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Suppose I have drawn 21 samples from a population which I assume to be normal, where the sample has mean 3.8 and sample standard deviation 0.7.

What is the probability that the mean of the next 210 samples will be less than 3.5?

I can calculate the probability that the next draw will be less than 3.5, as $\Phi(-0.3/0.7)$, which gives 33.4%.

I can calculate the probability that the population mean will be less than 3.5, via a t-test with 20 degrees of freedom for $t=-0.3/(0.7/\sqrt{21})$, which gives 3.2%.

I assume that the probability for the mean of the next 210 samples will be close to the probability for the population mean, but I don’t know an exact approach.

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    $\begingroup$ Note that although you can estimate the probability that the next draw will be less than $3.5$ as $\Phi(-0.3/(0.7\sqrt{12}))\approx 0.0248,$ you do not know that probability for sure--so you cannot actually calculate it. Ordinarily, one therefore accompanies such an estimate with a statement of its uncertainty, such as a confidence interval for it. The same ought to apply to the estimated chance that the mean of the next $210$ observations will be less than $3.5.$ This is related to the theory of prediction intervals. $\endgroup$
    – whuber
    Commented Oct 30, 2018 at 20:42

1 Answer 1

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Rather than the $t$-statistic, we can start by considering the statistic $$u = \frac{m-m'}{s/\sqrt{21}}$$ where $m$ is the mean of the first 21 samples, $m'$ is the mean of the next 210 samples, and $s$ is the sample standard deviation of the first 21 samples.

We are looking for the probability that $u>(3.8-3.5)/(0.7/\sqrt{21})$. Fortunately, the distribution of this statistic remains unchanged if we make linear transformations of the parent distribution. So we can analyze it assuming that the samples are drawn from a standard normal. This gives:

\begin{align} m &\sim N(0,1/\sqrt{21})\\ m' &\sim N(0,1/\sqrt{210})\\ m-m' &\sim N(0,\sqrt{1/21+1/210})\\ \sqrt{10/11}\, (m-m') &\sim N(0,\sqrt{1/21})\\ v = \frac{\sqrt{10/11}\, (m-m')}{s/\sqrt{21}} &\sim t_{20} \end{align}

In this case $v=1.87$, and the probability that a $t$-statistic with 20 degrees of freedom is greater than $1.87$ is $4.13\%$.

We've effectively taken the usual $t$-statistic, and applied a correcting factor of $\sqrt{n_2/(n_1+n_2)}$.

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