Question 4 of 6: Dotplot, Boxplot and Skewness of Nitrate Data

Question 4: Dotplot, Boxplot and Skewness of Nitrate Data (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Dotplot, Boxplot, and Fences

Given. $n=13$ nitrate values (sorted): 2, 5, 8, 8, 9, 13, 16, 24, 45, 48, 57, 89, 120 (mg/L). Printed summary: Median $=16.0$, $Q_1=8.0$, $Q_3=52.0$.

Find. The dotplot and boxplot (with fences marked) and whether any outliers exist.

Approach. The standard (Tukey) boxplot fences sit $1.5\times IQR$ beyond each quartile; any data point outside a fence is flagged as an outlier and plotted individually, with the whisker itself drawn only to the most extreme non-outlier data value.

1. Interquartile range. $IQR=Q_3-Q_1=52.0-8.0=44.0$.
2. Fences. $\text{Lower fence}=Q_1-1.5\,IQR=8.0-66.0=-58.0$; $\text{Upper fence}=Q_3+1.5\,IQR=52.0+66.0=\boxed{118.0}$.
3. Outlier check. No data point lies below the (negative) lower fence, so the lower whisker runs to the minimum, 2. The maximum value, 120, exceeds the upper fence (118.0), so 120 is flagged as a single high outlier; the next-highest value, 89, sits below the fence, so the upper whisker is drawn only to 89 – never to the fence itself.

(b) Characteristics, Quartile Skew, and Coefficient of Skewness

Approach. Compare the mean to the median and compute both a robust (quartile-based) and a classical (moment-based) skewness measure to characterize the shape of the distribution.

1. Quartile (Bowley) skew. $S_Q=\dfrac{(Q_3-\text{Median})-(\text{Median}-Q_1)}{Q_3-Q_1}=\dfrac{Q_3+Q_1-2\,\text{Median}}{Q_3-Q_1}=\dfrac{52.0+8.0-2(16.0)}{44.0}=\dfrac{28.0}{44.0}$ $$S_Q=\boxed{0.636}$$
2. Coefficient of skewness (Pearson). $SK=\dfrac{3(\text{Mean}-\text{Median})}{StDev}=\dfrac{3(34.3-16.0)}{36.4}=\dfrac{54.9}{36.4}$ $$SK=\boxed{1.51}$$

Both measures are positive and substantial, confirming what the boxplot already shows visually: the upper whisker (89) is far longer than the lower whisker (down to 2), the box itself sits toward the low end of the whisker span, and the mean (34.3) lies well above the median (16.0) – the single high value pulls the mean upward while the median, being robust to extremes, stays close to the bulk of the data. The data are therefore strongly right (positively) skewed, with one flagged high outlier (120 mg/L) driving most of that skew; a lognormal or other right-skewed model, rather than a normal distribution, would be the appropriate choice for any further statistical inference on this data set, consistent with the "environmental data are typically right-skewed" characteristic identified in Question 1(b).