waldo7852p

Answered

2022-09-25

"... This means that if we used the same sampling method to select different samples and computed an interval estimate for each sample, we would expect the true population parameter to fall within the interval estimates 95% of the time. "

However, to my knowledge, isn't this one of the common misconceptions of a confidence interval? And that the actual interpretation is that if we gather n amount of these confidence intervals, there is a 95% probability that the n collected intervals all contain the true parameter?

My notes give this definition:Let $L:=L({X}_{1},\dots ,{X}_{n})$ and $U:=U({X}_{1},\dots ,{X}_{n})$ be such that for all $\theta \in \mathrm{\Theta}$,

$$\mathbb{P}(L<\theta \le U)\ge 1-\alpha $$

and then it says, that this is the probability $1-\alpha $ that the true parameter lies within this interval . Which one is correct?

Answer & Explanation

Katelyn Chapman

Expert

2022-09-26Added 13 answers

Step 1

No, the probability is for a single drawing.

Step 2

The probability that n drawings give a "right" answer would be $(1-\alpha {)}^{n}$, a much smaller number.

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