Chi-Squared Test of Independence for Tree Age and Wood Species

Math Problem Statement

A random sample of 104 logs felled at a specialty timber plantation has been selected to study diversity of undergrowth and timber quality. Some of the variables recorded for each log are listed below.

Variable Description subid Subject ID timber wood species numSpecies number of different species of plant observed in the undergrowth within 30m of the tree usable the percentage of the log producing usable timber ageTreeCat the age of tree

Part 1

Research question: we are interested in studying the association between the age of the felled tree and wood species in the population under consideration. Please complete the following tables, calculating the values of A and B.

Observed count | 1 (less than 15) | 2 (16-21) | 3 (22-26) | 4 (more than 26) | Sum blackwood | 13 | 11 | 14 | 16 | 54 sassafras | 14 | 10 | A=*** | 7 | 50 Sum | 27 | 21 | 33 | 23 | 104

Expected count | 1 (less than 15) | 2 (16-21) | 3 (22-26) | 4 (more than 26) blackwood | B=*** | 10.904 | 17.135 | 11.942
sassafras | 12.981 | 10.096 | 15.865 | 11.058

chi-squared | 1 (less than 15) | 2 (16-21) | 3 (22-26) | 4 (more than 26) blackwood | 0.074 | 0.001 | 0.573 | 1.379
sassafras | 0.08 | 0.001 | 0.619 | 1.489
A [integer]: Answer 1 Question 1 B [use at least 3dp]: Answer 2 Question 1

A chi-squared test of independence is run to answer the research question. Consider the output of the test below, and answer the questions:

Pearson's Chi-squared test

data: table(var1, var4) X-squared = 4.2164, df = 3, p-value = 0.239 What is the null hypothesis? Choose one among the following options: a. the age of tree felled at a specialty timber plantation is not distributed according to the given table of probabilities b. the age of tree is associated with wood species c. the age of tree and wood species are independent d. the age of tree felled at a specialty timber plantation is distributed according to the given table of probabilities Answer 3 Question 1

From the p-value as reported above, we can conclude that the decision of the test is: Answer 4 Question 1

What is the conclusion of the test? a. There is evidence that the age of tree and wood species are associated b. There is evidence that the age of tree felled at a specialty timber plantation is not distributed according to the given table of probabilities c. There is not enough evidence that the age of tree felled at a specialty timber plantation is distributed differently than the given table of probabilities d. There is evidence that the age of tree felled at a specialty timber plantation is not distributed according to the given table of probabilities e. There is evidence that the age of tree and wood species are independent f. There is not evidence that the age of tree and wood species are associated g. There is evidence that the age of tree felled at a specialty timber plantation is distributed according to the given table of probabilities Answer 5 Question 1

Solution

To solve these questions, let's go through each part in sequence.

Part 1: Calculating Values of $A$ and $B$

Calculate $A$ :
The sum for the column "3 (22-26)" is given as 33. For sassafras in this category, the formula for finding $A$ is: $A = \text{Sum of Column 3} - \text{Observed count for blackwood in Column 3}$ Given: $33 = 14 + A \Rightarrow A = 33 - 14 = 19$ Answer 1 (A): $A = 19$
Calculate $B$ :
For expected counts, use the formula: $\text{Expected Count} = \frac{(\text{Row Total} \times \text{Column Total})}{\text{Grand Total}}$ For blackwood in column "1 (less than 15)": $B = \frac{(54 \times 27)}{104} = \frac{1458}{104} \approx 14.019$ Answer 2 (B): $B = 14.019$ (rounded to at least 3 decimal places)

Part 2: Hypothesis Testing

Null Hypothesis Selection:
A chi-squared test of independence tests whether two categorical variables are independent. The null hypothesis in this context is: $\text{c. the age of tree and wood species are independent}$ Answer 3: $c$
Decision Based on p-value:
The p-value of 0.239 is greater than the typical significance level of 0.05, so we fail to reject the null hypothesis. Answer 4: Fail to reject the null hypothesis.
Conclusion of the Test:
Since we failed to reject the null hypothesis, we conclude that there is not enough evidence to suggest an association between the age of the tree and wood species. Answer 5: $f.$ There is not evidence that the age of tree and wood species are associated.

Would you like further clarification on any steps? Here are five additional questions that expand on these concepts:

How is the expected count calculated in chi-squared tests?
What would change in our calculations if the total number of logs was different?
What are the assumptions required for a chi-squared test of independence?
How does changing the significance level affect the conclusions of a hypothesis test?
What is the impact of sample size on chi-squared test results?

Tip: For hypothesis tests, always check the p-value relative to your significance level (often 0.05) to make your decision.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Statistics
Chi-squared Test of Independence
Categorical Data Analysis

Formulas

Expected Count = (Row Total × Column Total) / Grand Total
Chi-squared Value = Σ((Observed - Expected)^2 / Expected)

Theorems

Chi-squared Test of Independence

Suitable Grade Level

Undergraduate

Related Recommendation

Chi-Squared Test: Age of Felled Tree and Wood Species Association

Calculate Observed and Expected Counts Using Chi-Square Test

Chi-Squared Test for Tree Age Distribution at a Timber Plantation

Calculate Missing Values in Tree Species Contingency Table Using Chi-Squared Test

Chi-squared Test for Tree Age Distribution at Specialty Timber Plantation