Hypothesis testing and probability

Question

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.

Answer 1

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.

As for your second question,

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)

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.