Confidence interval example

Question

I read this example in Arthur Engel's Certainty of Chance book:

In Frankfurt, cabs are numbered $1, 2, ..., T$ but we don't know $T$. A cab driver claims that $T \geq 3000$. In the street we observe $4$ cabs with following numbers: $N_1 = 512$, $N_2 = 987$, $N_3 = 355$, $N_4 = 1200$. Let $M = \max(N_i)$, Hypothesis A = $T \geq 3000$ and Hypothesis B = $T < 3000$.

Under hypothesis A:

$$P(M\leq 1200) \leq \left(\frac{2}{5}\right)^{4} = 2.56\%$$

So with a $5\%$ confidence threshold we can reject hypothesis A and say $T<3000$

Ok I get it that our observations seemed to be too rare under hypothesis A and we reject it in favor of hypothesis B. But why necessarily $T<3000$ constitutes a better hypothesis. What can we say about $P(M\leq 1200)$ under hypothesis B?

The example continues with how to build an estimate $\hat{T}$ of T and a confidence interval:

Suppose we observe $n$ cabs. Let's sort their numbers:

$$X_1 \leq X_2 \leq X_3 \leq ... \leq X_n = M $$

The unknown distance $T-M$ can be estimated by the mean of the distances $X_1 - 0$, $X_2 - X_1$, ..., $X_n - X_{n-1}$, namely $\frac{M}{n}$.

Hence:

$$\hat{T}-M = \frac{M}{n}$$ $$\hat{T} = \left(1-\frac{1}{n}\right)M$$

With previous example $\hat{T} = \left(1-\frac{1}{4}\right)1200 = 1500$.

Let's now build a confidence interval by using the $5\%$ confidence threshold, i.e build an interval such that the probability that it contains $T$ is $95\%$. The probability that all cabs have a number $\leq M$ is:

$$ P = \left(\frac{M}{T}\right)^{n} $$

If we impose $P>5\%$ we get: $$\left(\frac{M}{T}\right)^{n} \geq \frac{1}{20}$$ $$T\leq M\sqrt[n]{20}$$

Hence: $$P(M\leq T \leq M\sqrt[n]{20}) = 95\%$$

With the previous problem ($M=1200$, $n=4$) we obtain: $$1200\leq T\leq 2538$$

I'm fine with the $\hat{T}$ estimate and included it just because it might interests others. But he lost me with the confidence interval: how the $5\%$ confidence threshold translates to a $95\%$ confidence interval? I suspect the book to be poorly translated and that doesn't help the understanding but still, what is your take on that example?

Answer 1

Why necessarily $T<3000$ constitutes a better hypothesis?

Well, it's the only other possibility, right? Either $T\ge 3000$ or $T<3000...$ there are no other possibilities.

What can we say about $P(M≤1200)$ under hypothesis B?

Well, they computed $P(M\le 1200) = 2.56\%$ for $T=3000,$ so since this probability will only get bigger as $T$ decreases, we can say $P(M\le 1200) >2.56\%$ under hypothesis $B.$ This is a little confusing since both hypthesis $A$ and $B$ have many different possible values of $T$ each of which has a different value of $P(M\le 1200).$

But he lost me with the confidence interval: how the 5% confidence threshold translates to a 95% confidence interval?

A 95% confidence interval is an interval $I(M)$ constructed from the test statistic $M$ with the property that $$ P(T\in I(M)) \approx 95\%.$$ Notice such a thing is not unique.

Here they have decided to set the lower limit of $I(M)$ at $M$ and to choose the upper limit large enough so that $P(T\in I(M))=95\%.$ They calculate $$ P(M\le T \le M\sqrt{20}) = P(T\le M\sqrt[n]{20}) \approx 95\%,$$ since first off, it's always the case that $M\le T,$ and second, we have $$ P(M\ge T/\sqrt[n]{20}) = 1-P(M< T/\sqrt[n]{20}) = 1 - \left(\frac{T/\sqrt[n]{20}}{T}\right)^n = 1- \frac{1}{20} = 95\%$$ So by the above definition, $[M, M\sqrt{20}]$ is a $95\%$ CI for the parameter $T.$