How heavy a load (pounds) is needed to pull apart pieces of Douglas fir 4 inches long and 1.5 inches square? Students doing a laboratory exercise sample 20 pieces of wood and find an average load capacity of 30,841 pounds. We are willing to regard the wood pieces prepared for the lab session as an SRS of all similar pieces of Douglas fir. Engineers also commonly assume the characteristics of materials vary normally. Suppose that the strength of pieces of wood like these follow a normal distribution with a population standard deviation of 3,000 pounds (As you can see all three necessary assumptions are met). a. Assuming that all evergreen wood has a known "load" capacity average of 30,000 pounds. Make a two-sided hypothesis (null and alternative statement) about Douglas Fir "load" capacity compared to the overall average.
How heavy a load (pounds) is needed to pull apart pieces of Douglas fir 4 inches long and 1.5 inches square? Students doing a laboratory exercise sample 20 pieces of wood and find an average load capacity of 30,841 pounds. We are willing to regard the wood pieces prepared for the lab session as an SRS of all similar pieces of Douglas fir. Engineers also commonly assume the characteristics of materials vary normally. Suppose that the strength of pieces of wood like these follow a normal distribution with a population standard deviation of 3,000 pounds (As you can see all three necessary assumptions are met). a. Assuming that all evergreen wood has a known "load" capacity average of 30,000 pounds. Make a two-sided hypothesis (null and alternative statement) about Douglas Fir "load" capacity compared to the overall average.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![**Text for Educational Website:**
**Understanding Load Capacity Testing of Douglas Fir Wood**
How heavy a load (in pounds) is needed to pull apart pieces of Douglas fir measuring 4 inches long and 1.5 inches square? In a laboratory exercise, students sampled 20 pieces of wood and found an average load capacity of 30,841 pounds.
These wood pieces, prepared for the lab session, are considered a Simple Random Sample (SRS) of all similar pieces of Douglas fir. Engineers often assume such material characteristics vary normally. It is assumed that the strength of the wood follows a normal distribution with a population standard deviation of 3,000 pounds. (All necessary assumptions are met for this analysis).
**Hypothesis Testing:**
a. **Statement of Hypothesis:**
- Null Hypothesis (H0): The average "load" capacity of all evergreen wood is 30,000 pounds.
- Alternative Hypothesis (H1): The "load" capacity of Douglas Fir differs from the overall average of 30,000 pounds.
b. **Calculating the Test Statistic:**
- We use the z-statistic formula:
\[
z = \frac{\overline{x} - \mu}{\sigma / \sqrt{n}}
\]
- In this formula:
- \(\overline{x}\) = sample mean = 30,841 pounds
- \(\mu\) = population mean = 30,000 pounds
- \(\sigma\) = population standard deviation = 3,000 pounds
- \(n\) = sample size = 20
- By applying these values, we determine how many "standard errors" the Douglas Fir sample average is away from the overall average.
c. **Conclusion:**
- Using Table "C" from the textbook (referred on page 701), apply a two-tailed alpha level of 0.05.
- Consult the z-score associated with a two-sided P-value of 0.05. Determine if the calculated z-score from part b is greater or less than this critical value to decide on rejecting or failing to reject the null hypothesis.
This process will help understand how significantly the sample average deviates from the hypothesized population mean.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fae8e0e00-e909-4cd4-8a43-4c71501433ad%2F8c9fe1b7-86eb-4fca-9d2e-a0ef9a4c27bc%2Ft0y6wu_processed.png&w=3840&q=75)
Transcribed Image Text:**Text for Educational Website:**
**Understanding Load Capacity Testing of Douglas Fir Wood**
How heavy a load (in pounds) is needed to pull apart pieces of Douglas fir measuring 4 inches long and 1.5 inches square? In a laboratory exercise, students sampled 20 pieces of wood and found an average load capacity of 30,841 pounds.
These wood pieces, prepared for the lab session, are considered a Simple Random Sample (SRS) of all similar pieces of Douglas fir. Engineers often assume such material characteristics vary normally. It is assumed that the strength of the wood follows a normal distribution with a population standard deviation of 3,000 pounds. (All necessary assumptions are met for this analysis).
**Hypothesis Testing:**
a. **Statement of Hypothesis:**
- Null Hypothesis (H0): The average "load" capacity of all evergreen wood is 30,000 pounds.
- Alternative Hypothesis (H1): The "load" capacity of Douglas Fir differs from the overall average of 30,000 pounds.
b. **Calculating the Test Statistic:**
- We use the z-statistic formula:
\[
z = \frac{\overline{x} - \mu}{\sigma / \sqrt{n}}
\]
- In this formula:
- \(\overline{x}\) = sample mean = 30,841 pounds
- \(\mu\) = population mean = 30,000 pounds
- \(\sigma\) = population standard deviation = 3,000 pounds
- \(n\) = sample size = 20
- By applying these values, we determine how many "standard errors" the Douglas Fir sample average is away from the overall average.
c. **Conclusion:**
- Using Table "C" from the textbook (referred on page 701), apply a two-tailed alpha level of 0.05.
- Consult the z-score associated with a two-sided P-value of 0.05. Determine if the calculated z-score from part b is greater or less than this critical value to decide on rejecting or failing to reject the null hypothesis.
This process will help understand how significantly the sample average deviates from the hypothesized population mean.
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