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I am a bit confused how we interpret confidence intervals.

For the formula $\bar{X}\pm z_{\large\frac{\alpha}{2}}\large\frac{\sigma}{\sqrt{n}}$ ,we substitute $\bar{x}_{obs}$ for $\bar{X}$, but isn't that observed sample mean $\bar{x}_{obs}$ only for one particular sample?

Also why is it incorrect to interpret a confidence interval as the probability that the actual $\mu$ lies in a interval (a,b) is $(1-\alpha)100\%$

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  • $\begingroup$ $\overline{x}_{obs}$ is the mean of a single set of samples. $\endgroup$ – Marconius Aug 9 '15 at 20:37
  • $\begingroup$ @Marconius I am still confused. Let's say the population is 20000, and we set the sample size to be 50. Then we find the average weight of it. Does $\bar {x}_{obs}$ mean the sample mean of repeatedly sampling 50 people from the population? $\endgroup$ – GalaxyVintage Aug 9 '15 at 23:37
  • $\begingroup$ I realise the wording may have been confusing. $\overline{x}_{obs}$ is just the mean for a single sample (of a set of observations/cases). $\endgroup$ – Marconius Aug 9 '15 at 23:47
  • $\begingroup$ why can we substitute $\bar{X}$ with $\bar{x}_{obs}$? $\endgroup$ – GalaxyVintage Aug 10 '15 at 0:17
  • $\begingroup$ $\overline{x}_{obs}$ is your best available estimate of the true population mean $\mu$, which you are inferring. The estimate is neither $\color{blue}{certain}$ nor $\color{red}{exact}$, hence the use of a $\color{blue}{confidence}$ $\color{red}{interval}$. $\endgroup$ – Marconius Aug 10 '15 at 1:03
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The reason why it is incorrect to interpret a confidence interval as "the probability that the actual μ lies in a interval $(a,b)$ is $(1−α)100$%" is becaue $\mu$ is fixed but unknown, not a random variable.

The actual probability would be $1$ or $0$ depending on whether $a<\mu<b$ or not.

It is the interval itself which is variable, since it depends on the actual sample obtained, so we refer to "the probability that the interval contains $\mu$" rather than the "probability $\mu$ lies in the interval".

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  • $\begingroup$ why is interval a variable? if $\bar{x}_{obs}$ is known, then shouldn't the 95% CI be fixed? $\endgroup$ – GalaxyVintage Aug 9 '15 at 21:42
  • $\begingroup$ @Lzy - I think the point is that if you took another sample, then you would most likely get a different interval, while the population wouldn't have changed. $\endgroup$ – Marconius Aug 10 '15 at 1:07
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There is something with a probability of $(1 - \alpha)$ or more, but that something has a more complex description than "is the estimate of $\mu$ correct".

The something is: "given the probability assumptions (such as normal distributions for the quantities of interest), the probability is at least that high that the estimation procedure we followed, applied to data generated in a way that satisfies the assumptions, will give an interval containing the correct answer".

In the standard accounts of probability, "is the estimate of $\mu$ correct" is not a random event to which a probability can be assigned, except in the trivial sense that the probability being $0$ or $1$ is another way of saying whether the estimate is correct or not.

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