Cfrgtkky

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

$$alpha = P_lambda(overline{X} ge k) = int_{k}^infty frac{1}{9}cdotGamma_{9, 1/lambda } dlambda$$

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

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 $bar X
sim mathsf{Gamma}(text{shape} = n, text{rate} = nlambda).$

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)

summary(x)

   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

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)

summary(y)

   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).