I have data with density $$ f(x;b) = \begin{cases} \dfrac{2 x}{b} & \text{if $ 0 < x < b $}, \\ \dfrac{2 (x - 1)}{b - 1} & \text{if $ b < x < 1 $}. \end{cases} $$ I have to test $ H_{0}: b = 0.5 $ vs. $ H_{1}: b < 0.5 $ at the $ 5 \% $-level. If your solution makes any assumption, please give an argument why it’s justified.
I’ve used a test with the rejection region $ \{ X < 0.5 \} $, so I’ve said $$ \alpha = \textbf{Pr}(X < 0.5 \mid b = 0.5) = 1 - \textbf{Pr}(X \leq 0.5 n \mid b = 0.5). $$ I’m not sure if I have to make an assumption for the value of $ n $ here.
I’m kind of lost here. Would this be the right approach?
Having now looked more closely at this problem, I can see that this function is a triangle with base vertices at $(0,0)$ and $(1,0)$ and an apex at $(b,2)$.
If $b=0.5$ then a sample drawn from this distribution is equally likely to take a value greater than $0.5$ as it is to take a value less than $0.5$
If $b>0.5$ then a sample drawn from this distribution is more likely to take a value greater than $0.5$
If $b<0.5$ then a sample drawn from this distribution is more likely to take a value less than $0.5$
These are not assumptions - they can be derived from the function.
We have no prior knowledge of the value of $b$, but have been given the null hypothesis $H_0: b=0.5$
There has arisen some doubt regarding this value of $b$ and we have been given the alternative hypothesis $H_1: b<0.5$ to test.
If $b<0.5$ the values are more likely to be less than $0.5$ so it is appropriate to conduct a one-tailed test. We will identify a critical value $x_c$ and then take a single sample $x$ from the distribution.
If $x<x_c$ then we will accept $H_1$, but otherwise we will continue to believe $H_0$.
To make this a $5\%$ test we want $P\left(x<x_c\right|b=0.5)=0.05$
$P\left(x<x_c\right|b=0.5)=\int_0^{x_c} f(x) dx=\int_0^{x_c} \frac {2x} {0.5} dx=\int_0^{x_c} 4x dx=\left[2x^2 \right]_0^{x_c}=2x_c^2$
We want $P\left(x<x_c\right|b=0.5)=0.05$
$2x_c^2=0.05$
$x_c^2=0.1$
$x_c=\sqrt{0.1}$
$x_c=0.3162$
If on the other hand, we are going to test by taking a sample of $n$ different measurements, then it is appropriate to use the Central Limit Theorem. Before we can do that, we need the variance of this distribution:
We know that $E\left(X\right)=0.5$
$E\left(X^2\right) =\int_0^1x^2 f(x) dx=\int_0^1 x^2 \frac {2x} {0.5} dx=\int_0^1 4x^3 dx=\left[x^4 \right]_0^1=1$
$Var\left(X\right)=E\left(X^2\right)-\left[E\left(X^2\right)\right]^2=1-0.25=0.75$
The central limit theorem says:
Suppose we have a distribution with mean $\mu$ and variance $\sigma^2$.
Take a sample of size $n$ measurements from the distribution and calculate the mean $\bar x$. Each time we do this we will get a different result, but the distribution of the sample mean is $N\left(\mu,\frac {\sigma^2} {n} \right)$
We have $\mu=E\left(X\right)=0.5$ and $\sigma^2=Var\left(X\right)=0.75$ and $n=100$
So the distribution of the sample mean is $N\left(0.5,0.0075\right)$
Our test now changes to considering the size of $\bar x$. If $\bar x<x_c$ then we will accept $H_1$, but otherwise we will continue to believe $H_0$.
The critical value $x_c$ will be different for this test.
We want $P\left(\bar x<x_c\right|b=0.5)=0.05$
$\Phi \left(\frac {x_c-0.5} {\sqrt {0.0075}} \right)=0.05$
$\frac {x_c-0.5} {\sqrt {0.0075}}=\Phi^{-1} \left(0.05\right)$
$\frac {x_c-0.5} {\sqrt {0.0075}}=-1.645$
$x_c-0.5=1.645 {\sqrt {0.0075}}=-0.1425$
$x_c=0.3575$
So take your sample of size 100, work out the mean and see how it compares to 0.3575