A recent study investigated tractor skidding distances along a road in a forest. The skidding distances (in meters) were measured at 20 randomly selected road sites. The data are given in the accompanying table. A logger working on the road claims that the mean skidding distance is at least 425 meters. Is there sufficient evidence to refute this claim? Use a = 0.10. E Click the icon to view the table. State the hypotheses to test if there is sufficient evidence to refute the claim that the mean skidding distance is at least 425 meters. Choose the correct answer below. YA Ho H=425 Hu<425 OB. Ho H= 425 H u4 425 O D. H,: u= 425 H u> 425 OC. H;: H#425 H u= 425

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### Study on Tractor Skidding Distances

**Objective:**
A recent study investigated tractor skidding distances along a road in a forest. The skidding distances (in meters) were measured at 20 randomly selected road sites. The data are given in the accompanying table. A logger working on the road claims that the mean skidding distance is at least 425 meters. The question is whether there is sufficient evidence to refute this claim. The significance level used for this test is \(\alpha = 0.10\).

**Hypotheses:**

- Null Hypothesis (\(H_0\)): \(\mu = 425\)
- Alternative Hypothesis (\(H_a\)): \(\mu < 425\)

The hypotheses listed in multiple-choice format:
- \( \text{A. } H_0: \mu = 425, \quad H_a: \mu < 425 \) (correct answer)
- \( \text{B. } H_0: \mu = 425, \quad H_a: \mu \neq 425 \)
- \( \text{C. } H_0: \mu \neq 425, \quad H_a: \mu = 425 \)
- \( \text{D. } H_0: \mu = 425, \quad H_a: \mu > 425 \)

**Calculation of Test Statistic:**

To determine whether to refute the logger's claim, we need to calculate the test statistic \( t \), which is to be rounded to two decimal places as needed. 

The formulas used will depend on whether the population standard deviation is known or not, and if not, the sample standard deviation is used.

Click on the icon to view the table data if needed.

**Conclusion:**

Based on the calculation of the \( t \)-value and comparing it to the critical value for \(\alpha = 0.10\), we can make a decision about whether there is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.
Transcribed Image Text:### Study on Tractor Skidding Distances **Objective:** A recent study investigated tractor skidding distances along a road in a forest. The skidding distances (in meters) were measured at 20 randomly selected road sites. The data are given in the accompanying table. A logger working on the road claims that the mean skidding distance is at least 425 meters. The question is whether there is sufficient evidence to refute this claim. The significance level used for this test is \(\alpha = 0.10\). **Hypotheses:** - Null Hypothesis (\(H_0\)): \(\mu = 425\) - Alternative Hypothesis (\(H_a\)): \(\mu < 425\) The hypotheses listed in multiple-choice format: - \( \text{A. } H_0: \mu = 425, \quad H_a: \mu < 425 \) (correct answer) - \( \text{B. } H_0: \mu = 425, \quad H_a: \mu \neq 425 \) - \( \text{C. } H_0: \mu \neq 425, \quad H_a: \mu = 425 \) - \( \text{D. } H_0: \mu = 425, \quad H_a: \mu > 425 \) **Calculation of Test Statistic:** To determine whether to refute the logger's claim, we need to calculate the test statistic \( t \), which is to be rounded to two decimal places as needed. The formulas used will depend on whether the population standard deviation is known or not, and if not, the sample standard deviation is used. Click on the icon to view the table data if needed. **Conclusion:** Based on the calculation of the \( t \)-value and comparing it to the critical value for \(\alpha = 0.10\), we can make a decision about whether there is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.
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