enter image description here

command for the graph:


I generated this sequence in R:

c1<-rexp(n=1000, rate=lambda1)

and did a z.test with a significance value of 5%:

z.test(c1,mu=98, alternative = "two.sided", sigma.x = sd(c1), conf.level = 0.95)

this is the output:

data: c1

z = 0.51946, p-value = 0.6034
alternative hypothesis: true mean is not equal to 98
95 percent confidence interval:
  93.43883 105.85080
sample estimates:
mean of x 

What I need to find out is if indeed the values follow a exponential distribution but I'm not sure.

  • 1
    $\begingroup$ A z-test doesn't tell whether data follow any particular distribution. Use goodness-of-fit tests, such as chi-squared or Kolmogorov-Smirnov. $\endgroup$
    – whuber
    Dec 5, 2017 at 17:54
  • $\begingroup$ It's useful to keep in mind that even a goodness of fit test doesn't actually tell you whether data follow a particular distribution; they sometimes tell you that the data are not consistent with some distribution but failure to reject doesn't mean that distributional model is what the data were actually drawn from. In general data will be consistent with an infinite number of distributions. $\endgroup$
    – Glen_b
    Dec 6, 2017 at 1:18

2 Answers 2


You defined c1 as a sample from an exponentially distributed population (in the line c1<-rexp(n=1000, rate=lambda1)), so that population must indeed be exponentially distributed.

(A sample itself can't be exponentially distributed; the only distribution a sample can be said to have is its empirical distribution, which must be discrete.)

  • 1
    $\begingroup$ But the graph does not look like an exponential, looks more like a log normal $\endgroup$ Dec 5, 2017 at 17:42
  • $\begingroup$ @Joaoricardo Edit your question to include the command you used to draw the plot. $\endgroup$ Dec 5, 2017 at 17:47
  • $\begingroup$ I included teh code for the graph $\endgroup$ Dec 5, 2017 at 17:49
  • $\begingroup$ (+1) It looks mighty exponential to me. The density estimate you computed, on the other hand, is bogus because it doesn't understand that $0$ is a left limit. See stats.stackexchange.com/questions/65866 for what to do about it. (Gavin Simpson posted a nice solution.) $\endgroup$
    – whuber
    Dec 5, 2017 at 17:52
  • 1
    $\begingroup$ Do follow the link previously given by @whuber, as it is detailed and informative. $\endgroup$
    – Nick Cox
    Dec 5, 2017 at 18:14

I've got a dataset with similar data "waiting time in seconds", which belongs to an exponential distribution according to graphs. I have tried normal, lognormal as well, but it fits best an exponential distribution.


    #compare it with a theoretical normal distribution curve
    add=TRUE, col="blue", lwd=2)
    legend("topright",col=c("blue","red"),legend =c("estimated normal density curve","kernel density curve"),lwd=2, bty = "n")

enter image description here

Let's fit data to an exponential distribution to the data and check it graphically

    fit.exp <- fitdist(wtime, "exp")

enter image description here

The second and third graph look convincing

Let's fit now the histogram, density curve and exponential curve together

    fit.exp#get the estimated rate: 0.03482814
    hist(wtime,probability = TRUE)
    curve(dexp(x, rate = 0.034828136), col = 3, lty = 2,lwd=2,
    legend =c("estimated exponential density",
             "kernel density"),
    lwd=2, bty = "n")

enter image description here

  • $\begingroup$ I'm afraid your data don't behave even remotely like they are drawn independently from an exponential distribution. The histogram obscures the large spike at zero (over a quarter of the data are zeros), which provides very strong evidence against that exponential hypothesis. This problem is very clear in the P-P plot. $\endgroup$
    – whuber
    Jul 30, 2018 at 14:15
  • $\begingroup$ I see, I was nearly convinced they followed an exponential distribution. Data aren't normal, neither log normal, which are the other options? Sorry if I am adding more confusion to this post, non intended. $\endgroup$ Jul 31, 2018 at 18:26
  • $\begingroup$ Sometimes you don't need to fit a distribution at all. If you do, an attractive option in this case would be a "zero-inflated" distribution or (more generally) a mixture of simple distributions. $\endgroup$
    – whuber
    Jul 31, 2018 at 21:05
  • $\begingroup$ Thanks! Yes, that's a very good point, no need to get obsessed about fitting a distribution. Zero inflated distributions look challenging though. $\endgroup$ Aug 1, 2018 at 17:03

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