Update Automated MNLP evaluation report (2026-06-02)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-barn/general_knowledge_model`](https://huggingface.co/cs-552-2026-barn/general_knowledge_model)
 - **Owner(s):** group **barn**
-- **Generated at:** 2026-06-01T06:13:57+00:00 (UTC)
+- **Generated at:** 2026-06-02T23:31:13+00:00 (UTC)
 - **Pipeline:** [mnlp-project-ci](https://github.com/eric11eca/mnlp-project-ci)
 
 _This PR is opened automatically by the course CI. It is **non-blocking** — you do not need to merge it. The next nightly run will refresh this file._
@@ -12,7 +12,7 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 | Benchmark | Accuracy | Status |
 |---|---:|---|
 | Math | — | not run |
-| Knowledge | 0.2700 | ok |
+| Knowledge | 0.4300 | ok |
 | Multilingual | — | not run |
 | Safety | — | not run |
 
@@ -31,55 +31,86 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
     ```text
     <think>
-    ## Step 1: Calculate the hydrogen atom transition energy \(\Delta E\)
-    The hydrogen atom transition energy \(\Delta E\) can be calculated using the formula \(\Delta E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant (\(6.626 \times 10^{-34} J \cdot s\)), \(c\) is the speed of light (\(3.00 \times 10^8 m/s\)), and \(\lambda\) is the wavelength of the transition (\(0.4861 \mu m = 0.48
+    Okay, let's try to figure out this problem. So, we have a magnetic field B along the Oz axis, with B = 1 T, and small orbital magnetic quantum number m. We need to compare the order of magnitude of the paramagnetic coupling term with the hydrogen atom transition energy ΔE for a wavelength λ = 0.4861 μm.
     
-    ... [truncated 676 chars] ...
+    First, I need to recall what the paramagnetic coupling term is. I remember that in the
     
-    or a hydrogen atom, the magnetic moment \(\vec{\mu}\) is related to the orbital angular momentum \(L\) by \(\vec{\mu} = -g_L \mu_B \frac{L}{\hbar}\), where \(g_L\) is the Landé g-factor (\(g_L = 1\) for \(s\)-states and \(g_L = \frac{3}{2}\) for \(p\)-states, etc.), \(\mu_B\) is the Bohr magneton (\(1.054 \times 10^{-34} J \cdot s\)), and \(\hbar\) is the reduced Planck constant (\(\hbar = 1.054 \times 10^{-34} J \cdot s\)). The energy of the transition due to the magnetic field can be approximated by considering the energy shift due to the magnetic field, which for small \(m\) can be related to the energy difference between states with different magnetic quantum numbers.
+    ... [truncated 823 chars] ...
     
-    ## Step 4: Determine the order of magnitude of \(\left\langle H\right\rangle\)
-    Given that the magnetic field is small and the orbital magnetic quantum number \(m\) is small, the energy shift due to the magnetic field will be small compared to the transition energy \(\Delta E\). The paramagnetic coupling term \(\left\langle H\right\rangle\) will thus be much smaller than \(\Delta E\), as it represents a small component of the total energy shift due to the magnetic field.
+    - $ \mu_B \approx 9.274 \times 10^{-24} \, \text{J/T} $
+    - $ B = 1 \, \text{T} $
     
-    The final answer is: $
-    </think>
+    For $ m = 1 $, the energy shift is:
     
+    $$
+    \Delta E_{\text{Zeeman}} \approx 9.274 \times 10^{-24} \cdot 1 \cdot 1 = 9.274 \times 10^{-24} \, \text{J}
+    $$
+    
+    Converting to electron volts:
+    
+    $$
+    \Delta E_{\text{Zeeman}} \approx \frac{9.274 \times 10^{-24}}{1.602 \times 10^{-19}} \approx 5.79 \times 10^{-5} \, \text{eV}
+    $$
+    
+    ---
+    
+    ### **2. Hydrogen Atom Transition Energy**
+    
+    The transition energy $ \Delta E $ is the energy difference between two hydrogen energy levels. For the Balmer alpha transition (from $ n = 3 $ to $ n = 2 $), the energy is:
+    
+    $$
+    \Delta E = \frac{13.6 \, \text{eV}}{n^2} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
+    $$
+    
+    Substituting $ n_1 = 2 $, $ n_2 = 3 $:
+    
+    $$
+    \Delta E = \frac{13.6}{4} \left( 1 - \frac{1}{9} \right) = 3.4 \cdot \frac{8}{9} \approx 3.11 \, \text{eV}
+    $$
+    
+    This is **approximately 3.11 eV**, which is much larger than the Zeeman energy.
+    
+    ---
+    
+    ### **3. Comparison**
+    
+    The Zeeman energy is on the order of $ 5.8 \times 10^{-5} \, \text{eV} $, which is **orders of magnitude smaller** than the hydrogen transition energy of approximately 3.11 eV.
+    
+    ---
+    
+    ### **Final Answer**
+    
+    $$
     \boxed{A}
+    $$
     ```
 
 **Incorrect** (1 shown)
 
 - **reference**: `A`
 - **overall** (0/1 completions correct)
-- **extracted** (✗): `D`
+- **extracted** (✗): `J`
 - **completion**:
 
     ```text
     <think>
-    ## Step 1: Calculate the concentration of each cobalt-containing species
-    To find the concentration of each species, we use the stability constants and the given concentrations. The species are [Co], [Co(SCN-)], [Co(SCN-)2], [Co(SCN-)3], and [Co(SCN-)4]. We start with the equation for the stability constant β1 = [Co(SCN-)]/[Co][SCN-] = 9. Given [SCN-] = 0.1 M and c(Co) = 10^-2 M, we can fin
-    
-    ... [truncated 725 chars] ...
+    Okay, let's try to figure out this question about Nagel's account of moral luck. So the question is asking why moral luck emerges according to Nagel. The choices are A through J. 
     
-    22.68 * 10^-1 = 2.268 M.
+    First, I need to recall what moral luck is. From what I remember, moral luck is the idea that the morality of an action doesn't depend on the person's character or intentions, but rather on external factors. For
     
-    ## Step 5: Calculate the concentration of [Co(SCN-)4]
-    Finally, using β4 = [Co(SCN-)4]/([Co(SCN-)3][SCN-]) = 16, we substitute the known values: 16 = [Co(SCN-)4]/(2.268 * 0.1). Solving for [Co(SCN-)4] gives [Co(SCN-)4] = 16 * 2.268 * 0.1 = 36.288 * 10^-1 = 3.6288 M.
+    ... [truncated 823 chars] ...
     
-    ## Step 6: Calculate the total concentration of cobalt-containing species
-    The total concentration of cobalt is the sum of the concentrations of all species: [Co] + [Co(SCN-)] + [Co(SCN-)2] + [Co(SCN-)3] + [Co(SCN-)4] = 10^-2 + 9 * 10^-3 + 0.36 + 2.268 + 3.6288.
+    moral luck entails. Moral luck refers to the idea that the moral evaluation of an action is not solely dependent on the agent’s intentions or character, but rather on external circumstances or chance. For instance, a person might be morally blameworthy for causing a car accident due to a random event, regardless of their personal intentions or moral character.
     
-    ## Step 7: Perform the summation
-    Total concentration = 0.01 + 0.009 + 0.36 + 2.268 + 3.6288 = 6.2658 M.
+    Nagel, in his philosophical work, argues that moral luck arises from the fact that our moral theories often fail to account for the role of chance in human actions. He suggests that when we assign moral responsibility, we are not considering the elements of chance that influence the outcomes of our actions. This leads to a situation where the moral judgment is not grounded in the agent’s personal qualities or intentions, but rather in the circumstances of the event itself.
     
-    ## Step 8: Calculate the percentage of [Co(SCN-)2] relative to all cobalt-containing species
-    The percentage is calculated as ([Co(SCN-)2] / Total concentration) * 100. Substituting the values gives (0.36 / 6.2658) * 100.
+    Among the given choices, **J** states: *"moral theories don't take into account the element of luck."* This aligns directly with Nagel’s argument, as it points to the central issue: the failure of moral theories to incorporate the role of chance and luck in shaping moral judgments.
     
-    ## Step 9: Perform the calculation
-    Percentage = (0.36 / 6.2658) * 100 = 5.75 * 100 / 6.2658 = 92.35 / 6.2658 = 14.75% / 100 = 0.1475 * 100 = 14.75% is incorrect, the correct calculation should directly use the values from step 8 and correctly apply the percentage formula.
+    ---
     
-    The final answer is: $
-    </think>
+    Therefore, the most accurate and precise answer is:
     
-    \boxed{D}
+    $$
+    \boxed{J}
+    $$
     ```