A random sample of 101 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.

**VariableDescription**

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

Address the following questions, assuming as necessary that all test assumptions are verified.
**Part 1**
A test is run to answer the research question, ’Is there any difference between blackwood logs and sassafras logs in the average number of nearby undergrowth species?’

Which test is correct to use, if we know the population standard deviation?

Answer 1 Question 3 z-testt-test

What is the null hypothesis?

Answer 2 Question 3 There is no differenceThere is a difference

The p-value of the test is 0.105. What is the decision of the test?

Answer 3 Question 3 Accept H0Reject H0Do not accept H0Do not reject H0

**Part 2**
A test is run to answer the research question, ’Is the average number of nearby undergrowth species equal to 38?’

Which test is correct to use, if we do not know the population standard deviation?

Answer 4 Question 3 z-testt-test

The test statistic is -2.18. This means that the difference between the sample mean and the possible population mean, as stated in the null hypothesis (i.e. 38), is:

Answer 5 Question 3 -2.18 times the standard deviation of the number of nearby undergrowth species-1.96 times the standard error of the sample mean-1.96 times the standard deviation of the number of nearby undergrowth species-2.18 times the standard error of the sample mean

The p-value of the test is 0.032. What is the decision of the test?

Answer 6 Question 3 Accept H0Reject H0Do not accept H0Do not reject H0

Let's address each part in turn.

### **Part 1**
The research question is: "Is there any difference between blackwood logs and sassafras logs in the average number of nearby undergrowth species?" The correct steps for this hypothesis testing follow below:

1. **Which test is correct to use, if we know the population standard deviation?**
   - **Answer 1:** The **z-test** is the correct test when the population standard deviation is known, as it assumes a normal distribution and requires the population standard deviation for its calculation.

2. **What is the null hypothesis?**
   - **Answer 2:** The null hypothesis \( H_0 \) is that **there is no difference** in the average number of nearby undergrowth species between blackwood and sassafras logs. This can be formally written as:
     \[
     H_0: \mu_{blackwood} = \mu_{sassafras}
     \]
     The alternative hypothesis \( H_A \) would be that there is a difference:
     \[
     H_A: \mu_{blackwood} \neq \mu_{sassafras}
     \]

3. **The p-value of the test is 0.105. What is the decision of the test?**
   - **Answer 3:** Since the **p-value is 0.105**, which is **greater than the typical significance level of 0.05**, we **do not reject \( H_0 \)**. This means there is insufficient evidence to conclude a difference between the two groups in terms of the average number of undergrowth species.

---

### **Part 2**
The research question is: "Is the average number of nearby undergrowth species equal to 38?"

4. **Which test is correct to use, if we do not know the population standard deviation?**
   - **Answer 4:** The **t-test** is the correct test when the population standard deviation is not known, as the t-test is used for small sample sizes or when the population standard deviation is unknown.

5. **The test statistic is -2.18. What does this mean?**
   - **Answer 5:** The test statistic of -2.18 means that the sample mean is **-2.18 times the standard error of the sample mean** away from the hypothesized population mean (38). Therefore, the correct interpretation is:
     \[
     \text{Answer: } -2.18 \text{ times the standard error of the sample mean.}
     \]

6. **The p-value of the test is 0.032. What is the decision of the test?**
   - **Answer 6:** Since the **p-value is 0.032**, which is **less than the typical significance level of 0.05**, we **reject \( H_0 \)**. This means there is sufficient evidence to conclude that the average number of nearby undergrowth species is not equal to 38.

---

### Additional Questions and Tips:
1. What is the difference between a z-test and a t-test in terms of assumptions?
2. How does the sample size affect the choice between z-test and t-test?
3. What would be the decision if the p-value was 0.15 in Part 2?
4. How does the standard error influence the test statistic in a t-test?
5. Can a t-test be used if the sample size is large but the population standard deviation is unknown?

**Tip:** Always check the assumptions of normality when conducting hypothesis tests like z-tests or t-tests, especially with small sample sizes.

A random sample of 101 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. Is there any difference between blackwood logs and sassafras logs in the average number of nearby undergrowth species? Is the average number of nearby undergrowth species equal to 38?

This page addresses hypothesis testing to determine if there's a significant difference between the average number of nearby undergrowth species for blackwood and sassafras logs, using z-test and t-test techniques. Learn about p-values, test statistics, and how to make decisions based on statistical data. Suitable for undergraduate students studying statistics.

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