6

It appears that the stated distribution is for $x_t^{(j)} | \eta_t^{(j)}$ and the random variable $y_{i,t}$ is being ignored for now. It also appears that the authors are being a bit loose in their notation for the normal distribution, using the variance parameter in some statements and the precision parameter in others. (I will parameterise with the ...


2

Yes, basically you've just changed the objects: modem $\rightarrow$ doctor customer $\rightarrow$ patient And, since $n,c,p$ are the same, your probability is $\approx 0.04$ as well.


2

Data: I have put your data ($\pm$ typing errors) into R. x = c(1,2,3,3, 4,5,6,3, 10,11,12,19, 20,21,22) y = c(20.5,25.3,29.3,26.0, 32.8,35.2,41.2,26.0, 46.7,68.2,62.8,81.6, 80.4,63.5,100.9) Data summaries: summary(x); sd(x) Min. 1st Qu. Median Mean 3rd Qu. Max. 1.000 3.000 6.000 9.467 15.500 22.000 [1] 7.633261 # SD x ...


1

As @gunes (+1) has said: Taken exactly as stated, you have the same question either way. $X \sim \mathsf{Binom}(n=100, p=0.1)$ is the probability a modem/doctor will be busy at a given time. And you seek $P(X > 15) = 1 - P(X \le 15) = 0.0399.$ In R, where pbinom is a binomial CDF, the computation is shown below: 1 - pbinom(15,100,.1) [1] 0.03989053 ...


1

You seem to be talking about an intervention working. If it is only worth persisting with the intervention, if the intervention works, then how do we tell whether it works? This leads to the counter-factual of "What would have happened, had the person not received the intvention?" Otherwise, you might just attribute random fluctuation and regression to the ...


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