Why is such hypothesis testing is about the number of integers?

Question

I was trying to solve this problem:
$N$ real numbers are drawn independently from a continuous rectangular distribution on $[0,a]$. The random variable $x$ is maximum of the $N$ numbers. A hypothesis test with a significance level of 5% is carried out using the value, $x$, of $X$. The null hypothesis is that $a=1$ and the alternative hypothesis is that $a<1$. The form of the test is such that $H_0$ is rejected if $x<c$, for some chose number $c$.

Using the approximation $2^{10} \approx 10^3$, determine the smallest integer value of $N$ such that if $x \leq 0.8$, the null hypothesis is rejected.

I don't quite understand why is the number of data would determine whether the hypothesis test would be rejected. I am quite confused with what is going to be tested. The mark scheme stated that we should find:
$(0.8)^N \leq 0.05$. I don't quite understand why is that so, like why does it about the $N$. I thought that the null hypothesis could be rejected if the calculated probability is not equal to the probability if $a=1$.
Could you guys please explain this to me? Thank you very much.

Answer 1

The 5% significance level means that even if $H_0$ is factually true, our test will reject $H_0$ in 5% of all cases. Now note that the maximum of $N$ random numbers is $< c$ iff each of the $N$ numbers is $<c$, which happens with probability $\left(\frac ca\right)^N$ (provided $0\le c\le a$). Hence for the 5% level we want $c^N=0.05$.

The problem statement asks us to find the number $N$ for which this happens with $c=0.8$.

We want $\frac1{2\cdot 10}=c^N=\frac {8^N}{10^N}=\frac{2^{3N}}{{10}^N}$, or $10^{N-1}=2^{3N+1}$. Raise to the 3rd power to get $10^{3(N-1)}=2^{9N+3}$. With the suggested approximation, this becomes $2^{10(N-1)}\approx 2^{9N+3}$, and by equating the exponents we arrive at $10N-10\approx 9N+3$, so $N\approx 13$.

(Indeed, numerically

$$0.8^{13}=0.05497\ldots,$$

so the -- possibly mental! -- calculation is not far off)