1. **Stating the Problem:** We want to perform hypothesis testing and curve fitting on fatigue data (Stress vs. Cycles to Failure) and reliability data using Weibull distribution. 2. **Hypothesis Testing for Fatigue Data:** We test if stress significantly affects cycles to failure. - Null Hypothesis $H_0$: Stress has no effect on cycles to failure. - Alternative Hypothesis $H_a$: Stress affects cycles to failure. 3. **Model for Fatigue Data:** The S-N curve is often modeled as a power law: $$ S = a N^b $$ Taking logarithms: $$ \log S = \log a + b \log N $$ This is a linear regression problem with $\log S$ as dependent and $\log N$ as independent variable. 4. **Curve Fitting Steps:** - Take logarithms of stress and cycles data. - Fit linear regression: $y = m x + c$ where $y = \log S$, $x = \log N$. - Estimate parameters $a = e^c$, $b = m$. 5. **Example Data (from table):** | Stress (S) | Cycles (N) | |------------|------------| | 300 | 10000 | | 250 | 50000 | | 200 | 100000 | | 150 | 500000 | | 120 | 1000000 | 6. **Calculate $\log S$ and $\log N$:** $$ \log S = [\log 300, \log 250, \log 200, \log 150, \log 120] $$ $$ \log N = [\log 10000, \log 50000, \log 100000, \log 500000, \log 1000000] $$ 7. **Fit linear regression:** Use least squares to find slope $b$ and intercept $\log a$. 8. **Reliability Curve Fitting (Weibull):** Reliability function: $$ R(t) = e^{- (t/\eta)^\beta} $$ Taking logarithm twice: $$ \ln(-\ln R(t)) = \beta \ln t - \beta \ln \eta $$ Plot $\ln(-\ln R(t))$ vs. $\ln t$ to get a straight line with slope $\beta$ and intercept $-\beta \ln \eta$. 9. **Testing Hypothesis for Weibull Parameters:** - Null Hypothesis $H_0$: $\beta = 1$ (exponential distribution, constant failure rate). - Alternative $H_a$: $\beta \neq 1$ (failure rate changes over time). 10. **Summary:** - Fit S-N curve by linear regression on log-log data. - Fit Weibull parameters by linearizing reliability data. - Test hypotheses on parameters using t-tests or confidence intervals. **Final answers:** - Fatigue S-N curve: $S = a N^b$ with $a,b$ from regression. - Reliability: $R(t) = e^{-(t/\eta)^\beta}$ with $\beta, \eta$ from Weibull fit.