# Hypothesis Probability

We test 5 patients for a new drug. That is we give each patient the drug and count how many get cured with the drug. Without the drug the patients have a 10% probability to cure.We found that 4 out of five got cured. Test on the 95%-confidence level, the hypothesis that the new drug has no effect.

Let $X_i$ denote the $i^{th}$ person chosen is cured, then $E(X_i) = p$. Now let Z denote the total number of people in our experiment, so $Z = X_1 + ... + X_5$. We have that 10% is the probabilty for the patient to be cured without any drugs and we found out through experiment that 20% was the actual probability that a person will be cured without any treatment. I found 20% by, $(1 - \frac {4}{5})$. Now let $P_{20\%}$ denote the probability of actual percentage. Now testing that the drug has no effect we have that $P_{20\%} (X_1 + ... X_5 \ge 10\%)$ so$P_{20\%} (\frac {Z - n \mu}{\sigma \sqrt n} \ge \frac {10- n \mu}{\sigma \sqrt n} )$ , thus $P(N(0,1) \ge \frac {10- n \mu}{\sigma \sqrt n})$.

My question is how can I look up the probability using the z table or will $P(N(0,1) \ge \frac {10- n \mu}{\sigma \sqrt n})$ suffice?

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Isn't it 80% that got cured? But +1 for showing your thoughts. –  Ross Millikan Oct 24 '12 at 3:18
Yes, you are right 80% did get cured but the question asks to find the condfidence level that the drug had no effect. Which is why I put 20 since 1-.80=.2. Or am I wrong? And thanks for the +1 :) –  Q.matin Oct 24 '12 at 3:24
Can't we just say that if the drug had no effect, the chance of $4$ cures out of $5$ is $5 \cdot 0.1^4 \cdot 0.9 =0.00045$. Maybe you should count also the chance of $5$ out of $5$, which would raise it to $0.00046$. In any case, the fact that this is less than $5\%$ says we have disproved the hypothesis at the $95\%$ confidence level.
How did you get $5 \cdot 0.1^4 \cdot 0.9 =0.00045$ ? Is it from $\begin{pmatrix} n\\k \end{pmatrix}p^k\left(1-p\right)^{n-k}$ ?? –  Q.matin Oct 24 '12 at 3:35