# Hypothesis testing based on 5 samples

I am reading some books about hypothesis testing, but I am not sure if my following reasoning makes sense:

Assume I have a gaussian random variable $X \sim N(\mu, \sigma)$ with $\sigma=1$. Now I obtain 5 iid samples, $x_1, \cdots, x_5$.

I want to check if $\mu<0$. So I set up the null hypothesis and alternative hypothesis to be $H_0: \mu=0$ and $H_1: \mu<0$

Therefore, for each sample $x_i$, I can compute the p-value = $P(X_i\le x_i)$, and denote it by $\alpha$. Therefore, I have $(1-\alpha)$ confidence to reject $H_0$. Also, by using the rule $(X\le x_i)$ to reject $H_0$, I have type-I error equal to $\alpha$.

Now based on 5 samples, I have 5 p-values $\alpha_i$, ($i=1, \cdots, 5$). Therefore, I have $P(X_1\le x_1, \cdots, X_5\le x_5) = \alpha_1 \times \alpha_2 \times \cdots \times \alpha_5$. Therefore, using the decision rule that $(X_1\le x_1, \cdots, X_5\le x_5)$ to reject H0, I have a type-I error equal to $\times \alpha_1 \times \cdots \times \alpha_5$. And therefore, I have $(1-\times \alpha_1 \times \cdots \times \alpha_5)$ confidence to reject $H_0$.

Basically, I want to use the 5 $x_i$'s for future testing. Next time I obtain 5 samples, I'll compare them to $x_i$. And I want to see how much confidence level this decision rule gives me. It seems to me that textbooks usually compute the p-value for one sample xi. I am basically trying to compute the "p-value" for 5 samples.

Is the above reasoning correct?

• Hmmm...How would you interpret the negative confidence that arises if, perchance, $5! \alpha_1 \times \cdots \times \alpha_5 \gt 1$? (This will happen if, for instance, all five $x$'s exceed $-0.3\sigma$, which occurs almost $9$% of the time by chance.) – whuber Sep 9 '12 at 22:23
• @whuber, that's a good point. I don't know how to interpret that, but maybe I can change my decision rule to if $(X_1\le x_1, \cdots, X_5\le x_5)$, then reject $H_0$. In that case, I have a type-I error of $\alpha_1 \times \cdots \times \alpha_5$ which is always smaller than 1, and never leads to negative confidence. – calbear Sep 9 '12 at 23:02
• I'm confused about what you're doing. Didn't you indicate that the $x_i$ are your sample results? Doesn't that constitute your entire dataset? Then what does $X_i\le x_i$ mean? What do the $X_i$ represent? – whuber Sep 10 '12 at 15:43
• sorry for not explaining well. $x_i$ is the i-th observed sample value, $X_i$ is the random variable of the i-th sample. $X_i$'s are iid, and satisfies $N(\mu, \sigma)$. – calbear Sep 10 '12 at 16:17
• OK, you definitely need to edit your question to incorporate this last clarification: otherwise you will likely get answers that don't address your situation and so they might confuse you. The name for what you are doing, by the way, is "prediction limit." – whuber Sep 10 '12 at 16:41