06 sep 2018

**Observation**PatiÃ«nt is showing post traumatic symptoms**Induction**Intherapy works**Deduction**\(H_0\): P: fair coin â†’ C: patiÃ«nt is balanced**Deduction**\(H_A\): P: Unfair coin â†’ C: patiÃ«nt is unbalanced**Deduction**\(H_A\): P: data \(\neq\) EV â†’ C: is unbalanced**Testing**Choose \(\alpha\) and Power**Evaluation**Make a decision

Let's analyse the null distribution of the results.

\[ {n\choose k}p^k(1-p)^{n-k} \\ {n\choose k} = \frac{n!}{k!(n-k)!} \]

With values:

n = 10 # Sample size k = 0:10 # Discrete probability space p = .5 # Probability of head

I landed 2 times head. Can we conclude that the therapy worked?

As you can see from the distribution of healthy coins, we cannot conclude that by definition.

What we can do is indicate how rare 2 is in a healthy population.

We can see that a percentage of 5% is very rare.

- Based on the null distribution we can see that the expected value (EV is 5.)
- We can now define the \(H_0\) hypothesis: \(H_0 = 5\)
- What is the alternative hypothesis?
- The alternative hypothesis describes a situation where the therapy worked.
- We could say that the alternative hypothesis is not 5.
- \(H_A \ne 5\)
- We could also formulate our \(H_0\) and \(H_A\) more abstract:
- \(H_0:\) the patient is balenced
- \(H_A:\) the patient is unbalenced
- What criterium should we use to conclude that one would be unbalenced?
- In the social sciences this \(\alpha\) criteria is often 5%.
- I tossed 2 times head. That is more frequent than 5%.
- Therefore, we conclude that our patient is probably healthy but we can never be sure.
- My coin could still be part of the unbalenced population.

But we have no clue of what this distribution could look like.

For now let's assume the probability of landing heads for my coin is .25