answer Right-Tailed Two-Tailed Left-Tailed

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

Step 3. Compute the value of the test statistic: Select an answer t₀ χ₀² z₀ = (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 Reject the null hypothesis in favor of the alternative. Do not reject the null hypothesis in favor of the alternative.

Step 6. Interpretation:

At 5% significance level we Select an answer DO DO NOT have sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.