# Hypothesis Testing for Exponential Distribution Mean

I'm having some trouble answering the following question:

Suppose that the lifetime of batteries produced using certain materials is exponentially distributed with parameter $$\lambda$$ (density function $$f(x) = \lambda e^{− \lambda x}$$ and that the average lifetime of batteries has always been $$3$$ hours. A manufacturer claims that the average lifetime of batteries will increase if new materials are used. However, it is expensive to add these materials in the absence of a significant evidence that indeed the manufacturer’s claim is true. So, the manufacturer uses new materials for $$10$$ batteries and calculates the average lifetime that turns out to be $$4.5$$ hours

What are the null and alternative hypotheses?

A. $$H_0 : \lambda = 3$$ vs $$H_1 : \lambda > 3$$

B. $$H_0 : \lambda = 3$$ vs $$H_1 : \lambda < 3$$

C. $$H_0 : \lambda = 3$$ vs $$H_1 : \lambda \neq 3$$

D. $$H_0 : \lambda = {{1}\over{3}}$$ vs $$H_1 : \lambda < {{1}\over{3}}$$

Is it possible to reject the null hypotheses at the $$\alpha = 0.05$$ significance level?

My first thought was that the answer is D since we're dealing with the average lifetime of the batteries, and since the mean of an exponential distribution is $$1\over{\lambda}$$, the null hypothesis in D makes sense. However, I don't quite understand why I'd be limited to a one-sided alternative hypothesis. The lifetime of the batteries not being equal to three doesn't really lend evidence in either direction.

• It's one-sided because the manufacturer wants to show that the new average lifetime is more than 3 hours. A two-sided test would be appropriate if one was interested in whether the new average lifetime is different than 3 hours. – angryavian Apr 15 at 22:51
• I guess that's why I'm confused that the alternative hypothesis is $\lambda < {1\over{3}}$ rather than $\lambda > {1\over{3}}$. If the latter is what we're trying to establish, why wouldn't that be the alternative hypothesis. – jmars Apr 15 at 22:55
• Oh wow, seems I forgot how fractions work. Sorry, thank you – jmars Apr 15 at 22:56
• Any clue as to the possibility of rejection the null hypothesis a 0.05 significance? – jmars Apr 15 at 23:03

One hypothesis is that the new batteries will last longer, not just differently. We are looking for a change in a specific direction. If the new batteries are worse than the old ones, there would be no reason to switch. The only reason to switch will be if the new batteries are an improvement. So this is a one-tailed test.

If the default mean is $$3 = 1/\lambda$$, then $$\lambda = 1/3$$ is the null hypothesis. If the mean is greater than $$3$$, then $$1/\lambda > 3$$ so $$\lambda < 1/3$$, and that is the alternative hypothesis.

As for deciding about the hypothesis based on the data:

$$n = 10, \alpha = 0.05$$ and $$\bar x = 4.5$$, and the distribution is exponential, not normal. However we are taking a sampling distribution, which would eventually converge to a normal distribution regardless of the original distribution for individual values. Since we only have $$10$$ data, we cannot use a $$z$$ test and would need a $$t$$ test.

For an exponential distribution, $$1/\lambda^2 = \sigma^2$$ so the standard deviation can be estimated as $$1/\lambda = 3$$ for the null hypothesis. A sample of size $$10$$ would have a standard error of roughly $$3/\sqrt{10}$$. The t-score for this particular sample would be $$\frac{\bar x - \mu}{\sigma/\sqrt{n}} = \frac{(4.5-3)}{3/\sqrt{10}} = 1.58$$ and the critical statistic $$t_{critical} = 1.833$$ based on an upper tail area of $$0.05$$, $$n-1 = 9$$ degrees of freedom, and consulting a table of values for the student-t distribution. Since the test statistic is below the critical value, we do not have enough evidence to reject the null hypothesis.

Just for fun and to compare, we can calculate a more exact answer. It turns out that the sum of $$n$$ independent exponential random variables is a gamma random variable with parameters $$n$$ and $$\lambda$$. This distribution is known to have a mean of $$n/\lambda$$ and a variance of $$n/\lambda^2$$. To get an average of exponentials, we divide by $$n$$ to find $$1/\lambda$$ is still the mean for the sample mean. The variance can be found by dividing the variance by $$n^2$$, giving $$1/(n\lambda^2)$$. With the values in the problem, this variance is $$9/10$$ and the standard deviation is $$0.94868$$, exactly the same as in the normal approximation.

Now the probability of getting a value at least as extreme as the sample mean by chance if the null hypothesis is true is the $$p$$-value. To find this, consider that if $$\bar x = 4.5$$ then $$\Sigma x_i = 45$$, and we could integrate the gamma density function: $$\int_{45}^{\infty} \frac{\lambda^n}{\Gamma(n)} x^{n-1}\cdot e^{-\lambda x} dx$$ or $$\int_{45}^{\infty} \frac{(\frac{1}{3})^{10}}{9!} x^{9}\cdot e^{-x/3} dx$$ which when solved gives a $$p$$-value greater than $$0.05$$, again leading to the conclusion that we do not reject the null hypothesis.