91Ó°ÊÓ

Blood cocaine concentration (mg/L) was determined both for a sample of individuals who had died from cocaineinduced excited delirium (ED) and for a sample of those who had died from a cocaine overdose without excited delirium; survival time for people in both groups was at most 6 hours. The accompanying data was read from a comparative boxplot in the article "Fatal Excited Delirium Following Cocaine Use" (J. of Forensic Sciences, 1997: 25-31). $$ \begin{array}{lllllllllllll} \text { ED } & 0 & 0 & 0 & 0 & .1 & .1 & .1 & .1 & .2 & .2 & .3 & .3 \\ & .3 & .4 & .5 & .7 & .8 & 1.0 & 1.5 & 2.7 & 2.8 \\ \text { Non-ED } & 0 & 0 & 0 & 0 & 0 & .1 & .1 & .1 & .1 & .2 & .2 & .2 \\ & .3 & .3 & .3 & .4 & .5 & .5 & .6 & .8 & .9 & 1.0 \\ & 1.2 & 1.4 & 1.5 & 1.7 & 2.0 & 3.2 & 3.5 & 4.1 \\ & 4.3 & 4.8 & 5.0 & 5.6 & 5.9 & 6.0 & 6.4 & 7.9 \\ & 8.3 & 8.7 & 9.1 & 9.6 & 9.9 & 11.0 & 11.5 \\ & 12.2 & 12.7 & 14.0 & 16.6 & 17.8 & \end{array} $$ a. Determine the medians, fourths, and fourth spreads for the two samples. b. Are there any outliers in either sample? Any extreme outliers? c. Construct a comparative boxplot, and use it as a basis for comparing and contrasting the ED and non-ED samples.

Step by step solution

01

Calculate Medians

For each group, we arrange the cocaine concentrations in ascending order. The median is the middle value of this ordered list.- **ED group**: There are 19 data points. Thus, the median is the 10th value, which is 0.3 mg/L.- **Non-ED group**: There are 53 data points. Thus, the median is the average of the 26th and 27th values, which are 3.2 and 3.5 mg/L respectively. Therefore, the median is \((3.2 + 3.5)/2 = 3.35\) mg/L.

02

Calculate Fourths and Fourth Spreads

The lower fourth (Q1) and upper fourth (Q3) are determined as follows:- **ED group**: Q1 is the 5th value (0.2 mg/L) and Q3 is the 15th value (0.5 mg/L). Thus, the fourth spread (IQR) is \(0.5 - 0.2 = 0.3\) mg/L.- **Non-ED group**: Q1 is the 13.25th value (interpolated between 0.9 and 1.0 mg/L, so approximately 0.925 mg/L) and Q3 is the 39.75th value (interpolated between 9.1 and 9.6 mg/L, so approximately 9.425 mg/L). Thus, the fourth spread is \(9.425 - 0.925 = 8.5\) mg/L.

03

Identify Outliers and Extreme Outliers

Outliers are values outside of Q1 - 1.5 * IQR and Q3 + 1.5 * IQR. Extreme outliers are outside Q1 - 3 * IQR and Q3 + 3 * IQR.- **ED group**: - Lower boundary: \(0.2 - 1.5 \times 0.3 = -0.25\) - Upper boundary: \(0.5 + 1.5 \times 0.3 = 0.95\). No outliers in this group.- **Non-ED group**: - Lower boundary: \(0.925 - 1.5 \times 8.5 = -11.825\) - Upper boundary: \(9.425 + 1.5 \times 8.5 = 22.2\) No outliers in this group.

04

Construct Comparative Boxplot

To construct a boxplot for both samples, plot the medians, quartiles, and extreme values as whiskers. - For the **ED group**, the box ranges from Q1 (0.2 mg/L) to Q3 (0.5 mg/L) with a median line at 0.3 mg/L. - For the **Non-ED group**, the box ranges from approximately Q1 (0.925 mg/L) to Q3 (9.425 mg/L) with a median line at 3.35 mg/L. Compare these plots to observe: - **ED group**: Concentration values are more tightly clustered around lower values. - **Non-ED group**: There's a much broader range of cocaine concentrations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Median

The median is a key measure of central tendency in statistics. It represents the middle value of a dataset when the numbers are arranged in order. To find the median, you must first sort the values from smallest to largest. Then, if the number of data points is odd, the median is the middle value. If even, it is the average of the two middle values.

In this exercise, we have two groups. The ED (Excited Delirium) group has 19 individuals. Here, the 10th value, 0.3 mg/L, is the median since it is the middle of the ordered dataset. For the Non-ED group, there are 53 individuals. Thus, we take the 26th and 27th values, which are 3.2 mg/L and 3.5 mg/L respectively, and find their average to get the median: \[\text{Median} = \frac{3.2 + 3.5}{2} = 3.35\, \text{mg/L}.\]

• The ED group has a median of 0.3 mg/L, showing a concentration tightly centered at a lower value.
• The Non-ED group, with a higher median of 3.35 mg/L, indicates a wider spread and higher central value.

Interquartile Range (IQR)

The Interquartile Range (IQR) is a measure of statistical dispersion and helps to understand the spread and variability within a dataset. It is calculated as the difference between the third quartile (Q3) and the first quartile (Q1). This range tells us what the middle 50% of our data looks like, giving a clearer picture of the variability than range alone.

For the ED group in this exercise, we determine the quartiles as follows:- **Q1 (first quartile)** is the 5th value, which is 0.2 mg/L.- **Q3 (third quartile)** is the 15th value, which is 0.5 mg/L.Thus, the IQR is:\[\text{IQR} = Q3 - Q1 = 0.5 - 0.2 = 0.3\, \text{mg/L}.\]For the Non-ED group:- **Q1** is approximately 0.925 mg/L, through interpolation between 0.9 mg/L and 1.0 mg/L.- **Q3** is approximately 9.425 mg/L.So, the IQR is:\[\text{IQR} = 9.425 - 0.925 = 8.5\, \text{mg/L}.\]

• An IQR of 0.3 mg/L in the ED group indicates a more clustered set of values.
• An IQR of 8.5 mg/L in the Non-ED group indicates a much wider spread in data.

Comparative Boxplot

A boxplot is a graphical representation of the distribution of a dataset. It highlights the median, quartiles, and potential outliers, allowing for quick data comparison. A comparative boxplot places two or more boxplots side by side, facilitating the comparison of different datasets.

To construct a boxplot for the ED group in this exercise: - **The box extends from** Q1 (0.2 mg/L) **to** Q3 (0.5 mg/L), with a line at the median (0.3 mg/L). - **Whiskers** may be plotted that extend to the minimum and maximum values within the non-outlier range.

For the Non-ED group: - **The box extends from** approximately Q1 (0.925 mg/L) **to Q3** (9.425 mg/L) with a line at the median (3.35 mg/L).

• The ED box is narrow, indicating less variability, while the Non-ED box is much wider, showing significant spread.
• This visual allows for easy comparison between the two groups, highlighting the relative uniformity versus variability in cocaine concentrations.

Outliers

Outliers are unusual data points that deviate significantly from the rest of the dataset. They are determined using the IQR; specifically, they fall outside the range defined by:\[Q1 - 1.5 \times \text{IQR} \quad \text{to} \quad Q3 + 1.5 \times \text{IQR}.\]Exceptional outliers further fall outside:\[Q1 - 3 \times \text{IQR} \quad \text{to} \quad Q3 + 3 \times \text{IQR}.\]

In the ED group, the calculations are as follows:- **Lower boundary**: 0.2 - 1.5 \times 0.3 = -0.25 mg/L- **Upper boundary**: 0.5 + 1.5 \times 0.3 = 0.95 mg/LAll values are within this range, indicating no outliers.

For the Non-ED group:- **Lower boundary**: 0.925 - 1.5 \times 8.5 = -11.825 mg/L (Not practical since values can't be negative)- **Upper boundary**: 9.425 + 1.5 \times 8.5 = 22.2 mg/LAll data points are below 22.2 mg/L, hence no outliers.

• In cases where there are outliers, they could skew the data interpretation, highlighting the importance of identifying and understanding them.
• In this exercise, no outliers were detected, providing a clearer overall analysis of the datasets.