I have this data plotted as a scatter plot in Excel:

enter image description here

I had done a regression in Excel, and the p value was 2.14E-05 while the R- value was 0.32. I was told the R value was too low compared to the significance of the p value, and was told to control for the dispersion of the data by running it through R with GLM with quasipoisson error.

This gave me

glm(formula = encno ~ temp, family = quasipoisson(link = log), 
    data = encnotemp)

Deviance Residuals: 
   Min      1Q  Median      3Q     Max  
-6.008  -2.431  -1.021   1.353   9.441  

            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 2.005807   0.174628  11.486  < 2e-16 ***
temp        0.029065   0.006528   4.453 1.53e-05 ***
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for quasipoisson family taken to be 10.19898)

    Null deviance: 1807.4  on 171  degrees of freedom
Residual deviance: 1620.1  on 170  degrees of freedom

Number of Fisher Scoring iterations: 5

How do I analyse this output?

The problem is that the scatterplot data is too dispersed, and I would like to make a scatterplot from the quasipoisson GLM output that shows less dispersed (more fitted) data points. Will this be possible?

  • $\begingroup$ .Do you have exact zeros? $\endgroup$
    – Glen_b
    Oct 21, 2015 at 5:18
  • $\begingroup$ Simulation from the fitted glm indicates that the quasi model doesn't describe the spread of the data well; the variance function doesn't increase quickly enough. $\endgroup$
    – Glen_b
    Oct 21, 2015 at 8:45

1 Answer 1


"I was told the R value was too low compared to the significance of the p value" -- sounds like nonsense to me.

On the other hand, some form of glm may be a good idea (but it looks to me like the spread may be increasing more than you might expect with a quasipoisson).

Note that nothing about the glm changes the spread of the data -- it only models changing spread (in a particular way). The data are still the data and if you plot them will still look as they do.

You can change the appearance of the data via a transformation. One that approximately stabilizes variance when the Poisson parameter is not very small is $\sqrt{y}$. If the Poisson parameter can take small values, you may like to try $\sqrt{y+\frac{3}{8}}$ or $\sqrt{y}+\sqrt{y+1}$ instead (it looks to me like that might well be the case that you have small values).

On the other hand, one that would linearize your fitted model would be a log (but that's only suitable if you don't have exact zeros).


Although it won't be satisfying to you, you can plot the fitted curve via

newdat <- data.frame(temp=seq(9,48,.5))
encnoglm1 <- glm(formula = encno ~ temp, family = quasipoisson(link = log), 
                     data = encnotemp)
fit <- predict(encnoglm1,newdata=newdat,type="response")

Or if you want to look at what would be a nearly constant variance if the quasipoisson were suitable:

  • 2
    $\begingroup$ It would be useful if whoever thinks my answer is wrong could post a comment so that I can see about improving it. $\endgroup$
    – Glen_b
    Oct 21, 2015 at 6:50
  • $\begingroup$ Great thank you for your reply :) I think the r value being too low was meaning that the r value (slope of the trendline) was not significant enough (apparently anything under or around 0.3 is very weak correlation) even though the p value is high. So I am not really sure what is going on here. I have tried plotting a log transformed data in R but still got a plot showing weak trend and data points that were highly spread out everywhere, but then there are a few exact zeros in the dependent variable data which I should probably fix before trying to log transform again. $\endgroup$
    – Nieve K
    Oct 21, 2015 at 7:19
  • 1
    $\begingroup$ A low $R^2$ -- even a zero $R^2$ doesn't indicate a bad model. You can get zero $R^2$ with the exact model that generated the data. Conversely a very high $R^2$ doesn't indicate that the model is suitable. Actually Cosma Shalizi (among others, including many people who answer here) says pretty much the same stuff I tell people -- take a read from p17 here $\endgroup$
    – Glen_b
    Oct 21, 2015 at 8:23
  • $\begingroup$ If you have exact zeros I'd avoid the log. If you want to fix the spread in the plot one of the square root ones should work fairly well, and you can still plot the fitted curve from the glm on it. $\endgroup$
    – Glen_b
    Oct 21, 2015 at 8:24
  • 1
    $\begingroup$ It doesn't matter whether you're talking about the correlation between data and fit or its square, since the same arguments apply. But the slope of the line is not the same as the correlation (at least not unless you standardize your variables). $\endgroup$
    – Glen_b
    Oct 21, 2015 at 11:16

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