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| bio | website | |
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| age | ||
| visits | member for | 2 years, 9 months |
| seen | Jan 27 at 0:34 | |
| stats | profile views | 52 |
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Jan 27 |
comment |
Are all values within a 95% confidence interval equally likely? If you are not satisfied by my answer, please read Greg Snow answer. |
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Jan 27 |
comment |
Are all values within a 95% confidence interval equally likely? Imagine the theorem: All the points inside a whatever p CI are equally "likely". Now consider the case p=0.50. Given a point estimate x of a parameter $\theta$, build the following two p CIs: [x-c,x+c] and (-inf,x-c).(x+c,+inf) where c is such that $Pr(X-c<\theta<X+c)=p$. The latter CI is obviously the union of two intervals even if I call it a CI. Are you sure you don't prefer the former even if the theorem states they are equivalent? If you are annoyed by the fact that the second one does not contain the MLE point estimate, change their definition to [x-c).(x+c] and (-inf,x-c).{c}.(x+c,+inf) |
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Jan 17 |
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What's wrong with XKCD's Frequentists vs. Bayesians comic? You say: rejecting the null hypothesis does not mean that you are entitled to accept the alternate hypothesis, and: that doesn't mean H1 is true. So, what does it means rejecting the null hypothesis that the sun has not gone supernova? |
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Jan 15 |
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Are all values within a 95% confidence interval equally likely? @whuber, I consider the sentence you cite only a minor sin. The main error is that CLT does not involve t distribution. |
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Jan 15 |
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Are all values within a 95% confidence interval equally likely? I strongly disagree. Your "proof" by simulation is... silly. First, note than the question is not if one interval bound is "likely" as the other one, but if they are equally "likely" as other points inside the confidence interval. Please answer to the following question: if all the points inside a confidence interval are equally "likely", why the point estimate is always its median, and never one of its bounds? |
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Jan 15 |
awarded | Critic |
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Jul 24 |
awarded | Popular Question |
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Apr 10 |
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Discrepancy between number of expected and observed events in logistic regression You are right. I made a mistake in my calculations. I apologize. Thanks for your attention. |
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Apr 10 |
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Discrepancy between number of expected and observed events in logistic regression No, I didn't suppress the intercept. Are you sure that the equality always holds? |
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Apr 10 |
asked | Discrepancy between number of expected and observed events in logistic regression |
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Feb 28 |
asked | How to compute confidence interval for number to treat in a logistic regression? |
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Feb 19 |
awarded | Notable Question |
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Oct 22 |
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Is there a reference that legitimises the use of the unpooled z-test to compare two proportions? You state that unpooled $z$ hasn't asymptotic standard normal distribution. I claim that this statement is false. |
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Oct 21 |
accepted | Is there a reference that legitimises the use of the unpooled z-test to compare two proportions? |
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Oct 21 |
comment |
Is there a reference that legitimises the use of the unpooled z-test to compare two proportions? For small or moderate $n$, they can differ in terms of power level, as for example it happens when standardizing the difference betweence two means of normal variables using a pooled or an unpooled variance estimator. In such a case, you can state that only the first option has a $t$ distribution, you can argue that the second option has less power than the first one, but you cannot state that the second option does not have asymptotically the $z$ distribution. I'm sorry for splitting my comment, but the site has a pressing limit in the number of characters. |
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Oct 21 |
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Is there a reference that legitimises the use of the unpooled z-test to compare two proportions? Moreover: they are asymptotically equivalent. With large $n$, it is indifferent which one to use, and you did not explain why in a certain case (which one, precisely?) one leads to a standardized variable whose variance halves that of the standard normal variable (I asked for the computation, but I did see anything). |
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Oct 21 |
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Is there a reference that legitimises the use of the unpooled z-test to compare two proportions? I don't agree at all. Why don't say also that the building of the confidence interval for the difference between two proportions contradicts the normal distribution? Indeed, first: in any case $z$ cannot have the $t$ distribution, because it is not a mean (or sum or linear combination) of normal random variables. On the contrary, it converges directly to the normal distribution when $n$ diverges (or $n_1$ and $n_2$, if you prefer). Second: the pooled and unpooled estimators of variance are both correct and consistent. |
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Oct 20 |
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Is there a reference that legitimises the use of the unpooled z-test to compare two proportions? I don't want to bother you but really I do not undertand why if $z$ has unit variance by construction you state that its variance can be $1/2$. It seems to me that its variance is equal to $\hat{p}(1-\hat{p})\frac{2}{n}$ in a case and $\frac{\hat{p}_1(1-\hat{p}_1)}{n}+\frac{\hat{p}_2(1-\hat{p}_2)}{n}$ in the other. Sorry, I do not understand how these quantities have a 2:1 ratio. Indeed, in the case $\hat{p}_1=\hat{p}_2$ they are the same. |
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Oct 20 |
awarded | Commentator |
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Oct 20 |
comment |
Is there a reference that legitimises the use of the unpooled z-test to compare two proportions? Forgive me but I am unable to follow your example. Why should the variance of $z$ be 1? Which values are you assuming for $\hat{p}_1$ and $\hat{p}_2$? |
