I am stuck at the follwoing problem:

Consider the exponentially distributed RVs $X_1, X_2, \ldots, X_9$ with parameter $\lambda$. We reject $H_0: \lambda \ge 1$ in favor of $H_A: \lambda < 1$ if $\overline{X} \ge k$.

We want to reach the significance level $\alpha = 0.05$. Find $k$ and calculate the powerfunction at $\lambda = 0.5$.

Since the sum of $n$ exponentially distributed RVs (with parameter $\lambda$) is $\Gamma_{n, \ 1/\lambda}$ distributed the problem reduces itself to find a $k$ such that

$$\alpha = P_\lambda(\overline{X} \ge k) = \int_{k}^\infty \frac{1}{9}\cdot\Gamma_{9, \ 1/\lambda } \ \ d\lambda$$

if I am not mistaken. But I am haveing a hard time finding a suitable $k$. Could you help me ?

  • 1
    $\begingroup$ It depends on your tools. R's qgamma and pgamma functions will give you $k$ and then the power. Or you could use a normal approximation $\endgroup$ – Henry Dec 5 '18 at 10:42
  • $\begingroup$ I am supposed to do this by hand unfortunately $\endgroup$ – 3nondatur Dec 5 '18 at 10:58
  • $\begingroup$ By parameter $\lambda$, do you mean the pdf is $\lambda e^{-\lambda x}\mathbf1_{x>0}$? $\endgroup$ – StubbornAtom Dec 5 '18 at 13:18
  • $\begingroup$ @StubbornAtom $\lambda$ is likely to be the rate or mean of the exponential distribution. Given that the critical region is associated with low $\lambda$ and higher values, I think we can assume $\lambda$ is the rate $\endgroup$ – Henry Dec 5 '18 at 14:32
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    $\begingroup$ In that case you can get $k$ as a fractile of $\chi^2_{18}$ distribution using a chi-square table, because with the above pdf, $X_i\stackrel{\text{ i.i.d }}\sim\text{Exp}\text{ with mean }1/\lambda\implies 2\lambda X_i\stackrel{\text{ i.i.d }}\sim\text{Exp}\text{ with mean }2\equiv\chi^2_2\implies 2\lambda\sum_{i=1}^9 X_i\sim\chi^2_{18}$. In other words, $18\lambda\overline X\sim\chi^2_{18}$. $\endgroup$ – StubbornAtom Dec 5 '18 at 17:25

Hypothesis test for exponential rate. Suppose $n = 60,$ so that you have a random sample $X_1, X_2, \dots, X_{50}$ from $\mathsf{Exp}(\text{rate} = \lambda).$ Then $\bar X \sim \mathsf{Gamma}(\text{shape} = n, \text{rate} = n\lambda).$

If you are testing $H_0: \lambda \ge 1$ against $H_a:\lambda < 1,$ then you want to reject when $\bar X \ge c$ where $c$ cuts probability $0.05$ from the upper tail of the null distribution $\mathsf{Gamma}(50,50).$ [Notice that large values of $\bar X$ correspond to small values of $\lambda$ because the exponential mean $\mu = 1/\lambda.]$

In R one finds $c = 1.243421.$

qgamma(.95, 50, 50) 
[1] 0.7792947

[If you are allergic to software, then you could use the relationship between gamma and chi-squared distributions to get $c$ from printed tables of the chi-squared distribution.]

Example not leading to rejection. For example, suppose I have a sample x with $\bar X = 1.008.$

set.seed(2005)    # generate fake data with rate 1
x = rexp(50, 1)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
0.01236 0.27856 0.69518 1.00772 1.42997 6.01451 

Then you do not reject $H_0: \lambda \le 1$ because $1.008 < c.$ The p-value of this test is the probability $0.46$ under the null distribution that a mean greater than $1.008$ is observed.

 1 - pgamma(1.008, 50, 50)
 [1] 0.4587632

enter image description here

Power of the test. Suppose that in fact $\lambda = 1/2.$ Then the power of this test is the probability $0.9989$ of rejecting (getting $X \ge 1.2434)$ under the alternative distribution $\mathsf{Gamma}(n, n/2).$

Intuitively, with $n = 50$ observations, it is not difficult to tell the difference between an exponential rate of $\lambda = 1$ and an exponential rate of $\lambda = 1/2.$ (In the figure below, the two density curves have little area in common.)

1 - pgamma(1.2434, 50, 50/2)
[1] 0.9989138

Example leading to rejection As an example, suppose we have a sample y with $\bar Y = 1.573.$ Then we reject $H_0: \lambda \ge 1$ in favor of the alternative $H_0: \lambda < 1.$ because $\bar Y = 1.573 > 1.2434.$

set.seed)2018)    # generate fake data with rate 1/2
y = rexp(50, 1/2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
0.01971 0.47256 1.01072 1.57307 2.22232 8.82067 

The p-value for the Y-sample is very small (much below 5%).

1 - pgamma(1.573, 50, 50)
[1] 0.0002244243

[It is difficult to use printed distribution tables to find p-values.]

In the figure below, the p-value is the very small area under the grey null curve to the right of the vertical dotted line (at the observed value $\bar Y).$ The power of the test is the large area under the heavy black alternative curve to the right of the vertical red line (at the critical value).

enter image description here


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