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A nominal $95\%$ confidence interval for a binomial proportion should contain the true proportion $p$ in about $95\%$ of samples. Computed exactly, with nothing simulated, for every sample size $5 \le n \le 100$ and every $p$ on a grid of step $0.001$, the textbook Wald interval covers $p$ less than $0.93$ of the time at a fraction 0.463088 of the 95904 $(n, p)$ pairs, and averages 0.88228. The Wilson score interval falls below $0.93$ at a fraction 0.032981 of them, and the Agresti-Coull interval at 0.003681. The Clopper-Pearson interval never covers less than $0.95$, but averages 0.970942, so it is wider than it needs to be. Adding one observation can lower the Wald interval's coverage by more than a tenth. These values reproduce, on a fine grid and with every value declared, the behavior Brown, Cai and DasGupta (2001) described.

• Published, Reproduced, Reviewed, Replicated.Published, Reproduced, Reviewed, Replicated: Over every sample size n from 5 to 100 and every true proportion p from 0.001 to 0.999 in steps of 0.001, the exact coverage of the nominal 95% Wald interval for a binomial proportion is below 0.93 at a fraction 0.463088 of the 95904 (n, p) pairs and below 0.95 at a fraction 0.923027 of them.
• Published, Reproduced, Reviewed.Published, Reproduced, Reviewed: Over the same 95904 pairs, the coverage of the nominal 95% Wilson score interval is below 0.93 at a fraction 0.032981 of them and that of the Agresti-Coull interval at a fraction 0.003681, with mean coverages of 0.952036 and 0.958695 against the Wald interval's 0.88228.
• Published, Reproduced, Reviewed.Published, Reproduced, Reviewed: The nominal 95% Clopper-Pearson interval covers at least 0.95 at every one of the 95904 pairs, with a lowest coverage of 0.9502, and its mean coverage of 0.970942 exceeds the nominal level by about 0.021.

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Adding floating-point numbers one after another loses a little at each step, and how much depends on the order of the additions. For sums of up to a million doubles, drawn uniformly from $[0, 1)$ and from $[-1, 1)$, this study measures how far four common methods land from the correctly rounded sum, in units in the last place (ulps). Adding left to right, the error averages 196.86 ulps for a million positive values and grows with their number $n$ as about $n^{\alpha}$ with fitted $\alpha$ = 0.545, close to the square root that independent rounding errors predict. Pairwise summation stays within 2 ulps at every length. Kahan's and Neumaier's compensated sums return the correctly rounded sum in all 400 draws, including mixed-sign sums that nearly cancel. Every value is reproducible exactly, on any machine and Python version.

• Published, Reproduced, Reviewed.Published, Reproduced, Reviewed: Adding 1000000 doubles drawn uniformly from [0, 1) from left to right lands a mean of 196.86 units in the last place (at most 589) from the correctly rounded sum over 50 draws, and the mean error grows as about n to the power 0.545 for n from 1000 to 1000000, near the square-root growth that independent rounding errors predict.
• Published, Reproduced, Reviewed.Published, Reproduced, Reviewed: Pairwise summation of the same positive draws stays within 2 units in the last place of the correctly rounded sum in all 200 draws for n from 1000 to 1000000, with a mean error of 0.4 units at n = 1000000.
• Published, Reproduced, Reviewed.Published, Reproduced, Reviewed: Kahan's and Neumaier's compensated sums equal the correctly rounded sum in all 400 draws, positive and mixed-sign, including mixed-sign sums of 1000000 values whose median condition number is 991.

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