confidence intervals for 20 different parameters - distribution, probabilit and most probable value.

Question

I need help with the subexercise (c) in the following exercise.

A researcher is planning a study where she must calculate confidence intervals for 20 different parameters. The intervals are independent of each other and all have 95% confidence . Let N be the number of intervals that is containing it's parameter.

( a) Wat is the distribution of N ?
( b) What is the probability that all the intervals that is containing its parameter?
( c ) What is the most probable value of N ?

Solution:

(a): $N$~$bin(20,1-α)=bin(20,0.95)$. I did get this by just using the definition/prove of binomial distribution. n=20 because we have 20 parameters which gives us 20 intervals.

(b): using probability function for binomial distribution with k=20.

my question is if following is right

(c): The most probable value of N is the espected value of N, i.e $E(N)=np=20*0.95=19$. Hence the most proabable value is $N=19$

Answer 1

In general, the most probable value is NOT the same as the expected value. The expected value may not even be a possible value of the distribution. The most probable value is the $x$ that maximizes $P(x)$, while the expected value is defined as $E(x)=\sum xP(x)$.

As an example when this is not the case, consider a simple distribution with $P(1)=1/3$ and $P(2)=2/3$. Then the most probable value is $2$, while the expected value is $E(x)=1/3+2\cdot2/3=5/3$.

Another example is any binomial distribution where $Np$ is not integral.

Answer 2

Yes, the binomial distribution is unimodal so there is only one maximum, and intuitively this maximum must be the expected value provided that the expected value is an integer. Things get messier if the expected value is not an integer. But it's still doable -- what can be said is that the maximum probability value $x$ of the distribution satisfies $p(n+1) -1 \leq x \leq p(n+1)$.