8
The proper interpretation of the do notation is that the expression $\operatorname{do}(X=1)$ means you are forcing $X$ to have the value $1.$ You are intervening to make that happen. In the Directed Acyclic Graph (DAG) context, the notation $\operatorname{do}(X=1)$ means you do three things:
Delete all arrows going into $X.$
Replace the node $X$ with the ...
4
Unbiasedness of the within and between group mean squares depends only on second-moment assumptions and does not require normality. If the form of the ANOVA linear model is correct (first-moment assumption) and if the responses are uncorrelated and have equal variances (second-moment assumption) then the within and between group variance estimates are ...
3
The distribution for $\theta$ conditional on $s$ (and $n$) is
\begin{equation}
p(\theta|s) = \textsf{Beta}(\theta|s+1,n-s+1) .
\end{equation}
The problem is that we don't observe $s$.
Instead we are given $\pi = (\pi_1, \ldots, \pi_n)$, where
\begin{equation}
p(x_i|\pi_i) = \textsf{Bernoulli}(x_i|\pi_i)
\end{equation}
and
\begin{equation}
p(x|\pi) = \...
2
Here is a plot of PDFs of $\mathsf{Binom}(12, 0.4),$ corresponding to $H_0,$ and $\mathsf{Binom}(12, 0.6),$ corresponding to $H_a.$
R code for figure:
x = 0:20; PDF = dbinom(x,20,.4)
PDF.a = dbinom(x,20,.6)
hdr = "PDFs of BINOM(20, .4) [blue] and BINOM(20, .6)"
plot(x-.1, PDF, type="h", xlab="x", col="blue", lwd=2, ...
2
One simple example that makes this pretty clear is comparing a drug and a placebo on some continuous outcome. You can look at the null hypothesis that the means $\mu_\text{treatment}$ are the same
$$H_0: \mu_\text{drug} = \mu_\text{placebo}$$
vs. the alternative hypothesis that they are different
$$H_A: \mu_\text{drug} \neq \mu_\text{placebo}.$$
That's the ...
1
Yes, that’s correct.
Remember that the likelihood function varies the parameters, not the data. (See What is the difference between "likelihood" and "probability"?) The likelihood is a function of $\theta$ that assumes $\text{Data}$ is fixed, i.e.
$$
L(\theta \mid \text{Data}) = p(\text{Data} \mid \theta)\text{.}
$$
What you have is now ...
1
Your question raises a number of important issues about detecting when two datasets come from different distributions. I will discuss several
situations and several possible tests, but this answer is only an
introduction to a large and complicated topic.
(1) In addition to the tests mentioned in your Question, you could also test to see if the two samples ...
1
Two points here. First, as regards an "infinite" population, you are really referring to a data generating process, one which can produce an infinite number of potential (think "future" if it helps to conceptualize) female responses, and one which can produce an infinite number of human responses. In this setting, the fact that female ...
1
Despite its "nonparametric" flavor, you have to make some assumptions when you bootstrap. To get valid asymptotic results, you have to bootstrap in a way that mimics the true data-generating process. For example, standard bootstrapping is done by sampling independently, so you are making a strong assumption that the true data-generating process ...
1
I think for a more specific answer we probably need you to give us more details about the data, the intervention, other possible data sets, etc. However, based on the information in the question I will try to give you an idea of what an answer might look like.
TLDR; it depends on how strong assumptions you are willing to make. i.e. are you willing to write a ...
1
A few thoughts beyond what BruceET provided in an extensive answer (+1).
First, the calculator page you link itself warns with respect to their method:
If, the sample proportion is close to 0 or 1 then this approximation is not valid and you need to consider an alternative sample size calculation method.
A reason is that the normal-approximation formula ...
1
In your situation, with very small values of $p,$ you
need to be careful about using normal approximations--either for testing or for determining sample size to
achieve a desired power.
Suppose your null hypothesis is $H_0: p=.01$ and $H_a: p > .01$ and you want
power .8 or .9 against the specific alternative
$p_a = 0.04.$ Below are results from a ...
1
The statement "$X_n/n$ converges in probability to $X/n$" does not have any precise meaning, because the claimed limit, $X/n$, depends on $n$.
It is true that
$X_n/n\stackrel{p}{\to}0$ and $X/n\stackrel{p}{\to}0$ so the two sequences have the same limit (your first approach)
$X_n/n-X/n\stackrel{p}{\to}0$ (your second approach)
These aren't ...
1
In my understanding MC-Dropout is for testing only. Arguably, though, its a final step in inference, you do it once right at the very end to generate an ensemble. So, it can be included in inference in a backhand way as the final step.
MCMC and variational methods are for inference. The "posterior" of a variable is an abstract theoretical ...
1
The $R^2$ of the full model is greater than the one the short model; this holds by construction, so it is no so informative result. The F test, on the full model, about the parameters not included in the short, tell us a bit more because if the F test is not significant we have, at least in certain sense, evidence that full model is overfitted.
Indeed $R^2_{...
1
The argument of "non-repeatability" is a red herring promoted by those on the extreme Bayesian side. After all, both frequentists and Bayesians start at precisely the same place: a model for potentially observable data given by $P(Y|X, \theta)$. This model states that the data we see is but one realization of a (potentially infinite) set of ...
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