# Hypothesis testing and probability

I would like to know how to solve this problem correctly, and if there is something wrong.

The users can connect to 4 servers from 4 countries, we choose 200 users that act independently from each other and we see which servers they chose(to connect to a network). The obtained data is:

Servers :  A  |  B  |  C | D
Users :   100 |  20 | 30 | 50


1. Find the estimated probability that a user is going to connect to server A or B?

P(A ∪ B) = P(A) + P(B) - P(A ∩ B))

P(A ∩ B) = P(A) P(B|A)

The problem is will these formulas solve this exercice? and How??

2. We need to reject the hypothesis that the users choose a server randomly without having any preferences from any of them?(confidence level is 0.05)

H0(null hypothesis): The users choose a server randomly

P(A) = 100/200, P(B) = 20/200, P(C) = 30/200, P(D) = 50/200

H1(alternative hypothesis): The users dont choose a server randomly

P(A) ≠ 100/200, P(B) ≠ 20/200, P(C) ≠ 30/200, P(D) ≠ 50/200

Assuming H0:

U = Σ [ (Ni - Ei)2 / Ei ] ~ Chi-Square, where Ni = 100,20, 30, 50

E1 = N*P1 = 200*P(A) E2 = N*P2 = 200*P(B) ....

I just need to know if im going in the right path. I would really apreciate your help on solving this exercice.

This question is hypothesis testing for multinominal distribution.

First, the underlying assumption is that the users chose to connect to either one (and only one) of the servers. Knowing this, $P(A \cap B) = 0$. The answer to your first question would be just $P(A) + P(B)$, and those two probabilities are directly estimated from the observed data.

You might have set up the hypothesis incorrectly. The $H_0$ and $H_A$ are going to be:
$$H_0: \; P(A)=P(B)=P(C)=P(D)=0.25 \\ H_A: \; \text{not all of them are equal}$$
Under $H_0$, the test statistic $$X^2 = \sum_{i=1}^{4} \frac{(n_i - n\pi_{i0})^2}{n\pi_{i0}} \ \sim \chi^2_{3}$$ where $\pi_{i0} = 0.25, i=1, 2, 3, 4$. $n_i$'s are the observed counts.
The degrees of freedom is 3 because that under $H_0$ there is no degrees of freedom, and under $H_A$ there are 3 degrees of freedom (one constraint that all probabilities sum up to 1). So the difference of df = 3.