6
As Pohoua commented, your understanding is correct (but I would say not entirely*). Concepts like confidence intervals, p-values, and hypothesis tests are not calculated from the likelihood $f(\theta|x)$ with $x$ fixed, but instead with the pdf $f(x|\theta)$, where $\theta$ is fixed, which is a different slice of the joint distribution $f(x,\theta)$....
6
I think you can simplify your specific problem for the asymptotic case. The normal distribution is summarised by two sufficient statistics, so the data can be reduced to six numbers. These are the two sample sizes $n_y,n_x$ and the mean and variance given as
$$\overline{y}=\frac{1}{n_y}\sum_{i=1}^{n_y}y_i$$
$$s^2_y=\frac{1}{n_y}\sum_{i=1}^{n_y}(y_i-\...
6
For consistency, by the weak law of large numbers $\bar X_n \stackrel{\text p}\to \frac 1\lambda + \theta$ and $X_\min \stackrel{\text p}\to \theta$ so by Slutsky
$$
\bar X_n - X_\min \stackrel{\text p}\to \frac 1\lambda.
$$
By assumption $\lambda > 0$ so the map $x \mapsto x^{-1}$ is continuous, and the continuous mapping theorem finishes the job.
For ...
5
Statistics uses inductive inference while logic uses deductive inference.
If you've taken a logic class, you likely know about syllogisms. For instance
All men are mortal
Socrates is a man
Socrates is mortal
This is an instance of deductive inference; If the premises are true the conclusions must be true. However, in statistics (well......
5
We need to be precise about the terms "frequentist" and "Bayesian", because they are ambiguous. "Frequentism" can be understood as adhering to a specific interpretation of the meaning of probability, which doesn't necessarily imply that any specific methodology needs to be applied. In this sense one can be a frequentist without ever computing confidence ...
5
1) No, this is not an appropriate way to estimate the time to an event. Imagine you changed your scale from weeks to days, or seconds. You would automatically inflate how much data you have, and so the estimated probability would tend to zero. Your outcome shouldn't change based on what scale you measure it on in this way. Tools from survival analysis are ...
4
Survival models are applicable even if there is no censoring. (Though, at first thought, the game ending with the pitcher still playing is a censored event, no?)
Your dataset is going to be time-varying though, since the pitcher accumulates "wear & tear" over the game, and this determines when they exit. Something like:
id time cumulative_throws ...
4
Broadly speaking, Bayesian analyses satisfy the so-called likelihood principle, which means that all the information about parameters $\theta$ from an experiment that observed $X^\star$ is contained in the likelihood
$$
L(\theta) \equiv p(X^\star | \theta),
$$
which is crucially only evaluated at the observed $X^\star$.
Contrast this with the sampling ...
4
$p=1$ precisely is not impossible. Possible values of the Wilcoxon statistic are discrete, so it is possible (if unlikely, but not less so than $p=0.99999$ in standard continuous testing) that the statistic takes the value most compatible with the null hypothesis precisely, leading to $p=1$. (I haven't checked against your data whether this is indeed the ...
3
Because the posterior (or, more precisely, the posterior mean that you state here) is formed by weighting over the contribution from the prior and the observations.
Hence, if $n$ increases, the contribution of any particular "head" to the total estimate decreases. When you move from $n=1$ to $n=2$ the first observation still is pretty important to form the ...
3
Although you are also asking about the estimator $\hat{\lambda}$, I am going to note some things about $\hat{\theta}$. In this particular case it is quite easy to obtain the exact distribution of this estimator. Since you have a series of shifted exponential random variables, you can define the values $Y_i = X_i - \theta$ and you then have the associated ...
2
That is a large topic, and the quality of the approximation must be studied, generally, on a case-by-case basis. That's why simulation is a useful approach. So the best way to answer your question is by an example, which you can adapt for your data and models.
So I will simulate, in R, some data for a logistic regression, and I will simulate binomial ...
2
All current approaches to handling multiple comparisons focus on (read: are equivalent to) using significance tests and modifying thresholds based on ranked unadjusted p-value, the overall number of tests, and either the prespecified family-wise error rate or the number of false discoveries, etc. It seems necessary to invoke null hypothesis significance ...
2
It helps to write out the full equations of the models implied by the standard formula shorthand.
Model 1
For Q1, restricted to those taking thiazide:
$$\sf{Na_{fu}} = \beta_0 + \beta_1 \sf{S} + \beta_2 \sf{Na_{bl}} + \epsilon.$$
Here $\sf{Na}$ is the sodium concentration at follow-up (fu) or baseline (bl), $\sf{S}$ is a variable taking the value 0 for ...
2
1) Exhaustive? Nope.
2) $\bar{x}$ is the MOM estimator and happens to be unbiased. Being unbiased and MOM do not go together, however.
3) The $n-1$ adjustment comes from taking the expected value of the MLE and finding that there’s an easy correction to get an unbiased estimator. When you take the square root to get what you might think is an unbiased ...
2
Given a particular parametric assumption, and a suitable test statistic, if you can compute the distribution of the test statistic under the null, you can perform a hypothesis test.
(Simulation could be used to compute p-values where such a calculation is not tractable/convenient.)
The main issue is then "Given a parametric distributional assumption, how ...
answered Jun 2 at 14:55
Glen_b -Reinstate Monica
236k25 gold badges484 silver badges842 bronze badges
2
With enough data (whatever "enough" means), a t-test will suffice. Briefly, the central limit theorem says that the sampling distribution of the sample mean is normal with mean $\mu$ and variance $\sigma^2$. Because we have to estimate $\sigma$, that means we can use the t test to test for a difference in means.
The story is different when we don't have "...
2
The general English meaning is pretty broad
a belief or opinion that you develop from the information that you know
In both cases, logic and statistics, broadly speaking, it’s about “learning” something. Sometimes different disciplines just borrow same words from common language.
1
You can check it numerically from the following Mathematica code:
data = RandomVariate[WeibullDistribution[3.5, 2], 50]
{1.46493, 1.60199, 2.41101, 1.64718, 1.41811, 1.51449, 1.65843, \
1.07234, 2.09288, 1.51687, 1.89899, 2.34875, 2.46311, 2.53133, \
2.03461, 2.31118, 2.77079, 2.33359, 2.20265, 1.19708, 1.61877, \
1.27819, 2.01383, 2.54109, 1.2091, ...
1
Despite apparently different titles and focus, the issues this raises are very similar to those in the recent thread Deleting outliers based on diagnostic plots is not working as intended (regression model) - data added
If you smooth say the absolute residuals you may well find its pattern approximately flat, indicating rough homoscedasticity. (In Stata, ...
1
A random experiment is an experiment or a process for which the outcome cannot be predicted with certainty (e.g. http://aix1.uottawa.ca/~glamothe/mat2377/ProbabilityI.pdf).
In statistical inference a random sample is a random variable $X_1,\dots,X_n$ if $X_1,\dots,X_n$ are independent and identically distributed random variables (e.g. Casella and ...
1
As discussed in the comments, this is not exactly a multi-armed bandit problem. In multi-armed bandit you know the rewards only after you "pull the arm" of your slot machine. For example, if you are running online ad campaign and you want to test between different ads, then you can only one add to user at a time and you don't know the rewards that you would ...
1
... I want to know which characteristic make the product more prone to be sold.
...
And I want to know if the product being yellow increase the probability of selling it and if it does, if it's more important than the material.
Is that something I could do with random forest and other models? And is there anything I should look at that we ...
1
Random Forest Regressor/Classifier is an appealing option, because:
It is very fast and easy to setup and train (especially with the Sklearn package).
It handles both categorical and numerical features.
The "feature importance" property provides a really nice analysis tool.
it requires less preprocessing than other supervised algorithms.
I am not ...
1
A warning is not an error report. I don't think the low expected frequencies implied here are really problematic.
Let's first note what should be explicit, which is that the chi-square test result is based on omitting observations that are NA on either variable, which does seem a very good idea. However, the row and column percents presented include NAs, ...
1
No, your understanding is not correct.
First, frequentist statistics does not allow us to "test to see what is the % chance the population mean falls within some range, using the sampling distribution". More precisely, frequentist statistics do not make probability statements on the population mean --- they only make probability statements on the estimates ...
1
Using Likert scores, lots of people pretend to have interval data and test these scores using t tests. Using 'number of 10's minus number of below-7's leaves you with only one number per group of 1000.
Two ideas:
(1) Chi-squared homogeneity test. You might admit that the data are fundamentally categorical, make a contingency table with two rows A & B ...
1
You do not say how the entries in your table arose. I explore two scenarios.
Assuming categories 1 through 6 are bins into which 44 values have been sorted: You have $12$ observations below the 5th bin and $16$ above. If there is a 50-50 chance of falling above or below that bin, then the probability that $12$ or fewer out of $12 + 16 = 28$ fall below is $0....
1
Yes, it looks strange---but something can be done. If we are assuming exponential distribution, as you have, with density function $f(x;\theta)=\frac1\theta e^{-x/\theta}\quad (x>0)$, cdf $F(x;\theta)=1-e^{-x/\theta}$ and quantile function $Q(p;\theta)=-\theta\log(1-p)$ where $\theta>0$ is the expectation. So the quantile function is explicit, and for $...
1
A better notation is $\mathbb P(D|A=+a)$ to see the distinction between a RV and a specific value. Then, $D$ is a RV, not a specific value so, yes, it should be calculated for both $d$ and $-d$. In the end, a table with two rows will be produced. You don't have to make a full joint distribution in order to answer $\mathbb P(D|A=+a)$.
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