Why are Pearson's residuals from a negative binomial regression smaller than those from a poisson regression? I have these data:
set.seed(1)
predictor  <- rnorm(20)
set.seed(1)
counts <- c(sample(1:1000, 20))
df <- data.frame(counts, predictor)

I ran a poisson regression
poisson_counts <- glm(counts ~ predictor, data = df, family = "poisson")

And a negative binomial regression:
require(MASS)
nb_counts <- glm.nb(counts ~ predictor, data = df)

Then I calculated for dispersion statistics for the poisson regression:
sum(residuals(poisson_counts, type="pearson")^2)/df.residual(poisson_counts)

# [1] 145.4905

And the negative binomial regression:
sum(residuals(nb_counts, type="pearson")^2)/df.residual(nb_counts)

# [1] 0.7650289

Is anyone able to explain, WITHOUT USING EQUATIONS, why the dispersion statistic for the negative binomial regression is considerably smaller than the dispersion statistic for the poisson regression?
 A: For the Poisson model, if the expection for the $i$th observation $Y_i$ is $\mu_i$ its variance is $\mu_i$, & the Pearson residual therefore
$$\frac{y_i-\hat\mu_i}{\sqrt{\hat\mu_i}}$$
where $\hat\mu$ is the estimate of the mean. The parametrization of the negative binomial model used in MASS is explained here. If the expection for the $i$th observation $Y_i$ is $\mu_i$ its variance is $\mu_i + \frac{\mu^2}{\theta}$, & the Pearson residual therefore
$$\frac{y_i-\tilde\mu_i}{\sqrt{\tilde\mu_i+\frac{\tilde\mu'^2}{\theta}}}$$
where $\tilde\mu$ is the estimate of the mean. The smaller the value of $\theta$— i.e. the more extra-Poisson variance—, the smaller the residual compared to its Poisson equivalent. [But as @whuber has pointed out, the estimates of the means are not the same, $\hat\mu\neq\tilde\mu$, because the estimation procedure weights observations according to their assumed variance. If you were to make replicate measurements for the $i$th predictor pattern, they'd get closer, & in general adding a parameter should give a better fit across all observations, though I don't know how to demonstrate this rigorously. All the same, the population quantities you're estimating are larger if the Poisson model holds, so it shouldn't be a surprise.]
A: This is rather straightforward, but the "without using equations" is a substantial handicap.  I can explain it in words, but those words will necessarily mirror equations.  I hope that will be acceptable / still of some value to you.  (The relevant equations are not difficult.)  
There are several types of residuals.  Raw residuals are simply the difference between the observed response values (in your case the counts) and the model's predicted response values.  Pearson residuals divide those by the standard deviation (the square root of the variance function for the particular version of the generalized linear model that you are using).  
The standard deviation associated with the Poisson distribution is smaller than that of the negative binomial.  Thus, when you divide by a larger denominator, the quotient is smaller.  
In addition, the negative binomial is more appropriate to your case, because your counts will be distributed as a uniform in the population.  That is, their variance will not equal their mean.  
