Question 5 of 6: Dotplot, Boxplot and Skewness of Nitrate Data

Question 5: 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. $n=10$ nitrate values (sorted): 2, 5, 8, 9, 13, 16, 24, 45, 57, 120 (mg/L). Printed summary: Median $=14.5$, $Q_1=7.3$, $Q_3=48.0$.

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. $IQR=Q_3-Q_1=48.0-7.3=40.7$.
2. Fences. $\text{Lower fence}=Q_1-1.5\,IQR=7.3-61.05=-53.75$; $\text{Upper fence}=Q_3+1.5\,IQR=48.0+61.05=\boxed{109.05}$.
3. Outlier check. No data point lies below the (negative) lower fence. The maximum value, $120$, exceeds the upper fence ($109.05$) – $120$ is flagged as a single high outlier. The whisker is therefore drawn only to the largest non-outlier value, $57$.

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

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

1. Quartile (Bowley) skew. $S_Q=\dfrac{(Q_3-\text{Median})-(\text{Median}-Q_1)}{Q_3-Q_1}=\dfrac{Q_3+Q_1-2\,\text{Median}}{Q_3-Q_1}=\dfrac{48.0+7.3-2(14.5)}{40.7}=\dfrac{26.3}{40.7}$ $$S_Q=\boxed{0.646}$$
2. Coefficient of skewness (Pearson). $SK=\dfrac{3(\text{Mean}-\text{Median})}{StDev}=\dfrac{3(29.9-14.5)}{36.4}=\dfrac{46.2}{36.4}$ $$SK=\boxed{1.27}$$

Both measures are positive and substantial, confirming what the boxplot already shows visually: the upper whisker (57) is far longer than the lower whisker (down to 2), the box itself is right-shifted relative to the whisker span, and the mean (29.9) sits well above the median (14.5) – 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 (e.g., before applying a t-test).