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Math Help - Confidence Intervals and Hypothesis Testing - Are related or not?

  1. #1
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    Confidence Intervals and Hypothesis Testing - Are related or not?

    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:

    H_{0}:{\mu}=0

    H_{a}:{\mu}>0

    If our data show this numbers:

    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:

    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 {\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:

    \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 \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:

    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.
    Last edited by falcald; April 2nd 2008 at 07:27 AM. Reason: heathrowjohnny remark - many thanks
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  2. #2
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    you should divide by  \frac{s}{\sqrt{n}} for the t-statistic. You divided by  s .
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  3. #3
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    I corrected, but I couldn't colour the changes to show your remark.
    Hate when this happens...

    Federico.
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