1. **Problem Statement:** Test the hypothesis that the sample data \(0.521, 0.317, -0.282, 3.454, 0.521, 0.282, 0.317, 0.282, 0.944, 0.125, 0.282, 0.784, 1.044, 0.317, 0.521, 0.623\) comes from the uniform distribution defined by: $$F(x) = \begin{cases} 0 & x < 0 \\ x & 0 \leq x < 1 \\ 1 & x \geq 1 \end{cases}$$ at the 5% significance level. 2. **Method:** Use the Kolmogorov-Smirnov (K-S) test for goodness of fit. - The null hypothesis \(H_0\): The sample comes from the specified uniform distribution. - The alternative hypothesis \(H_a\): The sample does not come from the specified distribution. 3. **Steps:** - Sort the sample data in ascending order: $$-0.282, 0.125, 0.282, 0.282, 0.282, 0.317, 0.317, 0.317, 0.521, 0.521, 0.521, 0.623, 0.784, 0.944, 1.044, 3.454$$ - Remove or note values outside the support of the distribution (\(x < 0\) or \(x \geq 1\)): Values \(-0.282, 1.044, 3.454\) are outside the support. - For K-S test, only values within \([0,1)\) are considered. So the effective sample size \(n=13\). - Calculate empirical distribution function (EDF) \(F_n(x) = \frac{i}{n}\) for each ordered value \(x_i\). - Calculate theoretical CDF \(F(x) = x\) for each \(x_i\). - Compute the K-S statistic: $$D = \max \left( \max_i \left| F_n(x_i) - F(x_i) \right|, \max_i \left| F_n(x_{i-1}) - F(x_i) \right| \right)$$ where \(F_n(x_{i-1}) = \frac{i-1}{n}\). 4. **Calculations:** | i | \(x_i\) | \(F_n(x_i) = \frac{i}{n}\) | \(F(x_i) = x_i\) | \(|F_n(x_i) - F(x_i)|\) | \(|F_n(x_{i-1}) - F(x_i)|\) | |---|---------|----------------------------|-----------------|-------------------------|----------------------------| |1|0.125|1/13 \approx 0.0769|0.125|0.0481|0.125| |2|0.282|2/13 \approx 0.1538|0.282|0.1282|0.1282| |3|0.282|3/13 \approx 0.2308|0.282|0.0512|0.0512| |4|0.282|4/13 \approx 0.3077|0.282|0.0257|0.1282| |5|0.317|5/13 \approx 0.3846|0.317|0.0676|0.1923| |6|0.317|6/13 \approx 0.4615|0.317|0.1445|0.317| |7|0.317|7/13 \approx 0.5385|0.317|0.2215|0.3846| |8|0.521|8/13 \approx 0.6154|0.521|0.0944|0.4615| |9|0.521|9/13 \approx 0.6923|0.521|0.1713|0.5385| |10|0.521|10/13 \approx 0.7692|0.521|0.2482|0.6154| |11|0.623|11/13 \approx 0.8462|0.623|0.2232|0.6923| |12|0.784|12/13 \approx 0.9231|0.784|0.1391|0.7692| |13|0.944|13/13 = 1.0|0.944|0.056|0.8462| - The maximum absolute difference \(D\) is approximately \(0.2482\) (from row 10). 5. **Critical value:** For \(n=13\) and \(\alpha=0.05\), the critical value for K-S test is approximately: $$D_{\alpha} = 1.36 / \sqrt{n} = 1.36 / \sqrt{13} \approx 0.377$$ 6. **Decision:** Since \(D = 0.2482 < 0.377 = D_{\alpha}\), we fail to reject the null hypothesis. 7. **Conclusion:** There is insufficient evidence at the 5% significance level to reject the hypothesis that the sample comes from the specified uniform distribution. **Final answer:** The sample data is consistent with the uniform distribution \(F(x)\) at the 5% level.