I'm reading this pdf on statistics. I wanted to ask the difference between sections 3.5 and 4.2.

  • In section 3.5 it is described how to compute, from a sample, an estimate of the mean with a confidence interval, using the sample mean and sample variance. This was computed using the standard way of doing it.

  • In section 4.2 it makes a null hypothesis for the value of the mean, and develops a hypothesis test for rejecting or accepting the hypothesis. The hypothesis test was based simply on the fact that the standardized variable, under the assumption of having mean $m$ as the null hypothesis, should be nearly Gaussian for large $n$ (or having a $t$ distribution for small samples).

I did not understand actually the operational difference between the two approaches, which they seem to be quite similar to me, and the conceptual difference, if there is one.


closed as unclear what you're asking by Xi'an, John, Andy, Nick Cox, Stat Mar 10 '15 at 12:34

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  • 2
    $\begingroup$ You will need to copy whatever context is necessary for your question into the Q. People are not going to want to read the pdf to answer your question. $\endgroup$ – gung - Reinstate Monica Mar 5 '15 at 17:38
  • $\begingroup$ ok I'll try to put more information, even if I already put a short description. $\endgroup$ – Thomas Mar 5 '15 at 17:39
  • $\begingroup$ I put some more descriptions $\endgroup$ – Thomas Mar 5 '15 at 17:49
  • $\begingroup$ I don't see the contradiction here. If they are only suggesting the standard approaches, what is the relevance of that specific pdf? Can you clarify your confusion? $\endgroup$ – gung - Reinstate Monica Mar 5 '15 at 17:53
  • $\begingroup$ I do not see the difference between the two approaches, this is the point. It seems to me that the second one, fixing a m value as the null hypothesis, is simply a complicated way to do the usual stuff of computing a sample mean and a sample variance, that is the first approach. I'm trying to get used to this concept of null hypothesis I'm not a statistician, but in the past I computed sample mean and sample variances... $\endgroup$ – Thomas Mar 5 '15 at 17:57

Without having read the article, there isn't a contradiction between forming confidence intervals and traditional hypothesis tests, like using a one-sample $t$-test. If a $1-\alpha\%$ confidence interval does not include the null value of the parameter of interest, a $t$-test will reject that null at the $\alpha$ level. If you are more comfortable with one of the approaches than the other, go ahead and use the one you are more comfortable with. I, for example, would typically use a $t$-test, because I've done it so often and it's so simple. However, I will make one argument for confidence intervals, which is that they can be psychologically more informative than hypothesis tests: you not only see that the null value lies outside the interval, but you can see how far outside and how narrow the interval is.

  • $\begingroup$ Thanks a lot for the answer. Can we say that the two methods compute the same quantities (sample mean and sample variance) but extract (or represent) different information? The confidence interval fixes an $\alpha$ level and says that the mean is inside the interval with a probability $\alpha$ whereas, using the null hypothesis, we get information about the exact confidence that we have for the fixed value $\mu$. Told another way, with the null hypothesis analysys, we know how large we have to make the confidence interval in order for $\mu$ to be inside the interval. $\endgroup$ – Thomas Mar 6 '15 at 9:26
  • $\begingroup$ @Thomas, the confidence interval very much does not say that "the mean is inside the interval with a probability α". It may help you to read this. I suppose the p-value from a hypothesis test does tell you how big the CI would have to be to include the null value. $\endgroup$ – gung - Reinstate Monica Mar 6 '15 at 15:10
  • $\begingroup$ wow... what a long article I need to take my time to read it. But why am I wrong? If we have a set of equally distributed independent variables $X_i$, we know, calling $M$ the sample mean, $\mu$ the population mean and $s^2$ the sample variance, that $Z=\frac{\sqrt{n}(M-\mu)}{s}$ is gaussian for big n and therefore we know $P(|Z|>c)=\alpha$. The condition $|Z|>c$ is equivalent to $M-\frac{sc}{\sqrt{n}}<\mu<M+s\frac{sc}{\sqrt{n}}$ from which we know the interval that we have to choose in order that $\mu$ be there with probability $\alpha$. What's wrong with that? $\endgroup$ – Thomas Mar 6 '15 at 15:25
  • $\begingroup$ No we don't. You might spend more time w/ the article. The observed data are fixed. The observed sample mean is a fixed value. The observed CI is fixed. None of that varies. Calculate them, wait 5 minutes & then look again; they will all be the same numbers. The pop mean is an unknown constant; it doesn't vary either. It either is w/i the CI or it isn't. $\endgroup$ – gung - Reinstate Monica Mar 6 '15 at 15:29
  • $\begingroup$ I see... very interesting I begin to glimpse the point I think !!! I'll pdf the other article and read it! statistics seems to be fascinating and full of subtle questions :) $\endgroup$ – Thomas Mar 6 '15 at 15:35

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