Tarefa nist_anova.smls02.agent

Veredito

Pergunta feita ao analista

NIST Statistical Reference Datasets, one-way analysis of variance, problem SmLs02: generated data with 1 constant leading digits; one factor (treatment) with 9 levels and 1809 observations in a balanced design, model y_ij = mu + tau_i + e_ij. Dataset d holds the factor treatment (levels 1 to 9), the response response, and indicator columns treatment_2 to treatment_9 (1 when the row is at that level; level 1 is the reference). Compute the ANOVA F statistic for the hypothesis that all 9 treatment effects are equal: the between-treatment mean square divided by the within-treatment mean square.

Como o agente resolveu e como foi validado

Resumo passo a passo do que o agente executou, por que o caminho está correto e como cada verificação conferiu o resultado.

1. 1Preparação da base de dados
   d— NIST StRD SmLs02: 1809 observations; treatment = the treatment level (1 to 9), response = the response, treatment_2 … treatment_9 = 1 when the row is at that treatment level (level 1 is the reference)
   Transformação "dummies" aplicada à base "d" (1809 → 1809 linhas). {"column":"treatment","drop_first":true,"prefix":"treatment"}
2. 2Chamada que gerou a resposta finalinfer.linear_hypothesis(Teste de hipótese linear conjunta (estatística F / Wald))citada na respostacoincide com a referência

   Ver chamada call_54c732f7 nos detalhes

   With classical errors, the joint test that every indicator's coefficient is zero is the one-way ANOVA F.

   Por que este é o caminho de referência: With classical errors, the joint test that every indicator's coefficient is zero is the one-way ANOVA F.
   Argumentos principais

   dataset:

   "d"

   hypothesis:

   "treatment_2 = 0, treatment_3 = 0, treatment_4 = 0, treatment_5 = 0, treatment_6 = 0, treatment_7 = 0, treatment_8 = 0, treatment_9 = 0"

   se:

   "classical"

   x:

   ["treatment_2","treatment_3","treatment_4","treatment_5","treatment_6","treatment_7","treatment_8","treatment_9"]

   y:

   "response"

   Resultado calculado

   • F = 201 200.9999999999995
     F statistic of treatment_2 = 0; treatment_3 = 0; treatment_4 = 0; treatment_5 = 0; treatment_6 = 0; treatment_7 = 0; treatment_8 = 0; treatment_9 = 0
   • p_value = 0,00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000403714probabilidade 4.037141885758291e-243
     p-value of treatment_2 = 0; treatment_3 = 0; treatment_4 = 0; treatment_5 = 0; treatment_6 = 0; treatment_7 = 0; treatment_8 = 0; treatment_9 = 0
   Diagnósticos: statistic = "F" (F = W/m against F(m, n − k)) · df_numerator = 88 (number of restrictions) · df_denominator = 1.8001800 (residual degrees of freedom n − k) · se_type = "classical" (classical (homoskedastic) standard errors) · n_obs = 1.8091809
3. 3Conferência de lastro e resposta final

   Todos os 1 números escritos na resposta foram conferidos contra as ferramentas e o enunciado (nenhum número sem lastro).

   Números encontrados: 201,0
   Figuras citadas: call_54c732f7:F
   • ✓
     #0resposta com lastro

     Garante que todo número citado no texto veio de uma ferramenta ou do enunciado (sem cálculo mental do modelo).
   • ✓
     #1resposta citainfer.linear_hypothesis:F

     Exige que a resposta referencie explicitamente a figura infer.linear_hypothesis:F.
   • ✓
     #2ferramenta chamadainfer.linear_hypothesis

     Exige que o resultado venha da ferramenta infer.linear_hypothesis.
   • ✓
     #3argumento igualse

     the ANOVA F is the classical-variance Wald F (the tool's default se is HC1)
   • ✓
     #4diagnósticodf_numerator

     certified between-treatment df
   • ✓
     #5diagnósticodf_denominator

     certified within-treatment df
   • ✓
     #6dígitos corretos (LRE)F

     certified F statistic

Valores esperados e de onde vêm

Verificações

Caminho de referência e caminho do agente

A chamada de referência ao lado da chamada que deu a resposta, argumento por argumento.

Trajetória

Cada evento na ordem em que aconteceu: turnos do modelo, chamadas de ferramentas, guardas, transformações e pontuação.

1. 0Início

   NIST Statistical Reference Datasets, one-way analysis of variance, problem SmLs02: generated data with 1 constant leading digits; one factor (treatment) with 9 levels and 1809 observations in a balanced design, model y_ij = mu + tau_i + e_ij. Dataset d holds the factor treatment (levels 1 to 9), the response response, and indicator columns treatment_2 to treatment_9 (1 when the row is at that level; level 1 is the reference). Compute the ANOVA F statistic for the hypothesis that all 9 treatment effects are equal: the between-treatment mean square divided by the within-treatment mean square.

   d (agent)

   NIST StRD SmLs02: 1809 observations; treatment = the treatment level (1 to 9), response = the response, treatment_2 … treatment_9 = 1 when the row is at that treatment level (level 1 is the reference)

   file 9c23f2c51935740a123d6882ee41b874b697b3618bdfe2915d1b76370817a080 · frame —

   1. dummies {"column":"treatment","drop_first":true,"prefix":"treatment"}
2. 1Transformação
   d · dummies {"column":"treatment","drop_first":true,"prefix":"treatment"}

   1809 → 1809 linhas

   Frame antes → depois: 6cb2bb68727f → 40265bb6a424
3. 2Turno do modelo
   call_c5d6ecc5referenceFim: tool_use0 ms0 de entrada · 0 de saída · 0 de raciocínio · 0 em cache

   Texto visível

   With classical errors, the joint test that every indicator's coefficient is zero is the one-way ANOVA F.

   Chamadas pedidas

   • infer.linear_hypothesis

     {
       "dataset": "d",
       "hypothesis": "treatment_2 = 0, treatment_3 = 0, treatment_4 = 0, treatment_5 = 0, treatment_6 = 0, treatment_7 = 0, treatment_8 = 0, treatment_9 = 0",
       "se": "classical",
       "x": [
         "treatment_2",
         "treatment_3",
         "treatment_4",
         "treatment_5",
         "treatment_6",
         "treatment_7",
         "treatment_8",
         "treatment_9"
       ],
       "y": "response"
     }

   Mensagens (formato neutro)Instruções do sistemaFerramentas oferecidas
4. 3Ferramenta
   infer.linear_hypothesis— Teste de hipótese linear conjunta (estatística F / Wald)v0.1.0call_54c732f7decidido pelo modeloreexecutável

   Motivo: With classical errors, the joint test that every indicator's coefficient is zero is the one-way ANOVA F.

   Argumentos

   {
     "cluster": null,
     "dataset": "d",
     "hypothesis": "treatment_2 = 0, treatment_3 = 0, treatment_4 = 0, treatment_5 = 0, treatment_6 = 0, treatment_7 = 0, treatment_8 = 0, treatment_9 = 0",
     "se": "classical",
     "subset": null,
     "x": [
       "treatment_2",
       "treatment_3",
       "treatment_4",
       "treatment_5",
       "treatment_6",
       "treatment_7",
       "treatment_8",
       "treatment_9"
     ],
     "y": "response"
   }

   Figuras call_54c732f7

   Chave	Valor	Valor bruto	Erro-padrão	IC
   F F statistic of treatment_2 = 0; treatment_3 = 0; treatment_4 = 0; treatment_5 = 0; treatment_6 = 0; treatment_7 = 0; treatment_8 = 0; treatment_9 = 0	201	200.9999999999995	—	—
   p_value p-value of treatment_2 = 0; treatment_3 = 0; treatment_4 = 0; treatment_5 = 0; treatment_6 = 0; treatment_7 = 0; treatment_8 = 0; treatment_9 = 0	0,00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000403714probabilidade	4.037141885758291e-243	—	—

   Diagnósticos

   • statistic = "F" (F = W/m against F(m, n − k))
   • df_numerator = 88 (number of restrictions)
   • df_denominator = 1.8001800 (residual degrees of freedom n − k)
   • se_type = "classical" (classical (homoskedastic) standard errors)
   • n_obs = 1.8091809
   • n_clusters = — (errors not clustered)

   Derivação matemática e pressupostos

   F

   1. Estatística de Wald das 8 restrições, com a covariância classical (homoskedastic) standard errors.→ 1.608 (1607.999999999996)
   2. Estatística F: a de Wald dividida pelo número de restrições.F = W / m→ 201 (200.9999999999995)

   p_value

   1. Probabilidade de uma F(8, 1800) exceder a estatística observada.→ 0,00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000403714 (4.037141885758291e-243)

   Pressupostos do estimador

   • errors are independent and have the same variance for every observation

   Tabelas (2)

   {
     "coefficients": [
       {
         "ci_high": 1.4138338221356335,
         "ci_low": 1.3861661778643664,
         "coef": 1.4,
         "p": 0,
         "se": 0.007053456158586019,
         "t": 198.48425630260851,
         "term": "Intercept"
       },
       {
         "ci_high": -0.08043602111632987,
         "ci_low": -0.1195639788836703,
         "coef": -0.10000000000000009,
         "p": 4.7048753115723204e-23,
         "se": 0.009975093361076447,
         "t": -10.024968827881601,
         "term": "treatment_2"
       },
       {
         "ci_high": 0.11956397888367021,
         "ci_low": 0.08043602111632975,
         "coef": 0.09999999999999998,
         "p": 4.704875311573031e-23,
         "se": 0.009975093361076453,
         "t": 10.024968827881585,
         "term": "treatment_3"
       },
       {
         "ci_high": -0.08043602111633012,
         "ci_low": -0.11956397888367006,
         "coef": -0.10000000000000009,
         "p": 4.704875311566536e-23,
         "se": 0.009975093361076319,
         "t": -10.024968827881729,
         "term": "treatment_4"
       },
       {
         "ci_high": 0.11956397888366986,
         "ci_low": 0.0804360211163301,
         "coef": 0.09999999999999998,
         "p": 4.7048753115651787e-23,
         "se": 0.009975093361076279,
         "t": 10.024968827881759,
         "term": "treatment_5"
       },
       {
         "ci_high": -0.08043602111632998,
         "ci_low": -0.1195639788836702,
         "coef": -0.10000000000000009,
         "p": 4.7048753115697243e-23,
         "se": 0.00997509336107639,
         "t": -10.024968827881658,
         "term": "treatment_6"
       },
       {
         "ci_high": 0.11956397888367007,
         "ci_low": 0.08043602111632989,
         "coef": 0.09999999999999998,
         "p": 4.7048753115697243e-23,
         "se": 0.00997509336107638,
         "t": 10.024968827881658,
         "term": "treatment_7"
       },
       {
         "ci_high": -0.08043602111633008,
         "ci_low": -0.1195639788836701,
         "coef": -0.10000000000000009,
         "p": 4.704875311567541e-23,
         "se": 0.009975093361076343,
         "t": -10.024968827881706,
         "term": "treatment_8"
       },
       {
         "ci_high": 0.11956397888366994,
         "ci_low": 0.08043602111633001,
         "coef": 0.09999999999999998,
         "p": 4.704875311567068e-23,
         "se": 0.00997509336107632,
         "t": 10.024968827881716,
         "term": "treatment_9"
       }
     ],
     "restrictions": [
       {
         "estimate": -0.10000000000000009,
         "hypothesized": 0,
         "restriction": "treatment_2 = 0"
       },
       {
         "estimate": 0.09999999999999998,
         "hypothesized": 0,
         "restriction": "treatment_3 = 0"
       },
       {
         "estimate": -0.10000000000000009,
         "hypothesized": 0,
         "restriction": "treatment_4 = 0"
       },
       {
         "estimate": 0.09999999999999998,
         "hypothesized": 0,
         "restriction": "treatment_5 = 0"
       },
       {
         "estimate": -0.10000000000000009,
         "hypothesized": 0,
         "restriction": "treatment_6 = 0"
       },
       {
         "estimate": 0.09999999999999998,
         "hypothesized": 0,
         "restriction": "treatment_7 = 0"
       },
       {
         "estimate": -0.10000000000000009,
         "hypothesized": 0,
         "restriction": "treatment_8 = 0"
       },
       {
         "estimate": 0.09999999999999998,
         "hypothesized": 0,
         "restriction": "treatment_9 = 0"
       }
     ]
   }

   Hashes SHA-256 e metadados de reprodução

   Digests dos dados

   d 40265bb6a4240c34fa765dd90a3fe6242ebe37ae0ae601c4865ceef0c45d0cd1

   sha256 do resultado

   6ab73cb1e801e4618823f52c940745782f7f2ed5a1e6eec82fab521678245fd7

   Resultado mostrado ao modelo
5. 4Turno do modelo
   call_39f78169referenceFim: tool_use0 ms0 de entrada · 0 de saída · 0 de raciocínio · 0 em cache

   Chamadas pedidas

   • submit_answer (meta-ferramenta)

     {
       "answer": "Estimativa (infer.linear_hypothesis): F = 201,0.",
       "caveats": [],
       "figure_refs": [
         "call_54c732f7:F"
       ]
     }

   Mensagens (formato neutro)Instruções do sistemaFerramentas oferecidas
6. 5Guarda
   resposta aceita

   Números encontrados

   201,0

   Sem lastro

   —

   Não citados

   —

   1 nova(s) tentativa(s) restante(s)
7. 6Resposta

   Estimativa (infer.linear_hypothesis): F = 201,0.

   Figuras citadas: call_54c732f7:F · Parada: answered

   Figura da resposta: call_54c732f7:F = 201200.9999999999995
8. 7Pontuação
   #0resposta com lastropassouobrigatóriaevery number backed
9. 8Pontuação
   #1resposta citapassouobrigatóriacited ['call_54c732f7:F']
10. 9Pontuação
    #2ferramenta chamadapassouobrigatóriaanswer call is infer.linear_hypothesis
11. 10Pontuação
    #3argumento igualpassouobrigatóriainfer.linear_hypothesis.se = 'classical'
12. 11Pontuação
    #4diagnósticopassouobrigatória8 vs 8
13. 12Pontuação
    #5diagnósticopassouobrigatória1800 vs 1800
14. 13Pontuação
    #6dígitos corretos (LRE)passouobrigatóriaLRE 14.59 (min 9.0)

Épocas × verificações

Tentativas

Cada tentativa é seu próprio arquivo e é publicada; só a última de cada época é pontuada.

• época 1, tentativa 1passoupontuada(exibindo)

Downloads desta tarefa