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A box plot compresses a statistical sample into a box spanning its lower and upper quartiles $Q_1,Q_3$, a line at its median $Q_2$, and two whiskers. It makes location and central spread easy to compare across groups on a common value axis. The box represents the central half in terms of quantile levels; with ties or interpolated quartiles, it need not contain exactly half the observations. Whisker endpoints require a stated convention: they can be the observed extremes, or they can stop short of separately marked observations. ^[ref]

Here is one explicitly toy dataset, reused under histogram: twelve invented, unitless values, with repeated values counted separately, already arranged as order statistics:For this box plot we choose the interpolated quartiles defined under quartile, using R's type-7 rule. This givesThe interquartile range is $I=Q_3-Q_1=3.5$. Our whisker convention uses the fencesThe whiskers end at the smallest and largest observations in the closed interval $[L,U]$: here $1$ and $8$. Values strictly outside the fences receive separate outlier markers: here only $20$. An observation exactly on a fence is retained within the whisker under this convention. The fences are thresholds, not observed whisker endpoints; the upper whisker is at $8$, not $11.5$. R's documented whisker rule likewise includes observations no more than $1.5$ box lengths away, although its box uses hinges, which can differ from our type-7 quartiles. ^[ref]

A separate outlier marker is a prompt to investigate, not a finding that the value is bad data and not an instruction to delete it. A distant observation can be a recording error, a legitimate unusual outcome, or evidence that the assumed statistical model misses some behavior. Check its provenance and context before deciding how to analyze it. No measurement-error claim is possible for our invented $20$: we deliberately put it there to demonstrate the rule. ^[ref]

The beauty of a box plot is its compact comparison of ordered positions rather than every observation. That compression also loses information: many different statistical samples have the same quartiles and whiskers, so the box alone cannot reveal every cluster or gap that a histogram might show. For example, replacing the two invented $6$ values by $5.5$ and $6$ leaves the type-7 quartiles, whiskers and separate marker unchanged. Displaying the individual values alongside the box plot is especially helpful for such a small statistical sample. Always state both the quartile rule and the whisker rule; changing either can change which values are marked.

The least-squares estimator minimizes $(Y-Xb{)}^T(Y-Xb)$. Differentiation gives $X^{T}X\hat{\beta }=X^{T}Y$, and the full column rank of $X$ makes $X^{T}X$ positive definite. HenceThe fitted values are $\hat{Y}=X\hat{\beta }=HY$, where the hat matrix is $H=X(X^{T}X{)}^{-1}{X}^T$. The regression residuals are $\hat{\epsilon }=Y-\hat{Y}=(I-H)Y$. Since $H$ is the orthogonal projection matrix onto the column space of $X$, $(I-H)X=0$ andThis is a singular multivariate normal distribution supported on the $(n-p)$-dimensional orthogonal complement of the column space: its components are generally correlated. The equality $\hat{Y}=X\hat{\beta }$ puts the fitted vector in that column space, while $X^T\hat{\epsilon }=0$ expresses fitted-residual orthogonality. No residual component lies in an explanatory direction that could improve the least-squares fit.

For the calibration model, write $R_i$ for a reading and $a_i$ for the known amount. The first fit assumes $R_i=\alpha +\beta a_i+{\epsilon }_i$ with independent $N(0,{\sigma }^2)$ errors. Its residual-versus-fitted plot has systematic changes in group means and an increasing within-group spread. These suggest both a nonlinear mean relationship and heteroscedasticity, rather than the random scatter expected from a satisfactory normal linear model.

The Box–Cox transformation plot is a profile log-likelihood for the power parameter $\lambda$. Its maximum is near $0.94$; the horizontal line gives an approximate 95% likelihood-ratio confidence interval, which excludes the untransformed value $1$. Since affine rescaling of the response does not change the fitted mean space, fitting the raw power $R_i^{0.94}$ is equivalent to using $(R_i^{0.94}-1)/0.94$ with an intercept. The second model isIts residual spread is more even, but observation 17 has a conspicuously negative residual near $-9.95$. This motivates checking for an outlier or recording error. The third fit omits that observation; omission should be justified by that investigation, since choosing exclusions from residuals can affect inference. Some group pattern still remains, so the transformation alone does not certify a perfect model.

At an amount of $3$ nanograms, the third fit predicts the transformed meanSimply applying the inverse power gives ${\hat{m}}^{1/0.94}=118.5187$. The requested expected reading needs bias correction after an inverse transformation, since $\mathbb{E}[g(Z)]\ne g(\mathbb{E}[Z])$ for nonlinear $g$. With $q=1/0.94$ and $s=2.832$, a second-order Taylor expansion givesHere the correction uses the residual variance, rather than the variance of the estimated fitted mean. The power-transformed Gaussian model is an approximation for positive readings; the fitted mean is over 31 residual standard deviations above zero, so the negative tail has negligible effect on this local approximation. The ordinary inverse-transformed prediction is therefore essentially $118.52$ at the displayed precision.