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Suppose we repeat an experiment identically and independently 100 times. Each time we construct a 99% confidence interval for $\mu$ via the t-distribution. Let X = the number of times the confidence interval fails to contain the true value of $\mu$. The distribution of X is:

Normal with $\mu=0$ and $\sigma^2 = 1$, is this correct? Since in Normal, the mean should be zero and the variance is always one.

Since this is a t-distribution, it should be Normal. I thought another answer might suit it being Normal with $\mu = 99$ and $\sigma^2 = 0.99$

Could someone explain which one could be correct, if correct at all?

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    $\begingroup$ Think about this, what values can $X$ take on? Can $X$ be negative? Can $X$ be a non-integer? $\endgroup$ Apr 19, 2014 at 1:15
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    $\begingroup$ @Cam.Davidson.Pilon So re-thinking about this, X can take on only integers that indicate where the CI succeeded or failed. It cannot be negative and must be an discreet variable. Since it's either success or failure, then this is a Binomial with $n=100$ and $p=0.01$, Does this make sense? $\endgroup$
    – GivenPie
    Apr 19, 2014 at 1:22
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    $\begingroup$ So perfect you should write it as an answer yourself! $\endgroup$ Apr 19, 2014 at 1:27

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So re-thinking about this, X can take on only integers that indicate where the CI succeeded or failed. It cannot be negative and must be an discreet variable because we essential are counting the number of successes which must be a whole integer. Since we are gauging whether success or failure, then this is a Binomial with n=100 and p=0.01. p = 0.01 since we that's the chance to fail in the 99% confidence interval.

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  • $\begingroup$ You're right, but it for completeness' sake it would help to state explicitly where you used the assumptions of "identically and independently." Can you construct counterexamples to your conclusion when either of those assumptions does not hold? $\endgroup$
    – whuber
    Aug 20, 2018 at 11:36

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