Let's go through each part of the question systematically.

Step 1: Calculate A and B

For A:

We know that the row sum for Blackwood is 46, and the sum for the first column is 28. To find $A$, we subtract the counts for the known entries in Blackwood's row:

$A + 10 + 11 + 14 = 46$ $A = 46 - (10 + 11 + 14) = 46 - 35 = 11$

Thus, A = 11.

For B:

The expected count $B$ in Sassafras for the second age group (16-21) can be calculated from the total and row sums.

The expected count formula is: $\text{Expected Count} = \frac{(\text{Row Total} \times \text{Column Total})}{\text{Overall Total}}$

For $B$: $B = \frac{(69 \times 32)}{115} = \frac{2208}{115} \approx 19.200$

Thus, B = 19.200.

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Step 2: Null Hypothesis

The correct answer to the question on the null hypothesis would be: c. the age of tree and wood species are independent

This is because in a Chi-squared test of independence, we are testing whether there is an association between two categorical variables (in this case, age of the felled tree and wood species).

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Step 3: Conclusion Based on the p-value

The given p-value is 0.5909.

In hypothesis testing:

• If the p-value is greater than the significance level (commonly 0.05), we do not reject the null hypothesis.

Since 0.5909 > 0.05, we conclude: Answer: Do not reject H0

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Step 4: Interpretation of the Test Conclusion

With a high p-value and having not rejected the null hypothesis, the conclusion is that there is insufficient evidence to suggest an association between the age of tree and wood species.

The correct conclusion would therefore be: c. There is not evidence that the age of tree and wood species are associated

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Summary of Answers

1. Answer 6 Question 1: A = 11
2. Answer 7 Question 1: B = 19.200
3. Answer 8 Question 1: c. the age of tree and wood species are independent
4. Answer 9 Question 1: Do not reject H0
5. Answer 10 Question 1: c. There is not evidence that the age of tree and wood species are associated

Would you like further clarification on any of these steps?

Related Questions

1. What conditions must be met for the Chi-squared test of independence to be valid?
2. How is the Chi-squared test statistic calculated from observed and expected counts?
3. Why is a p-value of 0.5909 considered insufficient evidence to reject the null hypothesis?
4. How would the results change if the p-value had been below 0.05?
5. What are some limitations of the Chi-squared test when applied to small sample sizes?

Tip

When calculating expected counts, remember to use the formula $\frac{(\text{Row Total} \times \text{Column Total})}{\text{Overall Total}}$ accurately to avoid errors in the test's validity.