interpretation of betareg coef I have a data that where the outcome is the proportion of a species observed in an area by a machine on 2 separate days. Since the outcome is a proportion and does not include 0 or 1 I used a beta regression to fit the model. Temperature is used as  an independent variable. Here is some toy R code:
set.seed(1234)
library(betareg)
d <- data.frame(
  DAY = c(1,1,1,1,2,2,2,2),
  Proportion = c(.4,.1,.25, .25, .5,.3,.1,.1),
  MACHINE = c("A","B","C","D","H","G","K","L"),
  TEMPERATURE = c(rnorm(8)*100)
)
b <- betareg(Proportion ~ TEMPERATURE,
  data= d, link = "logit", link.phi = NULL, type = "ML")
summary(b)
## Call:
## betareg(formula = Proportion ~ TEMPERATURE, data = d, link = "logit", link.phi = NULL, type = "ML")
## 
## Standardized weighted residuals 2:
##     Min      1Q  Median      3Q     Max 
## -1.2803 -1.2012  0.3034  0.6819  1.6494 
## 
## Coefficients (mean model with logit link):
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.0881982  0.2620518  -4.153 3.29e-05 ***
## TEMPERATURE  0.0003469  0.0023677   0.147    0.884    
## 
## Phi coefficients (precision model with identity link):
##       Estimate Std. Error z value Pr(>|z|)  
## (phi)    9.305      4.505   2.066   0.0389 *
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 

Above you can see the TEMPERATURE coefficient is .0003469. Exponentiating, exp(.0003469) = 1.000347
Update incorporating replies and comments:
You can see here how increasing temperature by 1 unit from -10 to 10 increase the proportion
nd <- data.frame(TEMPERATURE = seq(-10, 10, by = 1))
nd$Proportion <- predict(b, newdata = nd)
nd$proportion_ratio <- nd$Proportion/(1 - nd$Proportion)
plot(Proportion ~ TEMPERATURE, data = nd, type = "b")

The interpretation is: A 1-unit change in TEMPERATURE leads to a relative change of 1.000347 ≈0.04%  in the Proportion: $$\frac{\mathrm{E}(\mathtt{Proportion})}{1-\mathrm{E}(\mathtt{Proportion})}$$ 
The key word there is relative change so when you compare exp(coef(b))[2] to nd$proportion_ratio[2] / nd$proportion_ratio[1]  you'll see they are the same
## ratio of proportion
nd$proportion_ratio[2] / nd$proportion_ratio[1] 
exp(coef(b))[2]
nd$proportion_ratio[-1] / nd$proportion_ratio[-20]

 A: Yes, the logit link can be interpreted like that. It's just not a change in "odds" (= ratio of probabilities) but a change in a ratio of proportions. More formally, the model equation for the expectation is the same as in logistic regression:
$$ \mathrm{logit}(\mu_i) = x_i^\top \beta $$
where $\mu_i = \mathrm{E}(y_i)$. For your setup this means:
$$
\begin{eqnarray*}
  \mathrm{logit}(\mathrm{E}(\mathtt{Proportion})) & = & -1.31 + 0.004 \cdot \mathtt{Temperature} \\
  \frac{\mathrm{E}(\mathtt{Proportion})}{1 - \mathrm{E}(\mathtt{Proportion})} & = & \exp(-1.31 + 0.004 \cdot \mathtt{Temperature}) \end{eqnarray*}
$$
Thus, an absolute 1-unit change in $\mathtt{Temperature}$ leads to a relative change of $\exp(0.004) \approx 0.4\%$ in $\mathrm{E}(\mathtt{Proportion})/(1 - \mathrm{E}(\mathtt{Proportion}))$.
With a bit of practice you can get a reasonable feeling for what this means in the actual expected $\mathtt{Proportion}$. If you don't have that feeling (yet), you can easily compute the effects of the changes in $\mathtt{Temperature}$, e.g.,:
nd <- data.frame(TEMPERATURE = seq(-150, 150, by = 50))
nd$Proportion <- predict(b, newdata = nd)
print(nd)
plot(Proportion ~ TEMPERATURE, data = nd, type = "b")

to check what the absolute changes in Proportion are for certain absolute changes in TEMPERATURE.
