I'd like to start a discussion related to Confidence Intervals (CI) and Hypothesis Testing (HT). I've read many books including the topic but I still sometimes find books and readings which establish a relationship between CI and HT.

They say that HT is based on CI to come to a conclusion. I don't agree.

Let's say, suppose we want to work out a problem like this: 30-data sample, IID Normal distributed variables, variance not know and hypothesis:

$\displaystyle H_{0}:{\mu}=0$

$\displaystyle H_{a}:{\mu}>0$

If our data show this numbers:

$\displaystyle n=30, \;\ \overline{x}=0.23, \;\ s=1.25, \;\ {\alpha}=0.05$

We would use a T-test and would write that under null hypothesis, we would have:

$\displaystyle P[T>\frac{0.23-0}{\frac{1.25}{\sqrt{30}}}]=p-value$

And this is a probability, not a CI, which have to be compared to the $\displaystyle {\alpha}$ we were given as the decision's rule. Under the same logic, if we were going to check our test through statistics, we would just write:

$\displaystyle \frac{0.23-0}{\frac{1.25}{\sqrt{30}}} \;\ ? \;\ t_{0.95,30}$

(where I wrote "?" to mean a comparison to resolve the test)

So, I come to the conclusion that CI are not related to HT because we compared the probability related to the $\displaystyle \overline{x}$ estimator and our Type-I error maximum accepted probability.

On the other hand, CI is just a tool to "presume", without any good reason, what the value of the parameter is. Despite it use all the sample, there are not a probability argument behind it. We just do a free-of-all-meaning calculation and asign some number as "confidence".

Using the same data as above, the 0.95 confidence intervel for the population mean would be:

$\displaystyle C[\overline{x}-t_{0.95,30}\frac{s}{\sqrt{30}}\;\ <= {\mu} <= \;\ \overline{x}+t_{0.95,30}\frac{s}{\sqrt{30}}]=0.95$

In this case, we do NOT work with probability. So, again, neither are CI and HT related through probabiltiy nor through same argument.

Am I wrong?

All the best,

Federico.