# Tag Info

107

Part of the issue is that the frequentist definition of a probability doesn't allow a nontrivial probability to be applied to the outcome of a particular experiment, but only to some fictitious population of experiments from which this particular experiment can be considered a sample. The definition of a CI is confusing as it is a statement about this (...

28

In frequentist statistics probabilities are about events in the long run. They just don't apply to a single event after it's done. And the running of an experiment and calculation of the CI is just such an event. You wanted to compare it to the probability of a hidden coin being heads but you can't. You can relate it to something very close. If your game ...

27

This is referred to as current status data. You get one cross sectional view of the data, and regarding the response, all you know is that at the observed age of each subject, the event (in your case: transitioning from A to B) has happened or not. This is a special case of interval censoring. To formally define it, let $T_i$ be the (unobserved) true event ...

22

Formal, explicit ideas about arguments, inference and logic originated, within the Western tradition, with Aristotle. Aristotle wrote about these topics in several different works (including one called the Topics ;-) ). However, the most basic single principle is The Law of Non-contradiction, which can be found in various places, including Metaphysics book ...

18

Mathematically, it's not the case that these are necessarily close. It would work if it was the case that $E(Y/X^2) = E(Y)/E(X)^2$ but this is false in general and in some particular situations it might be quite far out. However, for a fairly realistic set of bivariate height and weight data it looks like the impact will be small. For example, consider ...

17

The concept of significance or hypothesis testing is not relevant for a whole population. Hypothesis testing is based on the assumption that you deal with a sample from a (usually) infinite population, and asks the question: what is the probability that we have drawn the sample by chance from a population that fulfills the assumptions of the null hypothesis? ...

15

This question goes to the heart of what statistics is and how to to conduct a good statistical analysis. It raises many issues, some of terminology and others of theory. To clarify them, let's begin by noting the implicit context of the question and go on from there to define the key terms "parameter," "property," and "estimator." The several parts of the ...

12

Why does a 95% CI not imply a 95% chance of containing the mean? There are many issues to be clarified in this question and in the majority of the given responses. I shall confine myself only to two of them. a. What is a population mean? Does exist a true population mean? The concept of population mean is model-dependent. As all models are wrong, but some ...

11

I'm surprised that no one has brought up Berger's example of an essentially useless 75% confidence interval described in the second chapter of "The Likelihood Principle". The details can be found in the original text (which is available for free on Project Euclid): what is essential about the example is that it describes, unambiguously, a situation in which ...

11

It is not the case that one uses R-squared for entire populations and adjusted R-squared for samples. They each give different information. R-squared is the proportion of variability in the data accounted for by your model. Adjusted R-squared takes into account (i.e., adjusts for) the number of explanatory terms in your model. R-squared can never be ...

11

As with many questions on definitions, answers need to have an eye both on the underlying principles and on the ways terms are used in practice, which can often be at least a little loose or inconsistent, even by individuals who are well informed, and more importantly, variable from community to community. One common principle is that a statistic is a ...

10

It's not completely correct, but it will usually not make a huge difference. For example, suppose your population has weights 80, 90 and 100kg, and is 1.7, 1.8 and 1.9m tall. Then the BMIs are 27.68, 27.78 and 27.70. The mean of the BMIs is 27.72. If you calculate the BMI from the means of the weights and heights, you get 27.78, which is slightly different, ...

9

Generally, when one has only a fraction of the population, i.e. a sample, you should divide by n-1. There is a good reason to do so, we know that the sample variance, which multiplies the mean squared deviation from the sample mean by (n−1)/n, is an unbiased estimator of the population variance. You can find a proof that the estimator of the sample variance ...

9

A simple way to verify that the method matters is to choose particular probabilities for types of marbles, and calculate the chance of each subset according to some methods. This can't prove that the method doesn't matter, though. Suppose there are $3$ types and the chances of each type are $1/2$, $1/4$, and $1/4$, respectively. Suppose you are choosing $2$ ...

9

The first question is one that has no generally agreed upon answer. My own view is like yours, but others have argued that a population can be viewed as a sample from a "super-population" where the exact nature of a super-population varies depending on context: E.g. a census of all the people living in a building could be viewed as a sample from all the ...

9

I need to correct a number of mistaken or partly misplaced ideas in the question first (as well as some that aren't in the question but are commonly seen and may be indirectly influencing the way you ask your question), but I will return to the main issue. The answer you link to says: The t-test assumes that the means of the different samples are ...

8

I don't know whether this should be asked as a new question but it is addressing the very same question asked above by proposing a thought experiment. Firstly, I'm going to assume that if I select a playing card at random from a standard deck, the probability that I've selected a club (without looking at it) is 13 / 52 = 25%. And secondly, it's been stated ...

8

I think you are misreading the paper, they do not claim what you say. Their claims are not based on number of top players, but on their ratings. If the statistical distribution of strength is the same among men and women, then the expected number of women among the top 100 is 6, if their proportion of the total population is 6%. Some citations from the ...

8

The word "population" does not refer to all living people in the world. That is the generic understanding of the word, and not the statistical understanding. Statistically, population refers to the class/group of units (or individuals in this case) about whom you want to make some inference. If the group you want to make the inference on is "women in Math ...

8

The multivariate delta method has a heuristic justification here: https://en.wikipedia.org/wiki/Delta_method#Multivariate_delta_method. For the multivariate delta method you have a specific function $f$ that takes a vector argument which is $p$ dimensional and maps this to a $k$ dimensional space. In the case of a ratio estimator $p=2$ and $k=1$. The ...

7

Suppose only 2 out of 12 committee members are women. The proportion $\frac{1}{6}$ can be taken as a statistic descriptive of the whole population (the committee). Perhaps something ought to be done to correct the imbalance, regardless of how it arose. Or it can be taken as an estimate of the probability of a woman's being selected for the committee—...

7

You are very close... If $X_1, \dots, X_n$ is a sample of i.i.d normal observations with mean $\mu$ and variance $\sigma^2$, then the standardized mean $$\frac{\bar X_n-\mu}{\sigma/\sqrt{n}}$$ is standard normal. Now, as you pointed out, in reality we never know $\sigma$. So we replace $\sigma$ by its sample estimate $S$ and consider the "studentized" ...

6

There's an extensive literature on this area in spatial ecology; but it comes down to what kind of assumptions you're willing to make. A very common method for estimating spatial distributions based on abundance data is a maximum entropy approach, take a look at the software and publications There's a variety of other algorithms designed for this purpose ...

6

This exercise will be pretty useless unless the sample of the surnames is statistically sound, i.e., a random sample with known probabilities of selection. Otherwise, you are estimating the number of female drivers by first picking a color (say yellow), counting the fraction of female drivers in the yellow cars, and then obtaining the estimate of the ...

6

It depends on to whom you wish to generalize your final results. If your sole interest was just to see how these people react and you don't care about inference, they are your population. If you wish to use the results to somehow infer how other similar people may behave under influence, then they are samples. Most studies tend to do the latter. Also, for ...

6

I tend to think of parameters by analogy by thinking about the normal distribution: $$\text{pdf}=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{1}{2}\frac{(x_i-\mu)^2}{\sigma^2}}$$ What's important to recognize about this function is that, as ugly as it is, I pretty much know what most of the parts are. For example, I know what the numbers $1$ and $2$ are, what $\... 6 By selecting data within a certain range, it no longer has the original distribution. You have changed it so it is no longer exponential (It actually becomes a doubly-truncated exponential; equivalently an exponential that's shifted and truncated. No matter how you describe it, it's not exponential.)$\hspace{2cm}^\text{Histogram of large random sample, ...

6

The confidence interval with $\frac{1}{\sqrt{n}}$ is based on the same idea as the confidence interval with $1.96\sqrt{\frac{p(1-p)}{n}}$ but is more "conservative", in the sense that it is larger. The reason for that is that the function $$f(x) = x \left(1-x\right), x \in [0,1]$$ can be shown (with elementary calculus) to be maximal for $x= \frac{1}{2}$....

6

The random variable $Y$ describes a relationship between events and the corresponding probabilities of those events. In more practical terms, a random variable describes a data-generating process. When you generate a random data point that is described by the random variable $Y$, the probability distribution of $Y$ describes the probability distribution of ...

6

There is no way - the "population" of interest is part of the specification of the problem. Statistical problems involving inference to a "population" require specification of the group of interest, about which we are making an inference. Only a proper specification of the problem ---in this case, from a briefing from the client--- can give you this. Of ...

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