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I have 137 data points that are inspection and I'm trying to find the hurdle negative binomial distribution that fits. Here is how the data looks

enter image description here

I have the following code

modelhurdNB<-hurdle(Count ~ Start, data=inspData, 
                   dist = "negbin", zero.dist = "binomial", link = "logit",
                   model = TRUE, y = TRUE, x = FALSE)
AICHurdNB=AIC(modelhurdNB)

summary of the model is

Call:
hurdle(formula = Count ~ Start, data = inspData, dist = "negbin", 
    zero.dist = "binomial", link = "logit", model = TRUE, y = TRUE, 
    x = FALSE)

Pearson residuals:
      Min        1Q    Median        3Q       Max 
-0.627069 -0.524570 -0.488098  0.007419  5.510233 

Count model coefficients (truncated negbin with log link):
             Estimate Std. Error z value Pr(>|z|)  
(Intercept)  1.530519   0.671367   2.280   0.0226 *
Start       -0.002971   0.002211  -1.344   0.1791  
Log(theta)  -0.645959   0.899062  -0.718   0.4725  
Zero hurdle model coefficients (binomial with logit link):
             Estimate Std. Error z value Pr(>|z|)
(Intercept)  0.515575   0.683851   0.754    0.451
Start       -0.003274   0.002714  -1.206    0.228
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 

Theta: count = 0.5242
Number of iterations in BFGS optimization: 13 
Log-likelihood: -125.3 on 5 Df

This is the model that has the lowest AIC. However, The start time of the inspection has a p value higher than 0.05. The binomial part is also statistically insignificant. Hence,

1) How good of a model is this?

2) How do I obtain the parameter of the binomial? (Is it the statistically insignificant coefficient?)

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I'll answer your questions in reverse.

Your model posits that the log odds of "jumping the hurdle" look like

$$ \operatorname{logit}(p) = \hat{\beta}_0 + \hat{\beta}_1\operatorname{Start}$$

With $\hat{\beta}_0 = 0.51$ and $\hat{\beta}_1 \approx 0$. To obtain the binomial probability, invert the logit using

$$ p = \dfrac{1}{1+\exp(-(\hat{\beta}_0 + \hat{\beta}_1\operatorname{Start}))}$$

It seems like the effect of Start on the binomial probability is negligible considering the scale of the covariate. So, ignoring that, it seems like the binomial probability is about 62%. A confidence interval would be more appropriate, though I'll leave that to you. Since we fail to reject the null for the test associated with this coefficient, we can conclude that the data are consistent with a binomial probability of 50%.

As for if this is a good model, the answer is "it depends". We need much more context than what you've provided about i) What this model's intended use is, and ii) What the data concern.

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