How to Test the Mean of an Exponential Distribution

If you have a sample from an exponential distribution and want to test a proposed population mean, you can use an exact test rather than relying on a normal approximation. The method follows from the distribution of the sample sum.

Assume throughout that $X_1,\ldots,X_n$ are independent observations from an exponential distribution with mean $\mu$. Its density is

The exact test statistic

Suppose the null hypothesis is $H_0:\mu=\mu_0$. Under this null, each observation has mean $\mu_0$, and the sum $S=\sum_{i=1}^n X_i$ has a gamma distribution with shape $n$ and scale $\mu_0$. Equivalently, the scaled sum has a chi-square distribution:

The degrees of freedom are $2n$, not $n$. This is an exact result for exponential data; it does not require a large sample.

Choose the alternative before calculating the p-value:

• For $H_1:\mu>\mu_0$, unusually large sums count against the null. The p-value is $P(\chi^2_{2n}\ge T_{\mathrm{obs}})$.
• For $H_1:\mu<\mu_0$, unusually small sums count against the null. The p-value is $P(\chi^2_{2n}\le T_{\mathrm{obs}})$.
• For a two-sided alternative $H_1:\mu\ne\mu_0$, an ordinary exact choice is the equal-tail p-value: twice the smaller of the two tail probabilities, capped at 1.

There is more than one possible definition of a two-sided test for a discrete distribution. The chi-square distribution here is continuous, but different two-sided rules can still differ. State the rule you use; equal-tail testing is a common, straightforward choice.

Worked example

Suppose $n=5$ independent waiting times have a total of $S=24$ time units. Test $H_0:\mu=2$ against $H_1:\mu\ne 2$ at significance level $\alpha=0.05$.

First calculate the statistic:

Under the null, $T$ follows $\chi^2_{10}$. For an equal-tail test, the 5% critical values are approximately $3.247$ and $20.483$. Since $24>20.483$, reject $H_0$.

The upper-tail probability is approximately $0.0076$, while the lower-tail probability is about $0.9924$. The equal-tail two-sided p-value is therefore approximately $2(0.0076)=0.0152$. This is below $0.05$, so the data provide evidence that the mean differs from 2; because the observed sum is high, the sample points toward a larger mean. The sample mean is $24/5=4.8$.

Using a confidence interval instead

The same pivot produces an exact confidence interval for the mean. Let $\chi^2_{\nu,p}$ denote the $p$ quantile of a chi-square distribution with $\nu$ degrees of freedom (the probability to its left is $p$). A two-sided $100(1-\alpha)\%$ interval is

For the example, a 95% interval uses the 97.5th and 2.5th percentiles of $\chi^2_{10}$:

The null value $2$ is outside this interval, consistent with rejecting the two-sided null at the 5% level. A one-sided test corresponds to a one-sided confidence bound rather than this two-sided interval.

If the parameter is the rate

Some texts write the exponential density as $f(x;\lambda)=\lambda e^{-\lambda x}$, where $\lambda$ is the rate. The mean is $\mu=1/\lambda$. To test a proposed rate $\lambda_0$, use

Be careful about the direction: a larger rate means a smaller mean. Thus, testing $\lambda>\lambda_0$ corresponds to sums that are unusually small, not large.

Common mistakes

• Using the wrong degrees of freedom: the statistic based on the sum has $2n$ degrees of freedom.
• Mixing up mean and rate: the mean is $1/\lambda$ in the rate parameterization.
• Choosing the tail after seeing the data: specify whether the alternative is greater, less, or different before interpreting a p-value.
• Assuming the model fits automatically: the exact result assumes independent, identically distributed exponential observations with complete measurements. Censoring, dependence, or a poor exponential fit requires a model and test that account for those features.
• Treating a non-significant result as proof: failure to reject means the data did not provide enough evidence against the proposed mean; it does not establish that the mean equals $\mu_0$.