Is the following statement true? Any interval of the form [X̄ − (S/√n)t, ∞) with a 90% quantile t is a 90% confidence interval

No. t = t_{n−1,0.9} under the convention t_{m,α} = (1−α)-th quantile That t is the 10th percentile, so the interval has coverage 10%, not 90%.

Counterexample

Is this true?

Any interval of the form $[{\bar{X}} - (S/\sqrt{n})t, \infty)$ with a 90% quantile t is a 90% confidence interval

No — it is false.

The counterexample

t = $t_{n-1,0.9}$ under the convention $t_{m,\alpha}$ $= (1-\alpha)-$ th quantile

Why it fails

That t is the 10th percentile, so the interval has coverage 10%, not 90%.

Keep practicing

What the inference means

p-values, size, power and error types say specific things. This was not one of them.

Published 23 Aug 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.