the deterioration of many municipal pipeline networks across the country is a growing concern. One technology proposed for pipeline rehabilitation uses a flexible liner threaded through existing pe. An article reported the following data on tensile strength (psi) of liner specimens both when a certain fusion process was used and when this process was not used. No fusion 2748 2700 2655 2812 2501 3159 3230 2763 3247 3203 m = 10 x = 2901.8 3346 3359 3027 n=8 Fused = y = 3108.1 $₂= 201.6 oes the data suggest that the standard deviation of the strength distribution for fused specimens is smaller than that for not-fused specimens? Carry out a test at significance level 0.01. tate the relevant hypotheses. (Use o, for the non-fused specimens and ₂ for fused specimens.) Ho: 0₁ = 0₂ H₂:01 50₂ Ho: 01 = 0₂ H₂:0₁ <0₂ Ho: 01 = 0₂ Ha: 01 0₂ Ho: 01 0₂ H₂:01 > 0₂ Calculate the test statistic. (Round your answer to two decimal places.) What can be said about the P-value for the test? OP-value > 0.100 s₁ = 278.3 3287 3135 2910 2889 2912 0.050 < P-value < 0.100 0.010 < P-value < 0.050 0.001 < P-value < 0.010 OP-value < 0.001 State the conclusion in the problem context. Fail to reject Ho. There is not sufficient evidence that the true standard deviation of the strength distribution for fused specimens is smaller than that of not-fused specimens. Reject Ho. There is not sufficient evidence that the true standard deviation of the strength distribution for fused specimens is smaller than that of not-fused specimens. lard deviation of the strength distribution for fused specimens is smaller than that of not-fused specimens
the deterioration of many municipal pipeline networks across the country is a growing concern. One technology proposed for pipeline rehabilitation uses a flexible liner threaded through existing pe. An article reported the following data on tensile strength (psi) of liner specimens both when a certain fusion process was used and when this process was not used. No fusion 2748 2700 2655 2812 2501 3159 3230 2763 3247 3203 m = 10 x = 2901.8 3346 3359 3027 n=8 Fused = y = 3108.1 $₂= 201.6 oes the data suggest that the standard deviation of the strength distribution for fused specimens is smaller than that for not-fused specimens? Carry out a test at significance level 0.01. tate the relevant hypotheses. (Use o, for the non-fused specimens and ₂ for fused specimens.) Ho: 0₁ = 0₂ H₂:01 50₂ Ho: 01 = 0₂ H₂:0₁ <0₂ Ho: 01 = 0₂ Ha: 01 0₂ Ho: 01 0₂ H₂:01 > 0₂ Calculate the test statistic. (Round your answer to two decimal places.) What can be said about the P-value for the test? OP-value > 0.100 s₁ = 278.3 3287 3135 2910 2889 2912 0.050 < P-value < 0.100 0.010 < P-value < 0.050 0.001 < P-value < 0.010 OP-value < 0.001 State the conclusion in the problem context. Fail to reject Ho. There is not sufficient evidence that the true standard deviation of the strength distribution for fused specimens is smaller than that of not-fused specimens. Reject Ho. There is not sufficient evidence that the true standard deviation of the strength distribution for fused specimens is smaller than that of not-fused specimens. lard deviation of the strength distribution for fused specimens is smaller than that of not-fused specimens
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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![### Analysis of Pipeline Tensile Strength for Fusion Process
The deterioration of municipal pipeline networks is becoming a critical issue. An article presented tensile strength data (measured in psi) for liner specimens, comparing those with a fusion process to those without.
#### Data Summary
- **No Fusion Specimens:**
- Tensile strengths: 2748, 2700, 2655, 2812, 2501, 3159, 3247, 3203, 3230, 2763
- Sample size (m): 10
- Mean (x̄): 2901.8 psi
- Standard deviation (s₁): 278.3 psi
- **Fused Specimens:**
- Tensile strengths: 3027, 3346, 3359, 3287, 3135, 2910, 2889, 2912
- Sample size (n): 8
- Mean (ȳ): 3108.1 psi
- Standard deviation (s₂): 201.6 psi
#### Hypothesis Testing
Does the data suggest that the standard deviation of the tensile strength for fused specimens is smaller than for non-fused specimens? Conduct a test at a significance level of 0.01.
- **Null Hypothesis (H₀):** σ₁ = σ₂
- **Alternative Hypothesis (Hₐ):** σ₁ > σ₂
#### Steps to Calculate the Test Statistic
Calculate the test statistic \( f \) for the hypothesis test:
\[ f = \left(\frac{s₁}{s₂}\right)^2 \]
**Significance of P-value:**
- Identify the appropriate range where the P-value falls:
- P-value > 0.100
- 0.050 < P-value < 0.100
- 0.010 < P-value < 0.050
- 0.001 < P-value < 0.010
- P-value < 0.001
#### Conclusion
Interpret the results based on your calculated P-value:
- **Fail to reject H₀:** Insufficient evidence that the true standard deviation for fused specimens is smaller.
- **Reject H₀:** Sufficient evidence indicates the true standard deviation for fused specimens is smaller.
This test helps in assessing the effectiveness of the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F379782e3-36e2-4b06-9a33-e2601af30423%2F431b2412-6797-457d-92ca-2244810548a4%2Fql1q6nb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Analysis of Pipeline Tensile Strength for Fusion Process
The deterioration of municipal pipeline networks is becoming a critical issue. An article presented tensile strength data (measured in psi) for liner specimens, comparing those with a fusion process to those without.
#### Data Summary
- **No Fusion Specimens:**
- Tensile strengths: 2748, 2700, 2655, 2812, 2501, 3159, 3247, 3203, 3230, 2763
- Sample size (m): 10
- Mean (x̄): 2901.8 psi
- Standard deviation (s₁): 278.3 psi
- **Fused Specimens:**
- Tensile strengths: 3027, 3346, 3359, 3287, 3135, 2910, 2889, 2912
- Sample size (n): 8
- Mean (ȳ): 3108.1 psi
- Standard deviation (s₂): 201.6 psi
#### Hypothesis Testing
Does the data suggest that the standard deviation of the tensile strength for fused specimens is smaller than for non-fused specimens? Conduct a test at a significance level of 0.01.
- **Null Hypothesis (H₀):** σ₁ = σ₂
- **Alternative Hypothesis (Hₐ):** σ₁ > σ₂
#### Steps to Calculate the Test Statistic
Calculate the test statistic \( f \) for the hypothesis test:
\[ f = \left(\frac{s₁}{s₂}\right)^2 \]
**Significance of P-value:**
- Identify the appropriate range where the P-value falls:
- P-value > 0.100
- 0.050 < P-value < 0.100
- 0.010 < P-value < 0.050
- 0.001 < P-value < 0.010
- P-value < 0.001
#### Conclusion
Interpret the results based on your calculated P-value:
- **Fail to reject H₀:** Insufficient evidence that the true standard deviation for fused specimens is smaller.
- **Reject H₀:** Sufficient evidence indicates the true standard deviation for fused specimens is smaller.
This test helps in assessing the effectiveness of the
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